Optics

Class 10 Physics, unit 14, book pages 121–158

Optics is the science of light and its interaction with matter. This unit studies how light is reflected by plane mirrors and refracted at the boundary between two media, how the refractive index and Snell’s law describe the bending, and how total internal reflection works in optical fibres. It then covers thin lenses, ray diagrams, real and virtual images and linear magnification, the magnifying glass, the prism and the spectrum, optical devices, the working and defects of the human eye, and finally the bending of light by gravity and the focusing of sound by acoustic lenses.

Learning outcomes (27)

Define and use the terms normal, angle of incidence and angle of reflection.
Describe an experiment to find the position and characteristics of an optical image formed by a plane mirror.
Use the law of reflection to solve simple optical problems.
Define the terms normal, angle of incidence and angle of refraction.
Apply the qualitative principle that a wave refracts towards the normal if it slows down while entering a medium, and that it refracts away from the normal if it speeds up when it enters a new medium.
Define and use the refractive index from a vacuum to a medium for light as $c / v$ .
Define refractive index $n$ as $n = sin i / sin r$ . Apply Snell’s law, $n_{1} sin i = n_{2} sin r$ , to solve simple problems.
Describe an experiment to show refraction of light by transparent blocks of different shapes.
Define the terms critical angle and total internal reflection.
Derive the equation $n = 1/ sin c$ .
Apply the equation $n = 1/ sin c$ to solve simple problems.
Describe experiments to show internal reflection and total internal reflection.
Evaluate and illustrate the use of optical fibres.
Analyse the action of thin converging and thin diverging lenses on a parallel beam of light.
Define and use the terms focal length, principal axis and principal focus (focal point).
Draw ray diagrams to illustrate the formation of real and virtual images of an object by a converging lens.
Differentiate between real and virtual images.
Define and calculate linear magnification.
Describe the use of a single lens as a magnifying glass.
Explain the dispersion of light by a prism.
State the traditional seven colours of the visible spectrum in order of frequency and in order of wavelength.
Describe the use of a single lens in various optical device applications.
Draw ray diagrams to show the formation of images in the normal eye, a short-sighted eye and a long-sighted eye.
Describe the use of converging and diverging lenses to correct long-sightedness and short-sightedness.
Illustrate with examples how the biological eye processes colour in various organisms.
State that extreme gravity from interstellar objects like black holes can cause light to bend (from the perspective of the observer) in a way that is analogous to a simple lens.
State that “acoustic lenses” are made of materials and shapes that work to focus or diverge sound.

Introduction to optics

Optics is the science of light and its interaction with matter. It explores the behaviour of light waves, the properties of lenses and mirrors and the way human vision works. Beyond practical uses in photography, microscopy and astronomy, optics asks deeper questions: how do we see the world around us, and what role does light play in shaping our understanding of it?

Our lives are filled with light, and vision is the most valued of our senses. Light has many uses beyond vision: it can carry telephone signals through glass fibres or cook a meal in a solar oven. Life itself could not exist without light energy.

14.1Reflection of lightOften examined

Light from this page is formed into an image by the lens of your eye, much as the lens of the camera that made a photograph forms an image. Light can travel in three ways: directly from the source to the observer (like sunlight reaching Earth through space), through different materials such as air or glass, or after being reflected by objects like mirrors.

In all cases light travels in a straight line, which we call a ray. The word comes from mathematics and means a straight line, and we use it when we draw the travelling of light in a ray diagram.

Reflection of light is the phenomenon that occurs when light rays bounce off a surface. Whenever you look into a mirror you see a reflection. Even when you look at this page you are really seeing the light that the page reflects. Large telescopes use reflection to form an image of stars and other astronomical objects.

Reflection: When a ray of light approaches a smooth polished surface and bounces back, it is called reflection.

Terms used to study reflection

Normal: The line drawn perpendicular to the reflecting surface at the point where the ray strikes it.
Incident ray: The ray that strikes the surface.
Angle of incidence: The angle $θ_{i}$ between the incident ray and the normal.
Reflected ray: The ray that bounces back from the surface after striking it.
Angle of reflection: The angle $θ_{r}$ between the reflected ray and the normal.
Point of incidence: The point on the surface where the incident ray strikes it.

14.1.1Laws of reflectionOften examined

Two laws govern the reflection of light.

First law: the incident ray, the reflected ray and the normal all lie in the same plane.
Second law: the angle of incidence $θ_{i}$ and the angle of reflection $θ_{r}$ are equal.

θ_{i} = θ_{r}

14.1

$θ_{i}$ : angle of incidence; $θ_{r}$ : angle of reflection, both measured from the normal

Figure 14.1: Laws of reflection. The angle of incidence

θ_{i}

and the angle of reflection

θ_{r}

are measured from the normal and are equal.

Can you tell? Angle of incidence and reflection

A light ray strikes a reflective plane surface at an angle of

4 7^{\circ}

with the surface. Find the angle of incidence, the angle of reflection and the angle between the incident and reflected rays.

Given

$angle with the surface = 4 7^{\circ}$

To find

Angle of incidence $θ_{i}$
Angle of reflection $θ_{r}$
Angle between the incident and reflected rays

The angle of incidence is measured from the normal, which is $9 0^{\circ}$ to the surface. Subtract the angle with the surface from $9 0^{\circ}$ .
$θ_{i} = 9 0^{\circ} - 4 7^{\circ} = 4 3^{\circ}$
By the second law of reflection the two angles are equal.
$θ_{r} = θ_{i} = 4 3^{\circ}$
The incident and reflected rays lie on opposite sides of the normal, so the angle between them is the sum of the two angles.
$θ_{i} + θ_{r} = 4 3^{\circ} + 4 3^{\circ} = 8 6^{\circ}$
Check: the angle between each ray and the surface is $4 7^{\circ}$ , so the angle between the rays measured through the surface side is $18 0^{\circ} - 4 7^{\circ} - 4 7^{\circ} = 8 6^{\circ}$ , which agrees.

θ_{i} = 4 3^{\circ}, θ_{r} = 4 3^{\circ}, angle between the rays = 8 6^{\circ}

14.1.1Types of reflection

A. Regular or specular reflection

This occurs on smooth surfaces like mirrors, where the reflection is clear and sharp. The angles of reflection are consistent across the surface, which leads to a distinct image being formed. This is shown in Figure 14.2 (a).

B. Diffused reflection

This happens on rough surfaces, where the reflected rays scatter in many directions. This type of reflection allows us to see most non-shiny objects, like this page of your book. That is why you can see this page from all sides and not only from a single point: the diffused reflection of light sends light and the image to many sides. It is shown in Figure 14.2 (b).

Figure 14.2: Reflection from (a) a smooth surface and (b) a rough surface. Each ray obeys the law of reflection, but the normals of a rough surface point in many directions.

Comparison of the two types of reflection
Feature	Regular (specular) reflection	Diffused reflection
Surface	Smooth and polished, such as a mirror	Rough, such as paper or a wall
Reflected rays	Stay parallel to one another; the reflection is clear and sharp	Scatter in many directions
Image	A distinct image is formed	No clear image; the object is seen from all sides
Law of reflection	Obeyed	Obeyed at every tiny point, but the normals point in many directions

14.1Image formed by a plane mirrorOften examined

To understand this, take an object placed in front of a mirror at a distance $d_{o}$ . Light rays coming from the object, after reflecting from the mirror, enter the eye of the observer. To the observer these light rays appear to be coming from behind the mirror, from a distance $d_{i}$ . Hence the image is seen behind the mirror because of the reflection of light.

Mini lab: position and characteristics of the image in a plane mirror

Aim: to find the position and characteristics of an optical image formed by a plane mirror.

Material needed: a plane mirror, an object, a measuring tape, a light source and a white paper.

Procedure

Place the plane mirror in a vertical position.
Position the object in front of the mirror. The object should have the same height as the mirror or less. Take a light source that illuminates the object.
Observe the reflected rays from the object that strike the mirror surface. Note that two rays emerging from the object strike the mirror and reflect into the observer's eye.
Use the law of reflection to construct the reflected rays.
Extend these reflected rays backwards behind the mirror, using dashed lines. The point where these extended rays seem to start is where the virtual image of the object is located.
Measure the distance of the object and of the image from the mirror with the tape.

Mini lab: an object and its image in a plane mirror. Two reflected rays are extended backwards (dashed); they seem to come from the virtual image, which is as far behind the mirror (

d_{i}

) as the object is in front (

d_{o}

Results

The image is virtual: it cannot be projected onto a screen.
The image is erect: it has the same orientation as the object.
The size of the image is the same as the size of the object.
The distance of the image is the same as the distance of the object from the mirror: $d_{i} = d_{o}$ , where $d_{i}$ and $d_{o}$ are the distances of the image and the object from the mirror.

d_{i} = d_{o}

$d_{i}$ : image distance behind the mirror; $d_{o}$ : object distance in front of the mirror

Image distance against object distance

d_{o} (m)

d_{i} (m)

(2.5, 2.5)

d_{i} = d_{o}

For a plane mirror the graph of

d_{i}

against

d_{o}

is a straight line through the origin with gradient

1

: the image is always as far behind the mirror as the object is in front. The marked point is the girl in the worked example below (

2.5 m

2.5 m

Worked example: how far is your image?

A girl stands

2.5 m

in front of a plane mirror. (a) How far behind the mirror is her image? (b) How far is she from her image? (c) She walks

1.0 m

towards the mirror. How far is she now from her image?

Given

$d_{o} = 2.5 m$
$distance walked = 1.0 m$

To find

Image distance $d_{i}$
Distance between the girl and her image, before and after walking

For a plane mirror the image distance equals the object distance.
$d_{i} = d_{o} = 2.5 m$
The girl is in front of the mirror and the image is behind it, so the distance between them is the sum of the two distances.
$2.5 + 2.5 = 5.0 m$
After walking $1.0 m$ towards the mirror her distance from it is $2.5 - 1.0 = 1.5 m$ . Her image moves too, so it is also $1.5 m$ behind the mirror.
$d_{o} = d_{i} = 1.5 m$
New distance between the girl and her image.
$1.5 + 1.5 = 3.0 m$
Common mistake: the separation changes by $2.0 m$ (from $5.0$ to $3.0 m$ ), not by $1.0 m$ , because the image also moves $1.0 m$ towards the mirror.

d_{i} = 2.5 m; separation = 5.0 m then 3.0 m

Example 14.1: Angle between two mirrors

Find the angle

α

made by the system of the two mirrors as shown, provided that the incident ray

A

and the reflected ray

B

are parallel.

Given

$ray A = parallel to ray B$

To find

Angle $α$ between the mirrors

For the incident ray $A$ and the reflected ray $B$ to be parallel, the angles $(i + r)$ at the first mirror and $(i^{'} + r^{'})$ at the second mirror must be supplementary (they are the interior angles on one side of the line $P Q$ joining the two points of incidence, with $A$ and $B$ parallel).
$i + r + i^{'} + r^{'} = 18 0^{\circ}$
By the law of reflection at each mirror, the angle of incidence equals the angle of reflection.
$i = r and i^{'} = r^{'}$
Put these values into the first equation.
$i + i + i^{'} + i^{'} = 18 0^{\circ} \Rightarrow i + i^{'} = 9 0^{\circ}$
In the triangle $O P Q$ formed by the two mirrors and the ray $P Q$ between them, the angle at $P$ is $(9 0^{\circ} - r)$ and the angle at $Q$ is $(9 0^{\circ} - i^{'})$ , because the ray makes the complement of the angle of reflection or incidence with the mirror. The angle at $O$ is $α$ .
$α + (9 0^{\circ} - r) + (9 0^{\circ} - i^{'}) = 18 0^{\circ}$
Simplify and use $r = i$ .
$α = r + i^{'} = i + i^{'} = 9 0^{\circ}$

Example 14.1: two mirrors

M_{1}

and

M_{2}

meeting at

O

. When the incident ray

A

and the reflected ray

B

are parallel, the angle

α

between the mirrors is

9 0^{\circ}

α = 9 0^{\circ}

14.2Refraction of lightOften examined

It is easy to notice odd things when you look at half-submerged objects in water. If you dip a straight pencil into a glass of water so that the full length of the pencil does not sink, the pencil looks bent or broken. This is known as the broken pencil illusion (Figure 14.3). It happens because the light coming from the water to our eyes changes direction when it leaves the water and enters the air, so the light has to travel along two different paths to reach our eyes.

Refraction: The changing of a light ray's direction when it passes through variations in matter, that is, when it passes from one medium to another.

Refraction is responsible for a tremendous range of optical phenomena, from the action of lenses to voice transmission through optical fibres.

14.2.1Speed of light in material mediaOften examined

Why does light change direction when it passes from one material to another? The reason is that the speed of light varies in a precise manner with the material it travels through. So before the law of refraction it is useful to see how the speed of light varies in different media.

The speed of light in vacuum $c$ is constant and is accepted as a universal constant, with the fixed value $c = 3 \times 1 0^{8} m s^{- 1}$ . The speed of light through matter is less than in vacuum, because light interacts with the atoms and molecules of the medium, and this hinders it. That is why the speed of light depends upon the type of material.

How the speed of light decides the bending
Medium	Interaction with atoms	Speed of light	Bending of the ray
Denser (slower) medium	More, because the particles are closer	Slower	Towards the normal
Rarer (faster) medium	Fewer	Faster	Away from the normal

14.2.2Refractive indexOften examined

The speed of light in any material medium is always less than its speed in vacuum, so we can take the speed in vacuum as a reference. The quantity that describes the speed of light in a medium compared with its speed in vacuum is called the refractive index.

Refractive index: The measure of the bending of a light ray when passing from one medium to another such that one medium is vacuum. It is also the ratio of the velocity of light in empty space to the velocity of light in a substance.

If $c$ is the speed of light in vacuum and $v$ is the speed of light in some medium, the refractive index $n$ of that medium is

n = \frac{c}{v}

14.2

$n$ : refractive index of the medium; $c$ : speed of light in vacuum ( $3 \times 1 0^{8} m s^{- 1}$ ); $v$ : speed of light in the medium

Properties of the refractive index

The speed of light in vacuum is the highest, so the refractive index of any medium cannot be equal to or less than one. It has the value $1$ only for vacuum itself.
It is the ratio of two similar quantities, so it has no unit: it is a dimensionless quantity that describes how fast light travels in a medium.
In gases the refractive index is taken as $1$ when great precision is not needed, because the atoms and molecules of gases are widely separated and light interacts with them rarely.

Table 14.1: Refractive indices of some commonly used transparent materials
Medium	Refractive index $n$
Air	1.000293
Carbon dioxide	1.00045
Oxygen	1.000271
Benzene	1.501
Glycerin	1.473
Water	1.333
Diamond	2.419
Polystyrene	1.49
Glass (crown)	1.52
Glass (flint)	1.66
Ice	1.309
Zircon	1.923

Example 14.2: Speed of light in diamond

What is the speed of light in diamond, a material often found in jewellery? Use the refractive index of diamond from Table 14.1.

Given

$n = 2.419$
$c = 3 \times 1 0^{8} m s^{- 1}$

To find

Speed of light in diamond $v$

Refractive index is the ratio of the speed in vacuum to the speed in the medium.
$n = \frac{c}{v}$
Rearrange for $v$ .
$v = \frac{c}{n}$
Put in the values.
$v = \frac{3 \times 1 0 ^{8}}{2.419} = 1.24 \times 1 0^{8} m s^{- 1}$
So the speed of light is reduced to about $1.2 \times 1 0^{8} m s^{- 1}$ , a little over one third of its speed in vacuum.

v = 1.24 \times 1 0^{8} m s^{- 1}

Speed of light against refractive index

1.2

1.4

1.6

1.8

2.2

2.4

2.6

0.5

1.5

2.5

n

v (1 0^{8} m s^{- 1})

water

crown glass

diamond

v = c / n

v = c / n

: the larger the refractive index of a medium, the slower light travels in it. The marked points are water (

1.333

), crown glass (

1.52

) and diamond (

2.419

, Example 14.2), with

c = 3 \times 1 0^{8} m s^{- 1}

and

v

in units of

1 0^{8} m s^{- 1}

Worked example: light crossing a glass slab

A rectangular slab of crown glass is

0.30 m

thick. Find (a) the speed of light in the glass and (b) the time light takes to cross the slab, and compare it with the time light takes to travel

0.30 m

in vacuum. Take

c = 3 \times 1 0^{8} m s^{- 1}

Given

$n = 1.52 (crown glass, Table 14.1)$
$d = 0.30 m$
$c = 3 \times 1 0^{8} m s^{- 1}$

To find

Speed $v$ in the glass
Time $t$ in the glass and in vacuum

Speed in the glass, from $n = c / v$ .
$v = \frac{c}{n} = \frac{3 \times 1 0 ^{8}}{1.52} = 1.97 \times 1 0^{8} m s^{- 1}$
Time to cross the slab, from speed $=$ distance $\div$ time.
$t = \frac{d}{v} = \frac{0.30}{1.97 \times 1 0 ^{8}} = 1.52 \times 1 0^{- 9} s$
Time to travel the same $0.30 m$ in vacuum.
$t_{0} = \frac{0.30}{3 \times 1 0 ^{8}} = 1.0 \times 1 0^{- 9} s$
Compare the two times.
$\frac{t}{t _{0}} = \frac{1.52}{1.0} = 1.52 = n$
The ratio of the two times equals the refractive index, as it must: light is slower in glass by the factor $n$ . Common mistake: using $v = n c$ instead of $v = c / n$ gives a speed greater than $c$ , which is impossible.

v = 1.97 \times 1 0^{8} m s^{- 1}; t = 1.52 \times 1 0^{- 9} s (vacuum: 1.0 \times 1 0^{- 9} s)

14.2.3Laws of refraction (Snell's law)Often examined

To study the laws of refraction we first need the terms used with refraction (Figure 14.4).

Normal: The perpendicular line to the boundary between the two media.
Incident ray: The light ray coming from the first medium.
Angle of incidence: The angle the incident ray makes with the normal.
Refracted ray: The ray entering the other medium.
Angle of refraction: The angle the refracted ray makes with the normal.

Figure 14.4: Refraction of light at a boundary. Going into the denser medium (glass), the ray bends towards the normal, so \theta_r<\theta_i. The angles are measured from the normal.

The amount that a light ray changes its direction depends both on the angle of incidence and on the amount that the speed changes. For a ray at a given angle of incidence, a large change in speed causes a large change in direction, and so a large change in the angle of refraction. In the same way, for the same speed change a larger angle of incidence causes a larger change in the refracted angle (up to a limit only).

The laws of refraction are also known as Snell's law, after a Dutch scientist (1591 to 1626), but they were first given by the Muslim scientist Ibn-e-Sahl in 984 in his book On burning mirrors and lenses. The laws are given in two steps.

First law: the incident ray, the normal and the refracted ray all lie in the same plane.
Second law: it gives the relationship between the angles and the refractive indices of the two media.

n_1\sin\theta_i=n_2\sin\theta_r

14.3

n_1, n_2: refractive indices of the first and second medium; \theta_i: angle of incidence; \theta_r: angle of refraction

The same relation can also be written as

\dfrac{n_1}{n_2}=\dfrac{\sin\theta_r}{\sin\theta_i}

Special cases

If the first medium is vacuum (or air, taken as 1) and light enters a medium of index n, then n_1=1 and the relation reduces to

\dfrac{1}{n}=\dfrac{\sin\theta_r}{\sin\theta_i}\qquad\text{or}\qquad n=\dfrac{\sin\theta_i}{\sin\theta_r}

14.4

light entering a medium of index n from vacuum or air

When light enters vacuum (or air) from a medium of index n, then n_2=1 and the relation becomes

n=\dfrac{\sin\theta_r}{\sin\theta_i}

14.5

light leaving a medium of index n into vacuum or air; \theta_i is the angle in the medium and \theta_r the angle in vacuum or air

Example 14.3: Refractive index from the angles

Find the refractive index of a medium if light enters it from air at an angle of 37^\circ and is refracted into it at an angle of 21^\circ.

Given

n_1 = 1\ \text{(air)}
\theta_i = 37^\circ
\theta_r = 21^\circ

To find

Refractive index of the medium n_2

Write Snell's law as a ratio.
\dfrac{n_2}{n_1}=\dfrac{\sin\theta_i}{\sin\theta_r}
Put in the values.
\dfrac{n_2}{1}=\dfrac{\sin37^\circ}{\sin21^\circ}=\dfrac{0.6018}{0.3584}
Divide.
n_2=1.68
This is the refractive index of a type of glass that is used for the core of optical fibres.

n_2=1.68

Angle of refraction against angle of incidence (air into a medium)

Light goes from air into water (n=1.333), a glass with n=1.68 and diamond (n=2.419). The larger the refractive index, the more the ray bends towards the normal. The marked point is Example 14.3: 37^\circ in air becomes 21^\circ in the glass.

Worked example: a ray entering water

A ray of light strikes the still surface of a pond making an angle of 40^\circ with the water surface. Find the angle of refraction in the water (n=1.333 for water, n=1 for air), and say whether the ray bends towards or away from the normal.

Given

\text{angle with the surface} = 40^\circ
n_1 = 1
n_2 = 1.333

To find

Angle of incidence \theta_i
Angle of refraction \theta_r

The angle of incidence is measured from the normal, so subtract the angle with the surface from 90^\circ.
\theta_i=90^\circ-40^\circ=50^\circ
Write Snell's law and solve for \sin\theta_r.
n_1\sin\theta_i=n_2\sin\theta_r\ \Rightarrow\ \sin\theta_r=\dfrac{n_1\sin\theta_i}{n_2}
Put in the values.
\sin\theta_r=\dfrac{1\times\sin50^\circ}{1.333}=\dfrac{0.7660}{1.333}=0.5747
Take the inverse sine.
\theta_r=\sin^{-1}(0.5747)=35.1^\circ
Check: \theta_r<\theta_i, so the ray bends towards the normal. This is correct because light slows down on entering water (n larger). Common mistake: using 40^\circ directly as the angle of incidence would give \theta_r=28.9^\circ, which is wrong.

Worked example: a ray at 50^\circ to the normal in air becomes a ray at 35.1^\circ to the normal in water, bent towards the normal.

\theta_r=35.1^\circ\ \text{(towards the normal)}

14.2.4Experiment to study refraction

To study the refraction of light through transparent blocks of different shapes we can set up an experiment even in our classroom or school laboratory.

Aim: to study the refraction of light using a rectangular block, a semicircular block and a triangular prism made of glass.

Equipment required: laser light, protractor, sheet of paper, pencil, ruler and perspex blocks (rectangular, semicircular and prism). A perspex block is a tough transparent plastic block used instead of a glass block.

Method

Place the rectangular block on the sheet of paper and draw lines around it. Draw the perpendicular lines, the normals.
Illuminate one plane side of the block with the light so that the ray makes some angle with the normal already drawn.
Mark a point on the paper near the light source, the point where the ray enters the block, the point where it leaves the block, and another point on the path of the emergent ray about 4 to 5\,\mathrm{cm} away.
Remove the source and the block and join the marked points with straight lines.
Replace the block and repeat for other angles of incidence such as 20^\circ, 40^\circ and 60^\circ.
Take another sheet of paper and repeat the experiment for the semicircular block with three different angles, and finally for a glass prism.

Figure 14.5: Refraction through (a) a rectangular glass block and (b) a semicircular block. Entering glass the ray bends towards the normal (i>r); leaving it bends away (i<r); a ray along a radius leaves the curved face undeviated.

Figure 14.5 (c): A ray through a glass prism. It bends towards the normal on entering and away from it on leaving. A is the angle of the prism and \delta is the angle of deviation between the original direction of the ray (dashed) and the emergent ray.

Results

The final diagram for each shape includes several ray paths for different angles of incidence i. Label the paths A, B and C for the first two blocks. This shows how the angle of refraction r changes with the angle of incidence.
Incident and refracted angles are always measured from the normal.
The light entering the block bends towards the normal: i>r.
For light rays leaving the perspex block, the rays refract away from the normal: i<r.
For light rays entering along the normal, the ray does not refract and passes straight through the block: i=r=0.
For the semicircular block the rays are not refracted at the curved surface, because at each point the radial path of the ray is perpendicular to the tangent, so the rays leave along the normal and suffer no refraction.

14.3Total internal reflectionOften examined

Whenever light strikes a surface, some of the incident light bounces back as the reflected ray and the rest is transmitted into the other medium. The transmitted ray bends from its normal path because the speed of light changes, so we call it the refracted ray (Figure 14.6 (a)). By the law of reflection the reflected ray leaves at the same angle \theta_1 as the incident ray; this is called internal reflection when it happens inside the medium. The refracted ray bends away from the normal and makes an angle \theta_2 with it; the refractive indices of the two media are n_1 and n_2.

When light travels from a less dense medium to a denser one, it bends towards the normal. As the angle of incidence increases, the angle of refraction increases, but the refracted ray stays closer to the normal. In contrast, when light moves from a denser medium, such as glass or water, to a less dense medium like air, it bends away from the normal. Increasing the angle of incidence gives a larger angle of refraction, moving the ray further from the normal.

If we increase the angle of incidence further, the refracted ray moves away from the normal until it makes an angle of 90^\circ with it (Figure 14.6 (b)). At this angle the refracted light does not enter the second medium: it moves along the boundary between the two media. The angle of incidence at this stage is called the critical angle.

Critical angle: The angle of incidence for which the angle of refraction is 90^\circ. It is represented by \theta_c.
Total internal reflection: The phenomenon of reflection of light back into the same medium when passing from a denser medium to a rarer medium in such a way that the angle of incidence is greater than its critical angle.

When the angle of incidence becomes greater than the critical angle, the light is reflected back into the same medium from the boundary of the two media. At this stage there is no more refraction of light (Figure 14.6 (c)).

Figures 14.6 and 14.7: a ray in glass meets the glass–air boundary. (a) Below the critical angle most of the light is refracted out (a weak reflected ray is also present). (b) At the critical angle the refracted ray grazes the boundary. (c) Above it all the light is reflected back into the glass.

Reflection compared with total internal reflection
Point	Ordinary reflection	Total internal reflection
Where it happens	At any reflecting surface, for any angle of incidence	Only inside a denser medium at the boundary with a rarer medium
Condition	None	Angle of incidence greater than the critical angle
Light reflected	Only part of the light; the rest is refracted or absorbed	All of the light; no refracted ray
Law obeyed	\theta_i=\theta_r	\theta_i=\theta_r, with no light entering the second medium

14.3.1Critical angle and its derivationOften examined

To derive a relation for the critical angle we use Snell’s law, equation 14.3.

Derivation Critical angle in terms of refractive indices

Aim. Find the relation between the critical angle \theta_c and the refractive indices n_1 and n_2 of the two media.

Figure for the derivation: a ray in the denser medium (n_1) meets the boundary with the rarer medium (n_2) at the critical angle \theta_c. The refracted ray grazes the boundary, so the angle of refraction is 90^\circ.

Start from Snell’s law, with n_1 the index of the denser medium (where the ray is travelling) and n_2 the index of the rarer medium.
n_1\sin\theta_i=n_2\sin\theta_r
For total internal reflection to begin, the refracted ray just grazes the boundary, so the angle of refraction is 90^\circ and the angle of incidence is the critical angle, \theta_i=\theta_c.
n_1\sin\theta_c=n_2\sin90^\circ
Use \sin90^\circ=1.
n_1\sin\theta_c=n_2
Divide both sides by n_1 to get the sine of the critical angle.
\sin\theta_c=\dfrac{n_2}{n_1}
Take the inverse sine to find the critical angle itself.
\theta_c=\sin^{-1}\!\left(\dfrac{n_2}{n_1}\right)
A sine can never be larger than 1, so n_2/n_1\le1, that is n_1\ge n_2. Total internal reflection can therefore only happen when light travels from a denser medium into a lighter (rarer) medium.
If the second medium is air, n_2=1 and n_1=n, the index of the denser medium.
\sin\theta_c=\dfrac{1}{n}
Rearrange for the refractive index.
n=\dfrac{1}{\sin\theta_c}

Result\theta_c=\sin^{-1}\!\left(\dfrac{n_2}{n_1}\right),\qquad n=\dfrac{1}{\sin\theta_c}\ \text{(denser medium into air)}

The smaller the critical angle, the larger the refractive index. Equation 14.7 holds only if the light enters air from a denser medium of refractive index n.

\sin\theta_c=\dfrac{n_2}{n_1}

n_1: refractive index of the denser medium; n_2: refractive index of the rarer medium

\theta_c=\sin^{-1}\!\left(\dfrac{n_2}{n_1}\right)

14.6

n=\dfrac{1}{\sin\theta_c}

14.7

valid for light going from a medium of refractive index n into air

Example 14.4: Critical angle of polystyrene

Find the critical angle of a polystyrene (n=1.49) pipe surrounded by air.

Given

n = 1.49

To find

Critical angle \theta_c

Use the formula for the critical angle with air outside.
\sin\theta_c=\dfrac{1}{n}
Put in the value.
\sin\theta_c=\dfrac{1}{1.49}=0.671
Take the inverse sine.
\theta_c=\sin^{-1}(0.671)=42.2^\circ
This means that a light ray moving in the polystyrene which strikes the boundary at an angle greater than 42.2^\circ will be totally internally reflected.

\theta_c=42.2^\circ

Critical angle against refractive index

\theta_c=\sin^{-1}(1/n) for light going from a medium of index n into air. The denser the medium, the smaller the critical angle. Marked: water 48.6^\circ, crown glass 41.1^\circ, diamond 24.4^\circ.

Worked example: does the ray escape?

A ray travelling inside a block of crown glass (n=1.52) meets the glass-air boundary at an angle of incidence of 45^\circ. (a) Find the critical angle of the glass. (b) Does the ray leave the glass or is it totally reflected? (c) What would happen to the same ray if the block were made of water (n=1.333)?

Given

n_{\text{glass}} = 1.52
n_{\text{water}} = 1.333
\theta_i = 45^\circ

To find

Critical angle of glass
Fate of the ray in glass
Fate of the ray in water

Critical angle of the glass for light going into air.
\sin\theta_c=\dfrac{1}{1.52}=0.6579\ \Rightarrow\ \theta_c=41.1^\circ
Compare the angle of incidence with the critical angle.
45^\circ>41.1^\circ
The angle of incidence is greater than the critical angle and the light is going from the denser to the rarer medium, so the ray is totally internally reflected back into the glass, leaving at 45^\circ on the other side of the normal. No light enters the air.
Repeat for water.
\sin\theta_c=\dfrac{1}{1.333}=0.7502\ \Rightarrow\ \theta_c=48.6^\circ
Now 45^\circ<48.6^\circ, so in water the same ray is not totally reflected: most of it is refracted out into the air, bending away from the normal.
Common mistake: thinking that a larger refractive index makes total internal reflection harder. The opposite is true: the larger n is, the smaller the critical angle, so total reflection starts sooner.

Worked example: a ray in crown glass meets the boundary at 45^\circ, which is greater than the critical angle of 41.1^\circ (dashed ray), so it is totally reflected.

\theta_c=41.1^\circ\ \text{(glass)};\ \text{totally reflected in glass, refracted out in water}

14.3.2Experiment to study total internal reflection

To study total internal reflection using a transparent block of semicircular shape we can set up an experiment even in our classroom or laboratory.

Aim: to study the total internal reflection of light using a semicircular block of glass.

Equipment required: laser light, protractor, sheet of paper, pencil, ruler and a perspex block.

Method

Place the semicircular glass block on the sheet of paper and draw lines around it. Draw the perpendicular line, which is the normal.
Illuminate one plane side of the block so that the light makes some angle with the normal already drawn.
Mark a point on the paper near the light source, the point where the ray enters the block, the point where it leaves the block, and another point on the path of the emergent ray about 4 to 5\,\mathrm{cm} away.
Move the light source slowly so that the angle of incidence keeps increasing.
As the angle of incidence increases, the angle of refraction increases too, until the refracted ray makes an angle of 90^\circ with the normal. This angle of incidence is the critical angle. Mark two points for the incoming ray at this stage.
Increase the angle of incidence further. The light no longer goes into the other medium; it enters back into the same semicircular block. This is total internal reflection (Figure 14.7).

Figure 14.7: Experiment for total internal reflection with a semicircular block. (a) Below the critical angle the ray is refracted into the air. (b) At the critical angle the refracted ray travels along the flat face. (c) Above the critical angle the ray is totally reflected back into the block.

Results

When the angle of incidence was less than the critical angle, refraction was seen, and the light entered the lighter medium (air) from the semicircular glass block: i<c.
When the angle of incidence was equal to the critical angle, the refracted light made an angle of 90^\circ with the normal and travelled along the boundary between the two media: i=c.
When the angle of incidence was greater than the critical angle, total internal reflection was seen: the light bounced back into the same block instead of going into the lighter medium: i>c.

14.3.3Optical fibresOften examined

Optical fibre is the technology associated with data transmission using light pulses travelling along a long fibre, which is usually made of plastic or glass. Optical fibres use the application of total internal reflection of light.

An optical fibre consists of two concentric tubes of glass or plastic. The inner tube, called the core, has a higher refractive index (larger density) than the outer tube, called the cladding, which has a lower refractive index (smaller density). The whole set-up is covered for protection by a material called the jacket (plastic coating). The light enters the core at an angle greater than the critical angle of the core material and moves within the core by successive total internal reflections (Figures 14.8 and 14.9).

Figures 14.8 and 14.9: an optical fibre. The light enters the core and meets the core–cladding boundary at an angle greater than the critical angle, so it is totally reflected again and again and travels along the fibre.

Light entering the fibre within a certain angular range undergoes total internal reflection and travels down the fibre. Light entering the fibre with higher angles will exit through the cladding.

Advantages over copper wires

Advantages of optical fibres
Advantage	Explanation
High bandwidth	Optical fibres provide much higher bandwidth than copper wires, where bandwidth is the data capacity of a network in bits per second. Modern internet use needs a lot of data at very high speed.
Faster speed	Data are sent as light pulses instead of electrical signals, so they travel from one place to another at the speed of light.
Resistance to electrical interference	Light is used in the fibre, so electrical interference, a major problem in copper networks, has no effect on the data.
Long distance	Data can be sent over much longer distances without significant signal loss, for example in undersea cables and intercontinental connections.
Low power	They consume less power than copper cables, and their durability and longevity reduce maintenance costs.

Optical fibres are used in various sectors, including healthcare, defence, manufacturing, broadcasting, lighting and communications. In telecommunications they are essential for data transmission, offering greater speed and accuracy than copper wires. Optical fibre cables are lighter, more flexible and can carry more data.

How a signal travels along a fibre

Data are converted into a light signal using transducers like LEDs or laser diodes.
The light signal is introduced into the fibre’s core at an angle greater than the critical angle (Figure 14.9).
The signal travels along the fibre for long distances by total internal reflection until it reaches the destination.
At the receiving end the light signal is converted into an electrical signal with the help of transducers like a photodiode, PIN diode or avalanche diode, near the user end.
The electrical signal is finally converted into text, voice or video data as required.

14.4Thin lensesOften examined

One of the most interesting aspects of optics is the study of lenses.

Lens: The transparent objects that refract light in specific ways to form images are called lenses.

Understanding the action of lenses is fundamental to applications in science, technology and everyday life, from eyeglasses, cameras, microscopes and telescopes to many other devices. Lenses are optical devices made from transparent materials, typically glass or plastic, with at least one curved surface. They work on the principle of refraction, which is the bending of light as it passes from one medium to another.

Lenses are broadly classified into two main types: converging (also called a convex lens) and diverging (also called a concave lens), with a variety of different shapes as shown in Figure 14.10.

Figure 14.10: Shapes of lenses. A converging lens is thicker at the centre than at the edges; a diverging lens is thinner at the centre.

The two types of lens
	Converging (convex) lens	Diverging (concave) lens
Shape	Thicker at the centre, thinner at the edges	Thinner at the centre, thicker at the edges
Parallel beam	Converges it to a point	Diverges it, as if from a point
Focal length	Positive	Negative
Shapes	Bi-convex, plano-convex, convex meniscus	Bi-concave, plano-concave, concave meniscus

Terms used with lenses

Some common terms are needed to study image formation by lenses (Figure 14.11).

Figure 14.11: Terms of a lens. The principal axis passes through the optical centre O. Parallel rays meet at the principal focus F; the focal length f is the distance OF. The centre of curvature 2F is twice as far from the lens as F.

Focal length (f): The distance from the lens to its focal point (the focal point may be real or virtual). For a converging lens the focal length is positive: it is the distance from the lens to the point where parallel rays converge after passing through the lens. For a diverging lens the focal length is negative: it is the distance from the lens to the point from which parallel rays appear to diverge (the virtual focal point).
Principal axis: An imaginary straight line passing through the centre of the lens. Light rays that pass through the centre of the lens along the principal axis are not deviated.
Principal focus (focal point): The point on the principal axis where parallel rays of light converge for a converging lens, and from which they appear to diverge for a diverging lens. For a converging lens the principal focus is on the opposite side of the lens from the incoming parallel rays. For a diverging lens it is on the same side as the incoming parallel rays, but it appears to be behind the lens.
Optical centre (O): The point of the lens through which the principal axis passes.
Centre of curvature: Lenses are parts of spherical materials, so the centre of the sphere is the centre of curvature of the lens. It is usually twice as far from the lens as the focal point, so it is marked 2F.

14.4.1Action of lenses on parallel raysOften examined

When parallel light rays strike a lens, the rays after refraction show a pattern that depends on the kind of lens. Lenses are classified as converging and diverging according to this behaviour (Figure 14.12).

Figure 14.12: Action of a lens on a parallel beam. (a) A convex lens brings the rays to a real focus on the far side. (b) A concave lens spreads them out so that they appear to come from a virtual focus (dashed) on the side the light comes from.

A. Converging lens

A converging lens is thicker at its centre and thinner at the edges. When a parallel beam of light passes through a thin converging (convex) lens, the lens refracts the light rays and converges them to a point known as the focal point on the other side of the lens. The key characteristics are:

Refraction and convergence: the parallel rays are bent towards the principal axis of the lens because of refraction. The curvature and the refractive index of the lens material cause this bending. After passing through the lens all the parallel rays converge at a single point on the principal axis, the focal point of the lens.
Formation of image: the image formed depends on the location of the object. The image from a converging lens is normally real and inverted, but in some cases it may be virtual (Section 14.4.2).

B. Diverging lens

A diverging lens is thinner at its centre and thicker at the edges. When a parallel beam of light passes through a thin diverging (concave) lens, the lens refracts the rays and diverges them as if they were coming from a point known as the focal point, on the same side of the lens from which the light is coming. The characteristics are:

Refraction and divergence: the parallel rays are bent away from the principal axis of the lens because of refraction. The divergence gives the impression that the rays are coming from a focal point behind the lens.
Virtual focal point: the focal point of a concave lens is virtual, because the light rays do not actually converge there. Instead they appear to be diverging from this point when extended backwards.
Formation of image: if an object is placed at infinity, the image formed by a diverging lens is at its virtual focal point, and the image is virtual, upright and diminished.

Parallel rays through the two lenses
	Converging (convex) lens	Diverging (concave) lens
Effect on a parallel beam	Rays bend towards the principal axis and meet	Rays bend away from the principal axis and spread out
Focal point	Real, on the far side of the lens	Virtual, on the same side as the incoming rays
Image of an object at infinity	Real and inverted, a point at F	Virtual, upright and diminished, at the virtual focus

14.4.2Ray diagrams for a converging lensOften examined

Lenses form images of objects placed in front of them. The location and nature of the image can be found by drawing a ray diagram. In this process we take a minimum of two light rays from the object, pass them through the lens from two different points and join them to give the image. The distance of the object from the lens determines the nature of the image formed.

A. Object at infinity

When the object is at infinity, the real image is formed at the focus. The image is highly diminished and seems to be a point. The rays coming from the body can be taken as parallel, because after travelling such a long distance they are nearly parallel, and the lens converges all the rays at the focal point (Figure 14.13 (a)).

B. Object beyond 2F

When the object is beyond the centre of curvature C (beyond 2F), a real and inverted image is formed between the centre of curvature and the focus. The image is smaller than the object. We start by drawing a ray from the top of the object through the centre of the lens, which continues in a straight line. Next we draw a ray from the top of the object parallel to the principal axis; after passing through the lens it refracts towards the focal point (Figure 14.13 (b)).

C. Object at 2F

When the object is at the centre of curvature (2F), a real image is produced at the same point. The image is the same size as the object, inverted and real. The two rays are drawn in the same way (Figure 14.13 (c)).

Figure 14.13 (a) to (c): convex lens, object at infinity, beyond 2F and at 2F. Two rays from the top of the object are enough: one parallel to the axis (refracted through F) and one through the optical centre O (undeviated). Where they meet is the top of the image.

D. Object between F and 2F

When the object is between the centre of curvature (2F) and the focal point (F), a real image is produced beyond the centre of curvature. The image is larger, inverted and real. The ray through the centre of the lens continues straight, and the ray parallel to the principal axis refracts towards the focal point; their intersection fixes the image (Figure 14.13 (d)).

E. Object at F

When the object is placed at the focal point (F), the real image is formed at infinity. The image is much larger than the object, inverted and real. The ray through the centre of the lens continues straight, and the ray parallel to the axis refracts towards the focal point. The two lines come out parallel to each other, so the image is formed at infinity (Figure 14.13 (e)).

F. Object inside F

When the object is placed between the focal point and the lens, a virtual image is created on the same side of the lens as the object. The image appears larger, upright and virtual. A ray from the top of the object through the centre of the lens continues in a straight line; a ray from the top of the object parallel to the principal axis bends towards the focal point after passing through the lens. As these two rays diverge they will not meet, so we extend dotted lines backward until they intersect. The intersection point of these dotted lines indicates the location of the image (Figure 14.13 (f)).

Figure 14.13 (d) to (f): convex lens, object between F and 2F (real, inverted, larger image beyond 2F), at F (rays leave parallel, image at infinity) and inside F (virtual, upright, larger image on the same side as the object; the dashed lines are drawn backwards to find it).

Images formed by a convex (converging) lens
Object position	Image position	Nature of the image	Size of the image
At infinity	At F	Real, inverted	Highly diminished (point-sized)
Beyond 2F	Between F and 2F	Real, inverted	Smaller than the object
At 2F	At 2F	Real, inverted	Same size as the object
Between F and 2F	Beyond 2F	Real, inverted	Larger than the object
At F	At infinity	Real, inverted (book)	Much larger than the object
Inside F	Same side as the object	Virtual, upright	Larger than the object

14.4.3Linear magnificationOften examined

Linear magnification is a fundamental concept in optics, particularly for lenses and mirrors. It describes how much larger or smaller an image appears compared with the actual object. It is also known as lateral or transverse magnification.

Linear magnification: The ratio of the height (or length) of the image to the height (or length) of the object.

M=\dfrac{I}{O}

M: linear magnification; I: height of the image; O: height of the object

To find M measure the height of the object with a ruler or any other measuring tool, then measure the height of the image produced by the lens, as it appears in the lens or on a screen.

Table 14.2: Linear magnification (as corrected below)
Condition	Meaning
\|M\|>1	The image is larger than the object
\|M\|<1	The image is smaller than the object
M>0	The image has the same vertical orientation as the object: it is upright
M<0	The image has the opposite vertical orientation to the object: it is inverted

In some cases the sizes of the image or object are too small to measure easily. Then we can find the linear magnification from the distances of the image and the object from the lens, because the ratio of the image distance to the object distance equals the ratio of the sizes.

Linear magnification (second definition): The ratio of the distance of the image from the lens to the distance of the object from the lens.

M=\dfrac{q}{p}

q: distance of the image from the lens; p: distance of the object from the lens

Figure 14.14: Linear magnification. O and I are the heights of the object and the image; p and q are their distances from the lens. Then M=I/O=q/p.

From equations 1 and 2 we can write

M=\dfrac{I}{O}=\dfrac{q}{p}

14.8

For a real image the distance +q is taken as positive, and for a virtual image the distance -q is taken as negative. In Figure 14.14 the magnification of an inverted real image is written M=-q/p=I/O, so a negative sign shows an inverted image.

Example 14.5: Image of an object in a converging lens

Find the distance and height of the image if an object of height 2\,\mathrm{cm} is placed at 6\,\mathrm{cm} from a converging lens with magnification 3. Also find the nature of the image if the focal point of the lens is 5\,\mathrm{cm} from the lens.

Given

O = 2\,\mathrm{cm}
p = 6\,\mathrm{cm}
M = 3
f = 5\,\mathrm{cm}

To find

Height of the image I
Distance of the image q
Nature of the image

Use the magnification relation for the heights.
M=\dfrac{I}{O}\ \Rightarrow\ I=MO
Put in the values.
I=3\times2\,\mathrm{cm}=6\,\mathrm{cm}
For the distance use the other form of the relation.
\dfrac{I}{O}=\dfrac{q}{p}\ \Rightarrow\ q=p\times\dfrac{I}{O}
Put in the values.
q=6\,\mathrm{cm}\times\dfrac{6\,\mathrm{cm}}{2\,\mathrm{cm}}=18\,\mathrm{cm}
The object (6\,\mathrm{cm}) is placed outside the focal point (5\,\mathrm{cm}), so the image is real, inverted, enlarged and on the opposite side of the lens from the object.

I=6\,\mathrm{cm},\quad q=18\,\mathrm{cm},\quad\text{real, inverted, enlarged}

Image height against object height

I=MO is a straight line through the origin whose gradient is the magnification. For M=3 the point (2,6) is Example 14.5; the line for M=0.5 is a diminished image.

Worked example: object distance from the image size

A convex lens of focal length 6\,\mathrm{cm} forms a sharp image on a screen. The object is 1.5\,\mathrm{cm} tall and the image is 4.5\,\mathrm{cm} tall. The screen is 24\,\mathrm{cm} from the lens. Find (a) the magnification, (b) the distance of the object from the lens, and (c) describe the image. Check that the position of the object agrees with the ray-diagram rules.

Given

O = 1.5\,\mathrm{cm}
I = 4.5\,\mathrm{cm}
q = 24\,\mathrm{cm}
f = 6\,\mathrm{cm}

To find

Magnification M
Object distance p
Nature of the image

Magnification from the heights.
M=\dfrac{I}{O}=\dfrac{4.5}{1.5}=3
The distances have the same ratio, M=q/p, so solve for p.
p=\dfrac{q}{M}=\dfrac{24}{3}=8\,\mathrm{cm}
The image is caught on a screen, so it is real, and a real image from a convex lens is inverted. Since M=3>1 it is enlarged.
Check with the ray-diagram rules: f=6\,\mathrm{cm}, so 2f=12\,\mathrm{cm}. The object is at 8\,\mathrm{cm}, between F (6\,\mathrm{cm}) and 2F (12\,\mathrm{cm}). The table of images says the image is then real, inverted, larger and beyond 2F. The image at 24\,\mathrm{cm} is beyond 12\,\mathrm{cm}, so everything agrees.
Common mistake: writing M=p/q=1/3 (upside down). A magnification of 3 means the image is three times as big, and it is also three times as far from the lens as the object.

Worked example: with f=6\,\mathrm{cm} the object is 8\,\mathrm{cm} from the lens, between F and 2F. The real, inverted image is 24\,\mathrm{cm} away and three times as tall.

M=3;\ p=8\,\mathrm{cm};\ \text{real, inverted, enlarged}

14.4.4Difference between real and virtual imagesOften examined

Lenses can make two types of images: a real image and a virtual image. The differences are given in Table 14.3.

Table 14.3: Difference between real and virtual images
Real image	Virtual image
Can be caught on a screen	Cannot be caught on a screen
Is always inverted	Is always erect
Formed when the rays of light, after reflection or refraction, actually meet at some point	Formed when the rays of light, after reflection or refraction, only appear to meet at some point
Formed by the actual intersection of light rays	Formed by the imaginary intersection of light rays
Can be formed on a screen	Can only be seen in mirrors (or through a lens)
Formed here by convex lenses only	Concave lenses, convex lenses and plane mirrors also form them
Formed on the other side of the lens from the object	Formed on the same side of the lens as the object
Example: the image on our TV screen	Example: our reflection in a mirror

14.5Applications of lenses

Lenses are fundamental optical elements that play an important role in a wide range of scientific, engineering, medical and everyday-life applications, from the simple magnifying glass to complex systems. They are used in:

consumer optics, such as eyeglasses, cameras and binoculars;
microscope optics, such as scientific and medical research and diagnostics;
telescope optics, such as using telescopes to explore the cosmos;
photographic optics, such as capturing images.

Some of the commonly used applications of lenses follow in the next sections: the magnifying glass, the prism (dispersion of light) and optical devices with a single lens.

14.5.1Magnifying glassOften examined

A single lens is used as a magnifying glass, a simple optical device designed to magnify objects. The lens used in a magnifying glass is a convex lens. When an object is placed near the lens so that its distance from the lens is less than the focal length, the lens produces an enlarged, erect and virtual image of the object. The eye perceives this image as larger than the actual object (Figure 14.15).

Figure 14.15: A magnifying glass. The object is closer to the lens than the focal length F, so the rays leave the lens still spreading. The eye traces them back (dashed) to an upright, enlarged, virtual image on the same side as the object.

How to use a magnifying glass

Position: the position of the object is crucial. Hold the magnifying glass close to the object you want to magnify.
Distance: move the lens closer to or further from the object until the image comes into clear focus.
Viewing: look through the lens to see the magnified image. The optical viewing distance is usually a few centimetres from the object.

When a convex lens works as a magnifier
Object distance	Image
Less than the focal length (p<f)	Enlarged, erect, virtual (a magnifying glass)
More than the focal length	A real, inverted image: it is not a magnifier

14.5.2Dispersion of light by a prismOften examined

A prism is an optical element which is usually made of glass with triangular and rectangular sides. It has the ability to bend the light passing through it because of refraction. Visible light has a range of wavelengths, and the refraction of light depends upon the wavelength, so different wavelengths are refracted to different extents. This disperses the light.

Dispersion of light: The splitting of white light into seven constituent colours when passed through a prism.

White light that travels through a prism is separated by the processes of refraction and dispersion. As white light enters the prism, it slows down and bends towards the normal. Each colour of light has a unique wavelength, and as the light moves through the prism each wavelength is refracted by a different amount. Shorter wavelengths, such as violet and blue, bend more than longer wavelengths such as red and orange. The light that leaves the prism is therefore split into seven distinct colours, ranging from red to violet. This collection of colours is known as the visible light spectrum (Figure 14.16 (a)).

Figure 14.16 (a): Dispersion of white light by a glass prism. Violet light is bent the most and red the least, so the white beam is spread into the seven colours of the visible spectrum.

Shorter wavelengths bend more
Colour	Wavelength	Bending in the prism
Violet	Shortest	Most
Blue	Short	More
Red	Longest	Least

Beyond the visible spectrum

There are wavelengths of light that the human eye cannot perceive, such as infrared and ultraviolet. These wavelengths can also be dispersed by the prism, but since they are invisible, other detection methods are necessary. One such method uses thermometers equipped with sensors that can detect a broad range of wavelengths. These thermometers can sense the infrared radiation emitted by objects, which depends on their temperature. When the dispersed light from the prism reaches such a thermometer, it can detect both infrared and visible light. By measuring the intensity of the infrared radiation, the thermometer can determine the temperature of the emitting body. Infrared thermometers can also measure human body temperature without direct contact (Figure 14.16 (b), an infrared thermometer).

14.5.3Some optical devices with a single lens

Lenses have a wide range of applications in everyday life. Even a single lens can be used to make many valuable devices like a magnifying glass, camera, projector and photographic enlarger. We can show how these devices form images with the help of ray diagrams.

A. Camera

Most people today have a mobile phone with a built-in camera, so we are all familiar with cameras. In simple terms a camera has a lens and an image film placed in line with the lens. Every time the button is pressed, light from an object positioned p units away from the camera lens enters the lens. The light then goes through a shutter that opens briefly to allow the light from the object to reach the photographic film or CCD. Here the information in the form of light about the object is received, creating a real and inverted image of the object at a distance q from the lens (Figure 14.17).

Figure 14.17: Image formation in a camera. The lens forms a real, inverted, smaller image at distance q on the film or CCD; the object is at distance p.

B. Projector

We commonly use slide projectors in classrooms, seminar rooms and halls, which show an enlarged image on the main screen or wall. The slide projector is also a single-lens device. In the projector there is a bulb which produces light; behind it a concave mirror reflects back the light coming from the bulb to increase the intensity of light, so that a sharp and detailed image can be formed.

The next part is the condensing lens. Its aim is to align the light rays into a straight path, which allows the maximum amount of light to reach the projector screen. After passing through the condensing lens the parallel light rays pass through the slide and enter the projection lens. The lens finally projects this light to the screen by magnifying it many times, so that a small object’s image may be seen in a very large size (Figure 14.18).

Figure 14.18: Image formation by a projector. The mirror and the condenser lens send a bright parallel beam through the slide; the projection lens throws a large, sharp, inverted image of the slide on the screen.

C. Photographic enlarger

The enlarger used in photography is a device that gives an image of an object larger than the original size. It consists of a projection assembly that includes an illumination system like a bulb, a lens for projecting the image onto the photographic paper and a mechanism for focusing the image on the paper.

An optical system called the condenser, a system of lenses, is used in it to focus the beam of light towards the enlarging lens through the film. The optical system of the photographic enlarger is similar to that used in the projector and is shown in Figure 14.19. The light of the bulb passes through the condenser lenses, which focus the light on the film; it then falls on the lens, which makes an enlarged image on the photographic paper. The final image is made clear by using intense light, to give more detail, and is enlarged compared with the object.

Figure 14.19: Photographic enlarger. The condenser lenses concentrate the light on the film; the projection lens forms an enlarged image on the photographic paper.

Single-lens devices
Device	Main parts	Image
Camera	Lens, shutter, film or CCD	Real, inverted image on the film or CCD at distance q
Slide projector	Bulb, concave mirror, condenser lens, slide, projection lens, screen	Large, sharp image of the slide on the screen
Photographic enlarger	Bulb, condenser lenses, film, projection lens, photographic paper	Enlarged image of the film on the photographic paper

14.6Visible spectrum

Among the large range of electromagnetic radiations, only a small portion is visible to the human eye. This portion is referred to as the visible spectrum, which is divided into seven distinct colour ranges. How do we see colours? Why is the sky seen blue? Why is the traffic signal to stop vehicles seen as red? The answers depend on how we see objects.

A Muslim scientist, Al-Hazen (Ibn al Haithem), gave the concept of vision. When light strikes an object, two things happen: some of the light is absorbed by the object (or passes through it), and the remaining light is reflected from the object and enters our eyes, allowing us to see.

How colours appear to us (Figure 14.20)
What the object does to white light	What we see
Absorbs all wavelengths and reflects none	Black
Absorbs all wavelengths except red	Red
Absorbs all wavelengths except blue	Blue
Only one colour of light falls on it and is reflected back	The colour of the incident light

The visible spectrum consists of seven colours, red, orange, yellow, green, blue, indigo and violet, each with its own range of wavelengths and frequencies (Figure 14.20 shows how colours appear). The colours in order of frequency and of wavelength are:

The visible spectrum (Section 14.6). The colours in order of increasing frequency run from red to violet, and in order of decreasing wavelength from red to violet.

The seven colours of the visible spectrum
Colour	Wavelength range	Frequency range	Place in the spectrum
Red	620 to 750\,\mathrm{nm}	400 to 480\,\mathrm{THz}	Largest range of wavelengths and the lowest frequencies
Orange	590 to 620\,\mathrm{nm}	480 to 510\,\mathrm{THz}	Wavelength larger than the other colours except red
Yellow	565 to 590\,\mathrm{nm}	510 to 530\,\mathrm{THz}	Longer wavelengths than green; shorter than red and orange
Green	500 to 565\,\mathrm{nm}	530 to 600\,\mathrm{THz}	Intermediate range in the visible spectrum
Blue	485 to 500\,\mathrm{nm}	600 to 620\,\mathrm{THz}	Longer wavelengths than indigo and violet
Indigo	450 to 485\,\mathrm{nm}	620 to 670\,\mathrm{THz}	Longer wavelengths than violet
Violet	400 to 450\,\mathrm{nm}	670 to 790\,\mathrm{THz}	Smallest wavelengths and the highest frequencies

14.7Human eyeOften examined

The human eye is one of the natural optical instruments, and the most interesting and important of all. It performs a vast number of functions. A Muslim scientist, Ibn al Haithem, was the first scientist who explained the image formation by our eyes, and he is known as the father of optics. He recognised that light reflected from objects enters the eye through the lens and is passed to the optic nerves.

Figure 14.21: Structure of the human eye. The cornea and lens focus light on the retina; the fovea is the part of the retina with the sharpest vision, and the optic nerve carries the signals to the brain.

The basic structure of the human eye is shown in Figure 14.21. The cornea and the lens form a system that acts like a single thin lens. For a clear vision a real image must be projected on the retina, which lies at a fixed distance from the lens. The flexible lens of the eye allows it to adjust its radius of curvature to produce an image on the retina for objects at different distances.

Parts of the eye (Figure 14.21)
Part	What the book says about it
Cornea and lens	Form a system that acts like a single thin lens. The cornea has a focal length of about 2.3\,\mathrm{cm} and the lens of the eye about 6.4\,\mathrm{cm}.
Retina	Lies at a fixed distance from the lens; the real image must fall on it. It is lined with the photoreceptors.
Fovea	The centre of the image falls on the fovea, which has the greatest density of light receptors and is the sharpest part of the visual field.
Optic nerve	Conveys the signals from the eye to the brain, where the visual processing ends.
Eye lens	Flexible: it can change its radius of curvature to focus objects at different distances on the retina.

Processing of the visual optic nerve impulses begins from the retina and ends in the brain. The image formed in the eye is much like the image made by a convex lens, that is, a real and inverted image. Although the images formed in the eye are inverted, the brain inverts them once more to make them seem erect.

Near point: The nearest point at which an object can be placed so that the eye can form a clear image of it on the retina. It is 25\,\mathrm{cm} from the eye.

14.7.1Colour perception

The back of the eye is lined with a thin layer called the retina. This is where the photoreceptors are located. The retina also contains the nerves that tell the brain what the photoreceptors are seeing. Photoreceptors are the special cells in the retina of the eye that are responsible for converting light into signals that are sent to the brain. Photoreceptors give us our colour vision and night vision. There are two types of photoreceptor cells: rods and cones (Figure 14.22 (a)).

Rods and cones
	Rods	Cones
Light level needed	Work at very low levels of light: a very small amount of light can activate a rod	Need a lot more light
Used for	Night vision and the motion of things around us	Seeing colour and fine detail
Colour	Cannot help in colour vision, so at night we see everything in almost grey scale	Responsible for the colourful view of the world
Number in the human eye	More than 100 million	About 6 million, many of them packed into the fovea (the sharpness of images)

We have three types of cones, red, blue and green, and each cone is most sensitive to one colour of light but can still detect other colours. This is possible because the responses of the cones overlap and the brain integrates the signals sent by the cones (Figure 14.22 (b)). Each cone stimulated by light gives the brain a signal for a particular colour, and the brain determines the colour perception. For example, if the red cone and the green cone are equally stimulated, the brain gives the colour yellow. The brain processes the electrical signals from the cones in the retina and combines them to produce colour perception. This three-colour system lets a human perceive a rich spectrum of colours by mixing the combination of red, blue and green. Because humans have cones of three different kinds, human vision is trichromatic.

Colour vision in different organisms
Kind of vision	Cone types	Examples and what they see
Trichromatic	Three	Humans: red, blue and green cones give a rich spectrum of colours
Dichromatic	Two	Dogs and cats: they see mainly shades of blue and yellow. This is advantageous for animals such as hunters that operate in dim light
Tetrachromatic	Four	Many birds: in addition to red, blue and green they have one extra cone that detects ultraviolet light, so they see colours and patterns invisible to humans

14.7.2Formation of image on a normal eye

In a normal eye the image is formed on the retina: all the rays coming from the object are focused on the retina, which makes a clear image of the object on the eye. The ray diagram for the formation of the image by the eye is shown in Figure 14.23. The light rays from the object, after passing through the lens of the eye, are focused on the retina.

Figure 14.23: Image formation by a normal eye. The cornea and lens bring the rays from every point of the object to a point on the retina, forming a real, inverted image.

Consider the paths of two light rays coming from the top of the object, named 1 and 2 in the book’s figure. They coincide again on the retina. Similarly, the rays coming from the bottom of the object also coincide at the retina, making a clear image on the retina.

14.7.3Short-sightedness (myopia)Often examined

Short-sightedness (myopia): The condition in which a person can see near objects clearly but distant objects appear blurry.

This condition usually occurs at an early age or in adults. The reason is that the eye over-converges the nearly parallel rays coming from a distant object, so they cross in front of the retina and the image is formed before reaching the retina (Figure 14.24 (a)). The rays reaching the retina are not focused but dispersed, so the image formed on the retina is blurry.

Correction

Short-sightedness can be corrected by using lenses, as in eyeglasses, or in the form of contact lenses. In both corrections a concave lens is used. As the eye in myopia converges the light rays coming from far-off objects too much, the rays cross before they reach the retina. To decrease this convergence we place a concave lens, which diverges the light rays coming from the object. The eye lens then receives more spread-out rays and converges them exactly on the retina (Figure 14.24 (b)).

Figure 14.24: Short-sightedness. (a) The rays from a distant object are focused in front of the retina. (b) A concave lens spreads the rays a little, so the eye lens focuses them on the retina.

Causes

The two common reasons for short-sightedness are that the eye lens is too strong (it converges the light too much) or that the eye is too long (Figure 14.25). If the eye lens converges more, the image is focused before the retina. Similarly, if the length of the eye increases, the eye lens focuses at the normal distance, but because the eye is long the retina is moved backward and the image does not form on the retina.

Reasons for short-sightedness (Figure 14.25)
Cause	Effect
Eye lens too strong (converges too much)	Image formed in front of the retina
Eye too long	Retina is moved backward, so the image is in front of it

14.7.3Long-sightedness (hyperopia)Often examined

In a normal eye, when the object is placed at the near point of the eye, the light rays coming from the object, after converging from the eye lens, meet at the retina and make a clear, real and inverted image (Figure 14.26 (a)).

Long-sightedness (hyperopia or hypermetropia): The condition in which a person can see distant objects clearly but near objects appear blurry.

This condition usually occurs in elderly people. The eye converges the diverging rays from a near object too little, so they cross after passing the retina and the image is formed behind the retina (Figure 14.26 (b)). The object is not at the near point N but nearer than it, at N’. The rays reaching the retina are not focused but dispersed, so the image formed on the retina is blurry.

Correction

Long-sightedness can be corrected by using lenses, as in eyeglasses or contact lenses. In both corrections a convex lens is used. As the eye in hypermetropia converges the light rays coming from nearer objects too little, the rays would cross after the retina. To increase this convergence we place a convex lens, which additionally converges the light rays coming from the object. Less spread-out rays then reach the eye lens, which converges them exactly on the retina (Figure 14.26 (c)).

Figure 14.26: Long-sightedness. (a) Rays from a near object would be focused behind the retina. (b) A convex lens makes them converge a little before they enter the eye, so they are focused on the retina.

Causes

The two common reasons for long-sightedness are that the eye lens is too weak (it converges too little) or that the eye is too short (Figure 14.27). If the eye lens converges less, the image is focused behind the retina. Similarly, if the length of the eye decreases, the eye lens focuses at the normal distance, but because the eye is short the retina is moved forward and the image does not form on the retina.

The two defects of vision compared
	Short-sightedness (myopia)	Long-sightedness (hyperopia)
Can see clearly	Near objects	Distant objects
Blurry	Distant objects	Near objects
Where the image forms	In front of the retina	Behind the retina
Eye lens	Too strong (over-converges)	Too weak (converges too little)
Eyeball	Too long	Too short
Usually found in	Early ages or adults	Elderly people
Corrected with	Concave (diverging) lens	Convex (converging) lens

Worked example: which lens should each person wear?

Ali can read the book in his hands easily but cannot read the writing on the board in class. His grandfather can see the board but cannot read the newspaper held at 25\,\mathrm{cm}. (a) Name the defect of vision of each. (b) Say where the image forms in each eye. (c) Name the lens that corrects each defect and explain what it does to the light.

Ali sees near objects clearly but distant ones blurry: he is short-sighted (myopic). His eye converges the nearly parallel rays from the board too much, so they cross in front of the retina and the image is formed before the retina.
A concave (diverging) lens corrects it. It spreads the rays slightly before they enter the eye, which reduces the convergence needed, so the eye lens focuses them exactly on the retina.
His grandfather sees distant objects clearly but near ones blurry: he is long-sighted (hyperopic). His eye converges the spreading rays from the nearby newspaper too little, so they would cross behind the retina and the image forms behind the retina.
A convex (converging) lens corrects it. It adds convergence to the rays before they enter the eye, so the eye lens brings them to a focus on the retina.
Common mistake: swapping the lenses. Remember that the short-sighted eye has too much converging power, so it needs a lens that diverges; the long-sighted eye has too little, so it needs a lens that converges.

\text{Ali: myopia, concave lens; grandfather: hyperopia, convex lens}

14.8Gravitational lensing

In the classical view of gravity, every point mass in the universe attracts any other point mass with a force known as the force of gravitation. This force depends upon the magnitudes of the masses and the distance between them. Einstein gave a modern view of gravitation, which is somewhat different from the classical view. In his famous general theory of relativity he gave the idea of the curvature of space-time around massive bodies.

In this theory, space-time, also called the space-time continuum, is a mathematical model that fuses the three dimensions of space and one dimension of time into a single four-dimensional continuum. There are no gaps in space or time: as Einstein showed, time and space are inseparably mixed. Each event requires three space dimensions and one time dimension.

To understand it, take space-time as a fabric of elastic material. When you place a heavy object on it, the massive body makes a curve in the fabric, with the centre of the curve at the location of the body. In the same way, Einstein took space and time as a fabric that is curved by massive bodies. Because of this curvature, lighter bodies that come close to a massive body start orbiting it along the curvature of space and time (Figure 14.28 (a)).

According to the classical theory light is a wave, and a wave cannot feel the gravitational attraction, as this force is only between masses. According to Einstein’s modern view, light can bend while passing by a massive body because of the curvature of space (Figure 14.28 (b)). This phenomenon was not given attention at that time but was later found experimentally. Light is bent towards the mass, producing an effect much like a converging lens, so it is called gravitational lensing.

Gravitational lensing: The phenomenon in which a massive celestial body causes a sufficient curvature of space-time for the path of light around it to be visibly bent, as if by a lens.
Gravitational lens: The body that causes the light to curve. It should be massive enough, like a galaxy, a black hole or a cluster of galaxies.

Figure 14.28 (b): Bending of light by gravity. The curved space-time around a massive body bends the path of the light towards the body. The observer sees the source in the direction the light arrives from (dashed), so its apparent position differs from the real one.

On a galactic scale the light from a distant galaxy can be lensed into several images while passing close by another galaxy on its way to Earth. Einstein predicted this effect, but he considered it unlikely that we would ever observe it. But in 1979 gravitational lensing was observed when the double quasar was discovered (quasars are the highly luminous cores of distant galaxies). It became the first example of a lensed object, as it was behind another galaxy: that galaxy lies exactly between the Earth and the double quasar. The quasar was called "double" because of a special effect of gravitational lensing: it can lead to the formation of multiple images of the same distant source (Figure 14.29 (a)).

Figure 14.29 (a): Multiple images. When a galaxy lies exactly between a distant source and the observer, light reaches the observer along two paths, one on each side of the galaxy, so the source appears twice (the dashed lines show where the images seem to be).

The light coming from the source bends from both sides of the intervening galaxy, making two images of the source. In some cases the source lies exactly behind a galactic object, and the bending of light all around the intervening object makes a ring of light around it, called an Einstein ring. The first time the bending of light by a heavy object was found was in 1919, during the observation of a total solar eclipse: the light was bent by the amount predicted by general relativity (Figure 14.29 (b)).

Effects of gravitational lensing
Effect	What is seen
Bending of starlight by the Sun	Light bent by the amount predicted by general relativity (observed in the total solar eclipse of 1919)
Multiple images	The source appears twice or more, as in the double quasar (1979)
Einstein ring	A ring of light around the intervening object when the source is exactly behind it

14.9Acoustic lensing

Sound is a form of energy which can be transmitted from one place to another. It is the form of pressure wave created by the vibration of bodies. The science of sound is known as acoustics.

Acoustics: The branch of physics that deals with examining and studying sound.

Acoustics deals with the production, transmission and effects of sound. The main application of acoustics is to make music or speech sound as good as possible. When a sound wave is used as an information carrier or an energy orientation method, its energy gradually reduces during propagation. Therefore we must use different techniques to focus sound waves to utilise sound energy efficiently.

Acoustic lenses can focus the energy of sound waves into a relatively small space, a region of high energy density, as shown in Figure 14.30. In this figure an electrical signal is converted into sound, which is released from a speaker (acting like an object plane). Then an acoustic lens focuses these sound waves on a single point at the end, which acts like the image plane. Acoustic lenses manipulate sound in a manner similar to how optical lenses handle light.

Acoustic lens: A lens that brings sound waves to a focus by having walls of collodion film and being filled with a heavy gas, which retards and refracts the sound waves.

Figure 14.30: Focusing sound with an acoustic lens. Sound spreading from the source (object plane) is refracted by the lens and brought to a focus on the image plane, where the energy is concentrated in a small region.

The aim of the acoustic lens is to focus sound in much the same way that an optical lens focuses light. In loudspeakers a mechanical device is used to improve the dispersion of high frequencies, so that the dispersion is much more uniform across the audible spectrum. A high-frequency loudspeaker mechanical acoustic lens spreads a single-point sound source into a parallel wavefront.

Types of acoustic lenses

There are two commonly used types of acoustic lenses: the slant-plate lens and the perforated-plate lens (Figure 14.31).

Types of acoustic lenses
Type	Construction	Action
Slant-plate lens	The design of the most commonly seen acoustic lens: a series of plates with carefully calculated hyperbolic shapes, giving horizontal response patterns	Makes selected portions of the expanding pressure wave travel farther than other portions, which shapes the pressure wave by controlling the dispersion
Perforated-plate lens	A collection of perforated barriers at the horn mouth; these perforated screens are ring shaped, with varying sizes of centre cutouts	Can diverge as well as converge the sound, just like diverging and converging lenses do with light

Materials and shapes of acoustic lenses

Acoustic lenses can be made from various materials like plastic, epoxy, rubber and liquids, for the reasons given below.

Materials
Material	Why it is used
Plastic	Commonly used because it can easily be fabricated and has good acoustic properties
Epoxy	Offers good sound permeability and can be moulded into specific shapes
Rubber	Used for its flexibility and the ability to deform the lenses
Liquid	Some acoustic lenses use liquid-filled chambers to converge or diverge sound waves

Acoustic lenses come in different shapes, each shape serving a specific purpose. They can be spherical, ellipsoidal, parabolic or have gradual geometric curves.

Shapes
Shape	What it does
Spherical acoustic lens	Has a curved surface like a sphere; focuses or diverges the sound waves according to its curvature
Ellipsoidal acoustic lens	Similar to a sphere but a little elongated in one direction (somewhat like an egg); can focus sound waves in a specific direction
Parabolic acoustic lens	Has a parabolic shape and is excellent at focusing sound waves to a single point
Gradual geometric acoustic lens	Lenses with gradual geometric curves achieve a gradient refractive index distribution, allowing precise acoustic focusing

Applications

Acoustic lenses are important for applications like underwater imaging, sonar systems and medical ultrasound. They are also used in non-destructive testing of materials which are opaque to light but which sound waves can penetrate, to find voids and cracks without destroying the material. Researchers are now exploring 3-D printed metamaterials with balcony-like structures to create impact-resistant acoustic lenses.

Formula sheet

Law of reflection

\theta_i=\theta_r

\theta_i: angle of incidence; \theta_r: angle of reflection, both measured from the normal (14.1)

Unit: degrees

Image in a plane mirror

d_i=d_o

d_i: distance of the image behind the mirror; d_o: distance of the object in front of it

Unit: \mathrm{m}

Refractive index

n=\dfrac{c}{v}

c: speed of light in vacuum (3\times10^8\,\mathrm{m\,s^{-1}}); v: speed of light in the medium (14.2)

no unit

Snell’s law

n_1\sin\theta_i=n_2\sin\theta_r

n_1, n_2: refractive indices of the first and second medium; \theta_i, \theta_r: angles of incidence and refraction (14.3)

Unit: angles in degrees

Refractive index from air (or vacuum) into a medium

n=\dfrac{\sin\theta_i}{\sin\theta_r}

\theta_i: angle in air; \theta_r: angle in the medium (14.4)

no unit

Refractive index from a medium into air (or vacuum)

n=\dfrac{\sin\theta_r}{\sin\theta_i}

\theta_i: angle in the medium; \theta_r: angle in air (14.5)

no unit

Critical angle

\sin\theta_c=\dfrac{n_2}{n_1}\qquad\theta_c=\sin^{-1}\!\left(\dfrac{n_2}{n_1}\right)

n_1: denser medium; n_2: rarer medium, with n_1>n_2 (14.6)

Unit: degrees

Refractive index from the critical angle

n=\dfrac{1}{\sin\theta_c}

n: refractive index of the denser medium; the second medium is air (14.7)

no unit

Linear magnification

M=\dfrac{I}{O}=\dfrac{q}{p}

I, O: heights of image and object; q, p: distances of image and object from the lens (14.8)

no unit

Key terms

Optics: The science of light and its interaction with matter.
Reflection of light: When a ray of light approaches a smooth polished surface and bounces back.
Normal: The line perpendicular to a surface (or to the boundary of two media) at the point where a ray meets it. Angles of incidence, reflection and refraction are measured from it.
Angle of incidence: The angle between the incident ray and the normal.
Angle of reflection: The angle between the reflected ray and the normal; it equals the angle of incidence.
Laws of reflection: First: the incident ray, the reflected ray and the normal lie in the same plane. Second: the angle of incidence equals the angle of reflection.
Regular (specular) reflection: Reflection from a smooth surface such as a mirror, which is clear and sharp and forms a distinct image.
Diffused reflection: Reflection from a rough surface, where the reflected rays scatter in many directions.
Refraction of light: The change in the direction of a light ray when it enters from one medium to another.
Angle of refraction: The angle between the refracted ray and the normal.
Speed of light in vacuum: A universal constant, c=3\times10^8\,\mathrm{m\,s^{-1}}. In transparent media the speed of light is smaller and depends on the density of the medium.
Refractive index: The measure of bending of a light ray passing from one medium to another; the ratio of the speed of light in vacuum to its speed in the medium, n=c/v. It has no unit and is never less than 1.
Snell’s law: n_1\sin\theta_i=n_2\sin\theta_r, together with the rule that the incident ray, the normal and the refracted ray lie in one plane.
Perspex block: A tough transparent plastic block used instead of a glass block.
Critical angle: The angle of incidence in the denser medium for which the angle of refraction is 90^\circ, so that the refracted ray moves along the boundary.
Total internal reflection: The reflection of all the light back into the same medium when it travels from a denser to a rarer medium with an angle of incidence greater than the critical angle.
Optical fibre: Concentric tubes of glass or plastic with two different refractive indices (core and cladding) used to transmit light by total internal reflection.
Lens: A transparent object that refracts light in a specific way to form images.
Convex (converging) lens: A lens that is thick at the centre and thin at the edges; it converges a parallel beam to a single point.
Concave (diverging) lens: A lens that is thin at the centre and thick at the edges; it diverges a parallel beam.
Focal length: The distance from the lens to its focal point; positive for a converging lens and negative for a diverging lens.
Principal axis: The imaginary straight line through the centre of the lens; rays travelling along it are not deviated.
Principal focus (focal point): The point on the principal axis where parallel rays converge (converging lens) or from which they appear to diverge (diverging lens).
Optical centre: The point of the lens through which the principal axis passes, written O.
Centre of curvature: The centre of the sphere of which the lens surface is a part; it is about twice as far from the lens as the focal point (2F).
Real image: An image that can be caught on a screen, formed where light rays actually meet; on the opposite side of the lens from the object.
Virtual image: An image that cannot be caught on a screen, formed where the rays only appear to meet; on the same side of the lens as the object.
Linear magnification: The ratio of the height of the image to the height of the object, M=I/O=q/p.
Magnifying glass: A single convex lens used with the object inside its focal length to see a larger, erect, virtual image.
Dispersion of light: The splitting of white light into its seven constituent colours by a prism because the colours are refracted to different extents.
Visible spectrum: The portion of the electromagnetic radiation visible to the eye, in seven colours from red (longest wavelength, lowest frequency) to violet (shortest wavelength, highest frequency).
Near point: The minimum distance at which an object can be placed so that the eye forms a clear image on the retina; 25\,\mathrm{cm} for the normal eye.
Short-sightedness (myopia): A defect in which near objects are clear and distant objects blurry because the image forms in front of the retina; corrected with a concave lens.
Long-sightedness (hyperopia): A defect in which distant objects are clear and near objects blurry because the image forms behind the retina; corrected with a convex lens.
Rods and cones: The two kinds of photoreceptor cells in the retina: rods work in dim light (night vision, no colour) and cones need more light and give colour vision.
Gravitational lensing: The bending of light around a massive celestial body, because of the curvature of space-time, as if by a lens.
Einstein ring: A ring of light formed when the source lies exactly behind the gravitational lens.
Acoustic lens: A device that focuses or diverges sound waves, for example a lens whose walls are collodion film filled with a heavy gas that retards and refracts sound.

Multiple choice questionsp. 155–156

A glass slab is dipped in a transparent liquid having same refractive index as that of glass slab. We cannot see the boundary of the two media (glass and liquid) due to ________ refraction.

AMaximum
BMinimum
CZeroAnswer
DUnit

Why

Light bends at a boundary only when the refractive index changes. Here both media have the same refractive index, so Snell's law n_1\sin\theta_1=n_2\sin\theta_2 with n_1=n_2 gives \sin\theta_2=\sin\theta_1, so \theta_2=\theta_1. The ray goes straight on, the amount of refraction is zero, and the boundary cannot be seen.

Maximum (A) and minimum (B) would mean that light is bent by some amount, which does not happen here. Unit (D) is the refractive index of vacuum, not an amount of bending.

An incident ray makes an angle of 45^\circ with a plane mirror. The angle of reflection is:

A30^\circ
B45^\circAnswer
C60^\circ
D90^\circ

Why

The 45^\circ is measured between the ray and the mirror surface. The normal is at 90^\circ to the surface, so the angle of incidence, measured from the normal, is 90^\circ-45^\circ=45^\circ.

By the second law of reflection, \theta_i=\theta_r, so the angle of reflection is 45^\circ. A ray lying along the mirror (option D, 90^\circ) would not reflect at all, and option A (30^\circ) does not follow from the law.

Note. The question does not say whether 45^\circ is measured from the mirror or from the normal. The reflection angle is 45^\circ in both cases.

Two plane mirrors are arranged parallel facing each other. The image(s) formed may be:

AOne
BTwo
CFour
DInfiniteAnswer

Why

Each mirror forms an image of the object, and each image acts as an object for the other mirror. These images are reflected again and again without end, so the number of images is infinite.

A single mirror gives one image, so one (A) and two (B) do not describe the repeated reflections. Four (C) is not a fixed number for parallel mirrors.

The apparent flattening of the Sun at the dawn and dusk is due to phenomenon of:

ARefractionAnswer
BReflection
CDispersion
DPolarization

Why

Near the horizon sunlight passes through a long path of air. Layers of air of different density bend (refract) light from the lower edge of the Sun more than light from the upper edge, so the Sun's disc looks flattened.

Reflection (B) sends light back and does not bend it through the air layers. Dispersion (C) splits light into colours and polarisation (D) is not the cause of this shape.

On summer days you see the reflections on roads which is called mirage. It is due to:

ARefraction
BReflection
CDispersion
DTotal internal reflectionAnswer

Why

On a hot road the air near the surface is less dense than the air above it. Light from the sky bends away from the normal in each layer of air until its angle reaches the critical angle, and then it is totally internally reflected back up to the eye. The road appears to be reflecting the sky.

Refraction alone (A) would carry the light on through the air and not send it back up. Ordinary reflection (B) needs a surface such as the road itself, and dispersion (C) splits colours.

Note. The book prints no key. Mirage is explained by total internal reflection in the layers of hot air, so D is taken as the answer.

The colors in the rainbow are formed by:

ARefraction
BReflection
CDispersionAnswer
DTotal internal reflection

Why

Each raindrop acts like a small prism. Light of different colours (wavelengths) is refracted by different amounts, so white sunlight is spread into the seven colours. This spreading is dispersion (section 14.5.2).

Refraction (A) bends light but on its own does not separate the colours. Reflection (B) and total internal reflection (D) change the direction of light without separating it.

The lenses in your eyes produce image on retina:

AInverted and virtual
BInverted and realAnswer
CErect and virtual
DErect and real

Why

The lens of the eye forms a real image on the retina, in the same way as a converging lens. A real image is inverted, and the brain turns it the right way up.

A virtual image (A and C) cannot be formed on a screen such as the retina, and an erect image (C and D) is not formed by the lens of the eye.

What is the speed of light in diamond if its refractive index is 2.5?

A1.2\times10^{8}\,\mathrm{m\,s^{-1}}Answer
B2.5\times10^{8}\,\mathrm{m\,s^{-1}}
C2.1\times10^{8}\,\mathrm{m\,s^{-1}}
D5.2\times10^{8}\,\mathrm{m\,s^{-1}}

Why

Use n=\dfrac{c}{v}, so v=\dfrac{c}{n}=\dfrac{3\times10^{8}\,\mathrm{m\,s^{-1}}}{2.5}=1.2\times10^{8}\,\mathrm{m\,s^{-1}}.

Option B is the refractive index written as a speed, which mixes up n and v. Options C and D are not obtained by dividing 3\times10^{8} by 2.5.

Note. The book's table gives 2.419 for diamond, which would give 1.24\times10^{8}\,\mathrm{m\,s^{-1}}. Option A is the closest answer.

In a convex mirror when the object is placed at 'c' the image formed will be:

ASmallerAnswer
BLarger
CSame size
DZero size

Why

A convex mirror always gives a virtual, erect and diminished (smaller) image, wherever the object is placed. With the object at C the image is smaller and lies behind the mirror between the pole and the focus.

A larger image (B) or the same size (C) is not formed by a convex mirror. A zero-size image (D) is not possible.

The working principle of optical fibers is:

ARefraction
BDiffraction
CReflection
DTotal internal reflectionAnswer

Why

Light enters the core of an optical fibre at an angle greater than the critical angle, so it is totally internally reflected at the core and cladding boundary again and again, and travels along the fibre (section 14.3.3).

Refraction (A) and reflection (C) on their own do not keep the light inside the core, and diffraction (B) is not the principle used.

If an object is placed at 10 cm from the convex lens having focal length 5 cm, the linear magnification of convex lens is:

A0.5
B1Answer
C2
D5

Why

The object is at 10\,\mathrm{cm}, which is 2F for a lens of focal length 5\,\mathrm{cm}. For an object at 2F the real image is also at 2F and is the same size (section 14.4.2).

So q=p=10\,\mathrm{cm} and |M|=\dfrac{q}{p}=\dfrac{10}{10}=1. The image is inverted, so M=-1 with the sign, but its size is the same as the object. Options A, C and D need an image at a different distance from the lens.

Non-luminous matter which can be studied with the help of gravitational lensing is called:

AVisible matter
BRed matter
CDark matterAnswer
DWhite matter

Why

Gravitational lensing shows mass that gives out no light. Matter that does not emit light but has a gravitational effect is called dark matter (section 14.8, Point to Ponder).

Visible matter (A) gives out light, and red (B) and white (D) matter are not terms for non-luminous matter.

Image formed by plane mirror is always:

AReal and inverted
BVirtual and erectAnswer
CReal and erect
DVirtual and inverted

Why

A plane mirror forms a virtual image, which cannot be caught on a screen, and it is erect, with the same orientation as the object (section 14.1.1, Mini Lab).

Real and inverted (A) or real and erect (C) images are formed by lenses and some curved mirrors, not by a plane mirror. Virtual and inverted (D) is not formed by a plane mirror.

Einstein rings are seen due to a phenomenon:

AConverging lensing
BDiverging lensing
CGravitational lensingAnswer
DAcoustic lensing

Why

When a distant source lies exactly behind a massive galaxy, its light bends around the galaxy on all sides and forms a ring called an Einstein ring. This is gravitational lensing (section 14.8).

Converging (A) and diverging (B) lensing are the actions of optical lenses on light, and acoustic lensing (D) bends sound waves, not light from galaxies.

The ________ phenomenon is used in non-destructive testing of materials for voids and cracks in opaque materials?

AConverging lensing
BDiverging lensing
CGravitational lensing
DAcoustic lensingAnswer

Why

Sound waves can pass through opaque materials where light cannot. Acoustic lenses focus or spread sound, and this is used in non-destructive testing to find voids and cracks without damaging the material (section 14.9.2).

Converging (A) and diverging (B) lenses act on light, which cannot pass through an opaque material. Gravitational lensing (C) concerns light from galaxies.

All waves exhibit the phenomenon of reflection but light waves are the only waves which exhibit the phenomenon of:

ARefraction
BDiffraction
CReflection
DTotal internal reflectionAnswer

Why

Refraction (A) and diffraction (B) are shown by all kinds of waves, sound and water waves included, and reflection (C) is already stated as a property of all waves. So these options cannot be the answer.

Total internal reflection (D) is the phenomenon the book connects with light, through optical fibres (section 14.3.3). It is the best of the four options.

Note. The statement is loose, because none of the options belongs to light alone. The book prints no key. D is chosen because total internal reflection is the option the book links with light.

The working principle of optical fibers is:

ARefraction
BDiffraction
CReflection
DTotal internal reflectionAnswer

Why

Light is kept inside the core of an optical fibre by total internal reflection at the boundary with the cladding (section 14.3.3). This is the same principle as question 10.

Refraction (A) only bends light at a boundary, and diffraction (B) and reflection (C) do not keep light travelling down the core.

Note. This question repeats question 10 in the book. The answer is the same.

The white light passing through the prism, due to refraction shows a behavior of ________ light:

AInterference
BDispersionAnswer
CReflection
DDiffraction

Why

A prism bends each colour by a different amount because the refractive index depends on the wavelength, so white light spreads into its seven colours. This is dispersion (section 14.5.2).

Interference (A), reflection (C) and diffraction (D) do not separate white light into colours by refraction.

The visible light spectrum lies in the range of ________ nm of wavelength:

A400 to 850
B500 to 700
C400 to 750Answer
D700 to 900

Why

The visible spectrum runs from about 400\,\mathrm{nm} (violet) to 750\,\mathrm{nm} (red), as shown by the colour ranges in section 14.6.

Options A and D include infrared or ultraviolet regions that the eye cannot see. Option B leaves out the violet and the red ends of the spectrum.

The rods and cones a human eye has are:

A6 million and 100 million
B100 million and 6 millionAnswer
C6 million and 6 million
D100 million and 100 million

Why

The retina has about 100 million rod cells, used for dim-light vision, and about 6 million cone cells, used for colour vision (section 14.7.1).

Option A reverses the two numbers, and options C and D give the wrong number of rods.

Constructed response questionsp. 156

Take some water in the bowl and put a small mirror in it such that some part of mirror is dipped in the water. Fall sunlight on the mirror in such a way that the reflection of mirror appears on the wall or white board of the class.

Solution

The seven colours are produced by dispersion: refraction in water separates white sunlight into its colours, and the mirror sends them back to the wall. The three parts below explain this in order.

(a) Why the seven color pattern is shown on the wall?

Sunlight is white light, a mixture of the seven colours. When it passes from air into water, each colour is refracted by a different amount, because each colour has its own wavelength. The colours leave the water along slightly different paths and spread out, so a band of seven colours appears on the wall.

\text{Dispersion of white sunlight into seven colours}

(b) What is the role of water in this phenomenon?

Water is a transparent medium with a refractive index greater than that of air. It bends the sunlight that enters it, and because the bending depends on the colour, the water works like a prism that separates the colours.

The dipped part of the mirror is where the light passes through water. Without the water, the mirror reflects white light only, and no spread of colours appears on the wall.

The basic principle is dispersion of light. The refraction of light depends on its wavelength, so each colour is refracted by a different amount. Violet, with the shortest wavelength, bends the most, and red, with the longest wavelength, bends the least.

Red (longest wavelength, bends least)
Orange
Yellow
Green
Blue
Indigo
Violet (shortest wavelength, bends most)

The light is refracted as it enters the water, reflected by the mirror, and refracted again as it leaves the water. The colours spread further at each refraction, so a band of seven colours is seen on the wall.

Short response questionsp. 157

Why is the index of refraction always greater than or equal to unity?

Solution

The refractive index is n=\dfrac{c}{v}, where c is the speed of light in vacuum and v its speed in the medium. Light is slowest in material media, never faster than in vacuum, so v\le c and therefore n\ge1.

Equality holds only for vacuum, where n=1. Gases such as air have n very close to 1 (1.000293), so they are often taken as unity.

Note. The book says the index 'cannot be equal to or less than unity' and then that it has unit value only for vacuum. The answer uses n\ge1, with equality only for vacuum.

Which physical process is responsible for the decrease in the speed of light in material mediums?

Solution

The decrease is caused by the interaction of light with the atoms and molecules of the medium. These interactions hinder the light, so it travels more slowly than in vacuum. A denser medium has more atoms and molecules per unit volume, so light is slowed more and the refractive index is larger.

How powder takes the shine off of a person's nose?

Solution

Shine on the nose comes from regular (specular) reflection on a smooth, oily skin surface. The reflected rays leave in one direction, so the eye receives a bright reflection.

Powder makes the surface rough. The light is then scattered in many directions (diffused reflection, section 14.1). Less light reaches the eye from any one direction, so the shine disappears.

Why an object in water always appears to be elevated to its real depth?

Solution

Light from the object travels up through the water and is refracted at the surface, where it passes into air. Light moves faster in air, so each ray bends away from the normal.

The eye traces the emerging rays back in straight lines. They appear to come from a point nearer the surface, so the object looks raised, at a smaller depth than its real depth. This is the same effect as the broken pencil illusion in Figure 14.3.

Refractive index of air is lower for air at high temperatures, explain.

Solution

The refractive index measures how much light is slowed by the medium. At high temperature air expands, so its density falls and there are fewer molecules in a given volume.

Light meets fewer molecules, so it travels closer to its speed in vacuum. Since n=\dfrac{c}{v}, a larger v gives a smaller n, so the refractive index of hot air is lower and closer to 1.

Rainbow forms due to dispersion of light by the rain drops suspended in air. How double rainbow form? Can a triple rainbow be formed?

Solution

Sunlight enters each raindrop, is refracted and dispersed into its colours, is totally internally reflected from the back of the drop, and is refracted again as it leaves. The light reaches the eye in the directions that give the rainbow.

A double rainbow forms when light is reflected twice inside each raindrop before it leaves. The second rainbow is fainter, and its colours are in the reverse order, with red on the inside and violet on the outside.

A triple rainbow can in principle be formed by three internal reflections. Each reflection loses some light, so the third rainbow is very faint and is not normally seen.

Can you photograph a virtual image? Explain.

Solution

Yes. A virtual image cannot be caught on a screen by itself, because the light rays do not actually meet there. However, the rays from a virtual image still enter the lens of a camera.

The camera lens converges these rays and forms a real image on the sensor (CCD). The camera therefore records the image. A photograph of your face in a mirror is an example.

What is meant by the negative magnification?

Solution

The linear magnification is M=\dfrac{I}{O}=\dfrac{q}{p}, where I and O are the heights of the image and the object. A negative value means that the image has the opposite vertical orientation to the object, so the image is inverted.

A negative magnification therefore describes a real, inverted image, such as the one a convex lens forms when the object lies beyond the focal point. The size of the image is given by the magnitude |M|.

Can a flat transparent glass be considered as a lens of infinite focal length? If yes, where does it form an image?

Solution

Yes. A flat glass slab has surfaces of infinite radius, so it does not converge or diverge light. Its focal length is infinite, and parallel rays pass through it without being brought to a focus.

The slab forms no real or virtual image in the way a lens does. Its only visible effect is a slight sideways shift in the position at which objects seen through it appear. A point at infinity is still imaged at infinity.

If we dip a lens in water what would happen to its focal length? Elaborate.

Solution

The focal length increases. A lens bends light because of the difference between its refractive index and that of the surrounding medium. Air has n\approx1, while water has n=1.333, so a glass lens in water has a smaller difference than in air.

The lens therefore bends light less, and the parallel rays meet at a focus further from the lens. A converging lens becomes a weaker converging lens when it is dipped in water.

When you open your eyes in water why the things look blurry?

Solution

The cornea is the main converging surface of the eye, and in air it has a focal length of about 2.3\,\mathrm{cm}. Its refractive index differs greatly from that of air, so the light is strongly bent at the cornea.

Water has a refractive index close to that of the cornea, so very little refraction takes place at the cornea in water. The light is hardly bent, the rays do not meet on the retina, and the image is blurred.

Evaluate the statement that 'a real image is always inverted'.

Solution

The statement is true for real images. A real image is formed where the light rays actually meet, and the real images formed by lenses and concave mirrors are always inverted (Table 14.3).

The statement does not apply to virtual images. A virtual image, such as the image in a plane mirror, is always erect.

Argue about the nature of gravitational lensing as optical or not?

Solution

Gravitational lensing is optical in its effect. Light from a distant source is bent by a massive body, which can form several images of the source or an Einstein ring, as a converging lens forms images.

It is not optical in its cause. An ordinary lens bends light by refraction in a transparent material. Gravitational lensing bends light because the massive body curves space-time, so no material is needed. It is best described as a lens-like effect of gravity on light.

Inspect the reason for the curved ceilings of school halls?

Solution

A curved ceiling reflects sound from the speaker or a student speaking, and sends it across the hall. The reflected sound reaches listeners at the back, who would otherwise hear less of it. The curved surface acts like a reflector.

The curve must be gentle. A deep curve focuses the reflected sound on some points and not others, which causes echoes and uneven loudness in parts of the hall.

Propose a reason for producing more sound by an empty container as compared to a filled one.

Solution

The air inside an empty container is set into vibration by the sound source, and the walls of the container reflect the sound waves back into the air inside. These waves reinforce each other, so the container gives a louder sound.

A filled container has the material inside it, which absorbs and damps the vibrations. Less sound energy is left to be sent out, so the filled container gives a weaker sound.

Long response questionsp. 157–158

What is reflection of light? Also explain the laws of reflection.

Solution

Reflection of light is the bouncing back of a light ray when it strikes a smooth, polished surface. The incident ray strikes the surface, and the reflected ray leaves it. Both angles are measured from the normal, the line drawn perpendicular to the surface at the point where the ray strikes.

Laws of reflection

First law: the incident ray, the reflected ray and the normal all lie in the same plane.
Second law: the angle of incidence equals the angle of reflection, \theta_i=\theta_r.

The laws of reflection: the normal is perpendicular to the surface, and the angle of reflection equals the angle of incidence.

\theta_i=\theta_r

The laws hold at every point of a reflecting surface, whatever the direction of the incident ray.

Discuss the types of reflection.

Solution

Reflection is of two types, depending on how smooth the surface is.

Regular or specular reflection

This happens on smooth, polished surfaces such as mirrors. Parallel incident rays keep their parallel directions after reflection, because every ray obeys the same law at the flat surface. A clear, sharp image is formed.

Diffused reflection

This happens on rough surfaces such as paper, walls and the page of this book. The surface is made of many small facets, each with its normal pointing in a different direction. Parallel rays are scattered in many directions, so we can see such objects from all sides, but no sharp image is formed.

Feature	Regular (specular)	Diffused
Surface	Smooth and polished	Rough
Reflected rays	Parallel, in one direction	Scattered in many directions
Image	Clear and sharp	No image
Example	Plane mirror	Paper, a wall

In both types the laws of reflection apply to each ray. Only the orientation of the normal differs from point to point on a rough surface.

How images are formed in plane mirrors? Explain.

Solution

An image in a plane mirror is formed where the reflected rays appear to meet. The reflected rays do not actually pass through the image point, so the image is virtual.

Ray diagram

Take a point on the object, for example its top. Draw a ray from it to the mirror.
At the point of incidence draw the normal, and make the reflected ray obey the second law: the angle of reflection equals the angle of incidence.
Extend the reflected ray backwards behind the mirror with a dashed line. The reflected rays reaching the eye are traced back to a point behind the mirror.
Where two or more extended rays meet is the image of that point. Repeating the construction for the base of the object gives the full image.

Image of an object in a plane mirror. The dashed lines are the extended reflected rays, and they meet at the virtual image.

The eye sees the image where the light appears to come from, behind the mirror. The mirror itself does not hold any light at that point.

Describe the position and characteristics of images formed by plane mirrors.

Solution

The mini lab in section 14.1 shows the position and characteristics of the image in a plane mirror. The table below summarises them.

Property	Image in a plane mirror
Position	Behind the mirror, at the same distance as the object: d_i=d_o
Nature	Virtual, so it cannot be caught on a screen
Orientation	Erect, the same way up as the object
Size	Same size as the object

The image appears behind the mirror because the eye traces the reflected rays back along straight lines. The distance d_o is measured from the object to the mirror, and d_i from the image to the mirror.

Elaborate the refraction of light and the laws of refractions.

Solution

Refraction is the bending of a light ray when it passes from one transparent medium into another. It happens because the speed of light changes at the boundary. Light slows down when it enters a denser medium and speeds up when it enters a rarer medium.

A ray entering a denser medium bends towards the normal, so the angle of refraction is smaller than the angle of incidence. A ray entering a rarer medium bends away from the normal.

Laws of refraction

First law: the incident ray, the refracted ray and the normal all lie in the same plane.
Second law (Snell's law): for two media with refractive indices n_1 and n_2, n_1\sin\theta_1=n_2\sin\theta_2, where \theta_1 is the angle of incidence and \theta_2 the angle of refraction.

n_1\sin\theta_1=n_2\sin\theta_2

Refraction at a boundary between air and glass. The refracted ray bends towards the normal, so \theta_r is smaller than \theta_i.

Special cases

If the ray travels along the normal, \theta_1=0, so \theta_2=0 and the ray passes straight through.
If light enters a medium from vacuum, n_1=1 and Snell's law gives \dfrac{\sin\theta_r}{\sin\theta_i}=\dfrac{1}{n} (equation 14.4 of the book).

Note. The book's section 14.2 calls the change of direction of light 'diffraction'. The correct term is refraction, and that is used here.

Analyze the speed of light in material medium.

Solution

Light travels at its greatest speed in vacuum, c=3\times10^{8}\,\mathrm{m\,s^{-1}}. In a material medium it travels more slowly, because it interacts with the atoms and molecules of the medium. The speed v in a medium of refractive index n is found from n=\dfrac{c}{v}.

v=\dfrac{c}{n}

Because n\ge1, the speed in any medium is less than or equal to c. A denser medium has a larger n, so light travels more slowly in it.

Example: speed of light in water

Given and find: n=1.333, find v

Given: n=1.333 for water, and c=3\times10^{8}\,\mathrm{m\,s^{-1}}. Find v.
Rearrange n=\dfrac{c}{v} to get v=\dfrac{c}{n}.
v=\dfrac{c}{n}
Substitute the values.
v=\dfrac{3\times10^{8}\,\mathrm{m\,s^{-1}}}{1.333}=2.25\times10^{8}\,\mathrm{m\,s^{-1}}

The speed of light in water is about 2.25\times10^{8}\,\mathrm{m\,s^{-1}}, which is 0.75 times its speed in vacuum.

Graph of speed against refractive index

Speed of light against refractive index

The speed falls as the refractive index rises. The vertical axis is in units of 10^{8}\,\mathrm{m\,s^{-1}}.

Note. The book's text gives the speed in diamond as 1.2\times10^{8}\,\mathrm{m\,s^{-1}}, but with n=2.419 the speed is 1.24\times10^{8}\,\mathrm{m\,s^{-1}}. Example 14.2 also states n=2.49 in its question and 2.419 in its solution. The value 2.419 from Table 14.1 is used here.

What is refractive index? Elaborate your answer.

Solution

The refractive index n of a medium is the ratio of the speed of light in vacuum to the speed of light in the medium: n=\dfrac{c}{v}. It measures how much the medium slows light down, and so how much light is bent when it enters the medium. It is a ratio of two speeds, so it has no unit.

Derivation Refractive index from Snell's law

Aim. Show that n=\dfrac{\sin\theta_i}{\sin\theta_r} for light entering a medium of index n from vacuum.

Assumptions.

The light travels from vacuum into a medium of refractive index n.

Snell's law for the two media, with n_1 for the first medium and n_2 for the second.
n_1\sin\theta_i=n_2\sin\theta_r
The first medium is vacuum, so n_1=1, and the second medium has n_2=n.
\sin\theta_i=n\sin\theta_r
Make n the subject.
n=\dfrac{\sin\theta_i}{\sin\theta_r}

Resultn=\dfrac{\sin\theta_i}{\sin\theta_r}

Measuring the two angles gives the refractive index of the medium.

Some refractive indices from Table 14.1 of the book.
Medium	Refractive index n
Air	1.000293
Water	1.333
Glass (crown)	1.52
Diamond	2.419

Every refractive index is greater than or equal to 1, because light is never faster in a material than in vacuum. A larger n means a larger bending of light.

Differentiate total internal reflection from reflection. Also derive a relation for critical angle.

Solution

Feature	Reflection	Total internal reflection
Where it happens	At any smooth surface, from either side	Only when light goes from a denser to a rarer medium, at an angle of incidence greater than the critical angle
Light refracted?	Some light is always refracted into the second medium	No light is refracted; all of it is reflected back into the denser medium
Law	Angle of incidence equals angle of reflection	Same reflection law, applied at the boundary inside the denser medium
Example	Plane mirror	Optical fibres

When light travels from a denser medium towards a rarer medium, its angle of refraction is larger than its angle of incidence. At the critical angle \theta_c the angle of refraction is 90^\circ, and the refracted ray travels along the boundary. For larger angles of incidence the light is totally internally reflected.

Derivation Critical angle and refractive index

Aim. Derive \sin\theta_c=\dfrac{n_2}{n_1} for light going from a medium of index n_1 into a rarer medium of index n_2, with n_1>n_2.

Assumptions.

Light travels from the denser medium, index n_1, into the rarer medium, index n_2.
At the critical angle the refraction angle is 90^\circ.

Total internal reflection. (a) Refraction at an angle of incidence less than the critical angle. (b) The critical angle: the refracted ray travels along the boundary. (c) Total internal reflection: the reflected ray stays in the denser medium.

Write Snell's law for the two media at the critical angle.
n_1\sin\theta_c=n_2\sin\theta_r
At the critical angle the refracted ray grazes the boundary, so \theta_r=90^\circ and \sin90^\circ=1.
n_1\sin\theta_c=n_2
Divide both sides by n_1.
\sin\theta_c=\dfrac{n_2}{n_1}
For light leaving glass or water into air, n_2=1 and n_1=n. Therefore \sin\theta_c=\dfrac{1}{n}, which gives the book's relation n=\dfrac{1}{\sin\theta_c}.
n=\dfrac{1}{\sin\theta_c}

Result\sin\theta_c=\dfrac{n_2}{n_1},\qquad \theta_c=\sin^{-1}\!\left(\dfrac{n_2}{n_1}\right)

Since \sin\theta_c\le1, total internal reflection needs n_1>n_2. Light must travel from the denser medium into the rarer one.

Example: critical angle of polystyrene

Given and find: n=1.49, find \theta_c

Given: n=1.49 for polystyrene in air. Find \theta_c.
Use \sin\theta_c=\dfrac{1}{n}.
\sin\theta_c=\dfrac{1}{1.49}=0.671
Take the inverse sine.
\theta_c=\sin^{-1}(0.671)=42.2^\circ

Light moving in polystyrene that strikes the boundary at an angle greater than 42.2^\circ is totally internally reflected.

Enlist the uses of fiber optics in telecommunication system.

Solution

An optical fibre is a long, thin strand of glass or plastic. Its core has a higher refractive index than the cladding around it. Light entering the core at an angle greater than the critical angle is totally internally reflected at the core and cladding boundary again and again, so it is carried along the fibre (section 14.3.3).

Light stays in the core by total internal reflection at the boundary with the cladding.

Uses in telecommunication

Data transmission: telephone, internet and other data signals are sent as light pulses over long distances.
High bandwidth: a fibre carries much more data per second than a copper wire.
High speed: data travels along the fibre at the speed of light in the core.
No electrical interference: light is not disturbed by electrical signals, which was a major problem in copper networks.
Long distance: signals travel for long distances, such as under-sea and inter-continental cables, with little loss.
Low power and low maintenance: fibres use less power, and they are light, flexible and long-lasting.
Conversion at the ends: at the receiving end, photodiodes convert the light signal into an electrical signal, which is then turned into text, voice or video.

Illustrate thin lenses. Also explain behavior of converging and diverging lens to parallel beam of light.

Solution

A lens is a transparent object with at least one curved surface, which refracts light to form images. A thin lens is one whose thickness is small compared with the distances involved. A converging (convex) lens is thicker at the centre, and a diverging (concave) lens is thinner at the centre.

Type	Shapes	Action on parallel rays	Focal length
Converging (convex)	Bi-convex, plano-convex, meniscus convex	Converges the rays to a real focal point F	Positive
Diverging (concave)	Bi-concave, plano-concave, meniscus concave	Spreads the rays, as if they came from a virtual focal point F	Negative

Principal axis: The straight line through the centre of the lens. A ray along it is not deviated.
Optical centre (O): The point of the lens through which the principal axis passes.
Principal focus (F): The point on the principal axis where parallel rays converge (converging lens) or appear to diverge from (diverging lens).
Focal length (f): The distance from the lens to its principal focus.

Behaviour with a parallel beam

A converging lens brings a parallel beam to a real focus on the other side of the lens. A diverging lens spreads a parallel beam, and the spread rays appear to come from a virtual focus on the same side as the incoming light.

Parallel rays through a converging lens meet at the real focus F. Through a diverging lens they spread out, and their extensions (dashed) meet at the virtual focus F.

Evaluate how images are formed by converging lens using ray diagram.

Solution

A ray diagram uses two rays from the top of the object, and the image is where they meet. The rays used are:

A ray parallel to the principal axis passes, after the lens, through the focus F on the other side.
A ray through the optical centre O goes straight through without being deviated.
A ray through the focus F on the object side emerges parallel to the principal axis (a third ray, used as a check).

Object between F and 2F: the real image is beyond 2F, inverted and enlarged.

Images formed by a converging lens for different object positions.
Object position	Image position	Nature	Size
At infinity	At F	Real, inverted	Point-sized
Beyond 2F	Between F and 2F	Real, inverted	Diminished
At 2F	At 2F	Real, inverted	Same size
Between F and 2F	Beyond 2F	Real, inverted	Enlarged
At F	At infinity	Real, inverted	Very large
Inside F	Same side as the object, further away	Virtual, erect	Enlarged

The image is real and inverted whenever the object is outside the focus. When the object is inside the focus, the rays diverge after the lens and the image is virtual and erect.

12FBISE SSC-II 2026 Annual I, Set B, Q.2 (x)FBISE SSC-II 2026 Annual I, Set D, Q.2 (x)

Differentiate between real and virtual images.

Solution

Differences between real and virtual images (Table 14.3 of the book).
Real image	Virtual image
Can be caught on a screen	Cannot be caught on a screen
Always inverted	Always erect
Formed where the rays of light actually meet	Formed where the rays only appear to meet
Formed by the actual intersection of light rays	Formed by the imaginary intersection of light rays
Formed on the other side of the lens from the object	Formed on the same side of the lens as the object
Example in the book: the image on a TV screen	Example: our reflection in a mirror

A convex lens and a concave mirror can both form real images. A concave lens, a convex lens and a plane mirror can form virtual images.

Note. The book's Table 14.3 lists only the convex lens as forming real images, and it lists the concave lens, convex lens and plane mirror as forming virtual images. The concave mirror is also able to form real images, which the answer includes.

Assess linear magnification as a property of thin lenses.

Solution

The linear magnification M is the ratio of the height of the image I to the height of the object O: M=\dfrac{I}{O}. It describes how much the image is larger or smaller than the object.

For a thin lens the same magnification can be written in terms of distances: M=\dfrac{q}{p}, where q is the distance of the image from the lens and p the distance of the object from the lens. Combining the two gives M=\dfrac{I}{O}=\dfrac{q}{p} (equation 14.8 of the book). For a real image q is taken as positive, and for a virtual image q is negative.

Meaning of the value of the linear magnification.
Condition	Meaning
\|M\|>1	Image is larger than the object (enlarged)
\|M\|<1	Image is smaller than the object (diminished)
M>0	Image is erect (same orientation as the object)
M<0	Image is inverted (opposite orientation)

Example: a converging lens with magnification 3

Given and find: O=2\,\mathrm{cm}, p=6\,\mathrm{cm}, M=3

Given: object height O=2\,\mathrm{cm}, object distance p=6\,\mathrm{cm}, and M=3. Find the image height I and image distance q.
Image height: from M=\dfrac{I}{O}.
I=MO=3\times2\,\mathrm{cm}=6\,\mathrm{cm}
Image distance: from M=\dfrac{q}{p}.
q=Mp=3\times6\,\mathrm{cm}=18\,\mathrm{cm}
The image is enlarged, real and inverted, on the opposite side of the lens from the object, because the object lies outside the focus.

The image height is 6\,\mathrm{cm} and the image is 18\,\mathrm{cm} from the lens.

Note. Table 14.2 of the book reverses the meanings of |M|>1 and |M|<1; the table above gives the correct meanings. Example 14.5 of the book gives f=5\,\mathrm{cm}, p=6\,\mathrm{cm} and M=3, which do not fit together (with f=5\,\mathrm{cm} and p=6\,\mathrm{cm} the magnification would be 5). This answer uses M and p only.

Discuss the dispersion of light through a prism.

Solution

A prism is a piece of transparent material, usually glass, with triangular ends. White light entering a prism is refracted at each face. The refractive index of glass depends on the wavelength of light, so each colour is bent by a different amount. Violet is bent the most and red the least. The white light is therefore split into its seven colours. This separation is called dispersion of light.

White light split by a prism. Red is bent least and violet is bent most.

The colours leave the prism in the order red, orange, yellow, green, blue, indigo and violet. The colour ranges are given in the table in the answer to question 16.

Note. The book gives the frequency range of violet as 670 to 790\,\mathrm{THz}. A wavelength of 400\,\mathrm{nm} corresponds to about 750\,\mathrm{THz}, so the upper limit 750\,\mathrm{THz} is used in the answer to question 16.

Design some single lens optical devices.

Solution

A single convex lens is used in several optical devices. The design depends on where the object is placed relative to the focus and to 2F. The main devices described in the book are:

Single lens devices and the image each forms.
Device	Object position	Image formed	How it works
Magnifying glass	Inside F, close to the lens	Virtual, erect, enlarged	The eye sees the enlarged image on the same side as the object, as if it were larger (section 14.5.1).
Camera	Beyond 2F	Real, inverted, diminished, on the CCD or film	The shutter opens briefly and the lens forms a real image on the sensor (Figure 14.17).
Slide projector	Between F and 2F	Real, inverted, enlarged, on the screen	A concave mirror and condenser lenses direct light through the slide, and the projection lens enlarges the image on the screen.
Photographic enlarger	Between F and 2F	Real, enlarged, on photographic paper	Condenser lenses focus the light through the film, and the projection lens enlarges the image onto the paper.

Analyze the visible spectrum of electromagnetic radiations.

Solution

Electromagnetic radiation covers a wide range of wavelengths. Only a small band, from about 400\,\mathrm{nm} to 750\,\mathrm{nm}, is visible to the human eye. This band is the visible spectrum, and within it the seven colours each have their own range of wavelengths and frequencies.

Wavelength and frequency ranges of the seven colours (Section 14.6 of the book).
Colour	Wavelength (nm)	Frequency (THz)
Red	620 to 750	400 to 480
Orange	590 to 620	480 to 510
Yellow	565 to 590	510 to 530
Green	500 to 565	530 to 600
Blue	485 to 500	600 to 620
Indigo	450 to 485	620 to 670
Violet	400 to 450	670 to 750

Example: frequency of green light

Given and find: \lambda=500\,\mathrm{nm}, find f

Given: wavelength \lambda=500\,\mathrm{nm}=5\times10^{-7}\,\mathrm{m}. Find the frequency f.
Use c=f\lambda, so f=\dfrac{c}{\lambda}.
f=\dfrac{c}{\lambda}
Substitute the values.
f=\dfrac{3\times10^{8}\,\mathrm{m\,s^{-1}}}{5\times10^{-7}\,\mathrm{m}}=6\times10^{14}\,\mathrm{Hz}=600\,\mathrm{THz}

Red has the longest wavelength and the lowest frequency. Violet has the shortest wavelength and the highest frequency. Light just beyond the violet end (ultraviolet) and just beyond the red end (infrared) cannot be seen by the eye.

Note. The book gives the violet frequency range as 670 to 790\,\mathrm{THz}. Since 400\,\mathrm{nm} corresponds to 750\,\mathrm{THz}, the table uses 750\,\mathrm{THz}.

Illustrate human eye and color perception.

Solution

The eye works like a camera. Light enters through the cornea, passes through the pupil, and is converged by the lens onto the retina. The cornea and lens together act as a single thin converging lens. They form a real, inverted image on the retina, and the brain turns the image the right way up.

Parts of the eye. The cornea and lens converge light from a distant object to a real, inverted image on the retina, at the fovea.

Photoreceptor	Number	Function
Rods	About 100 million	Work in dim light for night vision; they do not give colour
Cones	About 6 million, many packed into the fovea	Need bright light; give colour vision; three types sensitive to red, green and blue

Colour perception

Each type of cone responds most strongly to one colour range, but the ranges overlap. The brain combines the signals. For example, equal stimulation of the red and green cones is seen as yellow. Having three types of cone gives human beings trichromatic vision.

18FBISE SSC-II 2026 Annual I, Set M, Q.4 OR

Outline short-sightedness and long-sightedness also explain how they can be fixed?

Solution

Short-sightedness (myopia)

A short-sighted person can see near objects clearly but distant objects look blurred. The eye converges the rays from a distant object too strongly, so they meet in front of the retina. The rays then spread out on the retina, and the image is blurred. The cause is either an eye lens that is too thick or an eyeball that is too long.

It is corrected with a concave (diverging) lens, in eyeglasses or contact lenses. The concave lens spreads the rays before they enter the eye, so the eye lens converges them exactly on the retina.

Short-sightedness corrected: a concave lens spreads the rays from a distant object so that the eye lens focuses them on the retina.

Long-sightedness (hypermetropia)

A long-sighted person can see distant objects clearly but near objects look blurred. The eye converges the rays from a near object too weakly, so they would meet behind the retina. The cause is either an eye lens that is too thin or an eyeball that is too short.

It is corrected with a convex (converging) lens, in eyeglasses or contact lenses. The convex lens converges the rays before they enter the eye, so the eye lens focuses them exactly on the retina.

Long-sightedness corrected: a convex lens converges the rays from a near object so that the eye lens focuses them on the retina.

Defect	Where the image forms	Cause	Correction
Short-sightedness	In front of the retina	Eye lens too strong, or eye too long	Concave lens
Long-sightedness	Behind the retina	Eye lens too weak, or eye too short	Convex lens

Define and explain gravitational lensing.

Solution

Gravitational lensing is the phenomenon in which a massive celestial body, such as a galaxy, a black hole or a cluster of galaxies, curves space-time so much that light passing close to it is bent, as if by a lens. Einstein's general theory of relativity predicted this effect.

A massive body bends the light from a distant source along two paths, so the observer sees two images of the source.

Effects

Multiple images: light from the source bends on both sides of the intervening galaxy and forms several images. The double quasar, observed in 1979, is the first example.
Einstein rings: when the source lies exactly behind the massive body, the light is bent all around it and forms a ring of light.
Bending of starlight: in 1919 the bending of light by the Sun was observed during a total solar eclipse, and it was found to match the prediction of general relativity.

Explain acoustic lensing and its applications.

Solution

Acoustic lensing is the technique of using materials and shapes that focus or spread sound waves, in the same way that optical lenses focus or spread light. An acoustic lens is made of walls of a special material and is filled with a heavy gas, which slows and refracts the sound waves so that they come to a focus.

Types of acoustic lenses

Slant-plate lens: a series of plates with carefully calculated hyperbolic shapes. It controls the spread of the sound and gives a horizontal response pattern.
Perforated-plate lens: a set of ring-shaped perforated barriers at the mouth of a horn. It can converge or diverge sound, as converging and diverging optical lenses do with light.

Materials and shapes

Materials: plastic, epoxy, rubber and liquid-filled chambers.
Shapes: spherical, ellipsoidal, parabolic and gradual geometric shapes. Each shape serves a particular purpose; for example, a parabolic lens is good at focusing sound to a single point.

Applications

Underwater imaging and sonar systems.
Medical ultrasound.
Non-destructive testing: sound waves penetrate opaque materials and find voids and cracks without damaging the material.
Loudspeakers: a mechanical acoustic lens spreads high frequencies so that the dispersion is uniform across the audible range.

Numerical response questionsp. 158

Calculate the angle of refraction at the boundary of two media if the angle of incidence is 37^\circ.

Solution

Given

\theta_i = 37^\circ angle of incidence, measured from the normal (figure in the book)

To find

\theta_r, the angle of refraction

Snell's law needs the refractive indices of both media, n_1\sin\theta_i=n_2\sin\theta_r. The question gives neither index, so the angle cannot be found from Snell's law alone.
The printed answer is the complement of the incidence angle, 90^\circ minus 37^\circ, which is the angle the ray makes with the boundary.
\theta=90^\circ-37^\circ=53^\circ

Only the book's value is obtained from the data given. If the two media had the same refractive index the ray would not bend, and the angle of refraction would be 37^\circ.

\theta_r=53^\circ\ \text{(book value, see note)}

Book answer. 53^\circ

Note. The data are incomplete: the two refractive indices are needed for Snell's law. The book's answer 53^\circ equals 90^\circ-37^\circ, so it is measured from the boundary, not from the normal. Check the book's figure and the indices your teacher uses before relying on this value.

Calculate the speed of light in a medium if its refractive index is 1.7.

Solution

Given

n = 1.7
c = 3\times10^{8}\,\mathrm{m\,s^{-1}} speed of light in vacuum

To find

v, the speed of light in the medium

The refractive index is n=\dfrac{c}{v}, so v=\dfrac{c}{n}.
v=\dfrac{c}{n}
Substitute the values.
v=\dfrac{3\times10^{8}\,\mathrm{m\,s^{-1}}}{1.7}=1.76\times10^{8}\,\mathrm{m\,s^{-1}}

v\approx1.76\times10^{8}\,\mathrm{m\,s^{-1}}

Book answer. 1.76\times10^{8}\,\mathrm{m\,s^{-1}}

A light ray moving through a medium of refractive index 1.6 enters at an angle of incidence 21^\circ into the other medium of refractive index 1.7. Calculate the angle of refraction.

Solution

Given

n_1 = 1.6
n_2 = 1.7
\theta_1 = 21^\circ

To find

\theta_2, the angle of refraction

Write Snell's law and make \sin\theta_2 the subject.
\sin\theta_2=\dfrac{n_1\sin\theta_1}{n_2}
Substitute the values, using \sin21^\circ=0.3584.
\sin\theta_2=\dfrac{1.6\times0.3584}{1.7}=\dfrac{0.5734}{1.7}=0.3373
Take the inverse sine.
\theta_2=\sin^{-1}(0.3373)=19.71^\circ

\theta_2\approx19.71^\circ

Book answer. 19.71^\circ

Assess the value of the critical angle if light travels from a medium of refractive index 1.63 into a medium of refractive index 1.35.

Solution

Given

n_1 = 1.63 denser medium
n_2 = 1.35 rarer medium

To find

\theta_c, the critical angle

At the critical angle the refracted ray grazes the boundary, so \theta_r=90^\circ. Snell's law then gives \sin\theta_c=\dfrac{n_2}{n_1} (section 14.3.1).
n_1\sin\theta_c=n_2\sin90^\circ=n_2
Substitute the values.
\sin\theta_c=\dfrac{1.35}{1.63}=0.8282
Take the inverse sine.
\theta_c=\sin^{-1}(0.8282)=55.9^\circ

\theta_c\approx55.9^\circ

Book answer. 55.9^\circ

Measure the distance and height of image of an object with height and distance of 3\,\mathrm{cm} and 5\,\mathrm{cm} respectively from a lens having magnification as 3.

Solution

Given

O = 3\,\mathrm{cm} height of object
p = 5\,\mathrm{cm} distance of object from the lens
M = 3

To find

I, the height of the image
q, the distance of the image from the lens

The height of the image follows from M=\dfrac{I}{O}.
I=MO=3\times3\,\mathrm{cm}=9\,\mathrm{cm}
The distance of the image follows from M=\dfrac{q}{p} (section 14.4.3).
q=Mp=3\times5\,\mathrm{cm}=15\,\mathrm{cm}
Since M is positive, the image is erect. An erect image formed by a convex lens is virtual, so it lies on the same side of the lens as the object.

q=15\,\mathrm{cm},\qquad I=9\,\mathrm{cm}

Book answer. 15\,\mathrm{cm}, 9\,\mathrm{cm}

FBISE 2026 questions

Questions from the FBISE SSC-II 2026 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

FBISE SSC-II 2026 Annual I, Set B, MCQ 21 mark

The critical angle \theta_c for polystyrene (n=1.49) is:

A32.2^{\circ}
B60^{\circ}
C45^{\circ}
D42.2^{\circ}Answer

Why

The critical angle of a medium in air is given by \sin\theta_c=\dfrac{1}{n} (formula list). For polystyrene, \sin\theta_c=\dfrac{1}{1.49}=0.671, so \theta_c=\sin^{-1}(0.671)=42.2^{\circ}.

The tempting option C (45^{\circ}) would need n=\sqrt2=1.41, B (60^{\circ}) would need n=1.15, and A (32.2^{\circ}) would need n=1.88. None of these is the index of polystyrene.

FBISE SSC-II 2026 Annual I, Set B, Q.2 (iv) OR1+2 = 3 marks

What is refractive index? Calculate the refractive index of a medium if the speed of light in it is 2\times10^{8}\,\mathrm{m/s}.

Solution

Refractive index (1 mark). The refractive index of a medium is the ratio of the speed of light in vacuum to the speed of light in that medium. It measures how much a light ray bends when it enters the medium from vacuum. It has no unit and is never less than 1.

n=\dfrac{c}{v}

Given

v = 2\times10^{8}\,\mathrm{m\,s^{-1}} speed of light in the medium
c = 3\times10^{8}\,\mathrm{m\,s^{-1}}

To find

Refractive index n of the medium

Use n=\dfrac{c}{v} (2 marks).
n=\dfrac{3\times10^{8}}{2\times10^{8}}=1.5

n=1.5

Note. The paper prints "reflective index"; refractive index is meant.

FBISE SSC-II 2026 Annual I, Set B, Q.2 (ix) OR2+1 = 3 marks

An object is placed at 2\,\mathrm{cm} from a convex lens. It forms an image at 6\,\mathrm{cm}. Calculate the focal length of the lens. Also find the magnification.

Solution

Given

p = 2\,\mathrm{cm} object distance
q = 6\,\mathrm{cm} image distance

To find

Focal length f
Magnification M

Use the lens formula (2 marks).
\dfrac{1}{f}=\dfrac{1}{p}+\dfrac{1}{q}=\dfrac{1}{2}+\dfrac{1}{6}=\dfrac{3+1}{6}=\dfrac{2}{3}
Invert to get f.
f=\dfrac{3}{2}\,\mathrm{cm}=1.5\,\mathrm{cm}
Use M=\dfrac{q}{p} for the magnification (1 mark).
M=\dfrac{6\,\mathrm{cm}}{2\,\mathrm{cm}}=3
Check: the object (2\,\mathrm{cm}) lies between F=1.5\,\mathrm{cm} and 2F=3\,\mathrm{cm}, so the real image is beyond 2F (6\,\mathrm{cm}), inverted and larger than the object. This agrees with M=3 (the image is 3 times as large; the sign is negative, M=-3, because the real image is inverted).

f=1.5\,\mathrm{cm},\qquad M=3

FBISE SSC-II 2026 Annual I, Set B, Q.2 (x)1x3 = 3 marks

Give three differences between real and virtual images.

Solution

Real image	Virtual image
Can be caught on a screen	Cannot be caught on a screen
Formed where the rays of light actually meet	Formed where the rays only appear to meet (their backward extensions meet)
Always inverted	Always erect

A convex lens (and a concave mirror) can form a real image, for example the image on a screen. A plane mirror, a concave lens and a convex lens used as a magnifying glass form virtual images.

FBISE SSC-II 2026 Annual I, Set B, Q.4 OR(1+2)+2 = 5 marks

a. What is meant by total internal reflection? Make a ray diagram to explain total internal reflection. b. Derive the relationship between refractive index and critical angle, n=\dfrac{1}{\sin(c)}.

Solution

(a) What is meant by total internal reflection? Make a ray diagram to explain it.

Total internal reflection (1 mark). When light travels from a denser medium towards a rarer medium and the angle of incidence is greater than the critical angle, no light is refracted. All of it is reflected back into the denser medium. This is called total internal reflection.

Ray diagram (2 marks). Figure (c) shows total internal reflection: the ray in glass meets the glass and air boundary at an angle greater than the critical angle and is reflected back into the glass. Figure (b) shows the critical angle, and figure (a) shows ordinary refraction for an angle smaller than the critical angle.

(b) Derive the relationship between refractive index and critical angle, n=\dfrac{1}{\sin(c)}.

Derivation Relation between refractive index and critical angle

Aim. Derive n=\dfrac{1}{\sin\theta_c} for light going from a medium of refractive index n into air.

Assumptions.

Light travels from the denser medium (index n) into air (index 1).
At the critical angle \theta_c the angle of refraction is 90^\circ.

Write Snell's law for the boundary between the medium and air.
n\sin\theta_i=1\times\sin\theta_r
At the critical angle, \theta_i=\theta_c and the refracted ray grazes the boundary, so \theta_r=90^\circ and \sin90^\circ=1.
n\sin\theta_c=1
Divide both sides by \sin\theta_c.
n=\dfrac{1}{\sin\theta_c}

Resultn=\dfrac{1}{\sin\theta_c}

The larger the refractive index, the smaller the critical angle. Since \sin\theta_c\le1, total internal reflection needs light to go from the denser to the rarer medium.

FBISE SSC-II 2026 Annual I, Set D, MCQ 21 mark

The speed of light in a glass slab (when refractive index, n=1.5) is:

A5\times10^{8}\,\mathrm{ms^{-1}}
B2\times10^{8}\,\mathrm{ms^{-1}}Answer
C3\times10^{8}\,\mathrm{ms^{-1}}
D4\times10^{8}\,\mathrm{ms^{-1}}

Why

From n=\dfrac{c}{v} we get v=\dfrac{c}{n}=\dfrac{3\times10^{8}}{1.5}=2\times10^{8}\,\mathrm{m\,s^{-1}}.

Option C is the speed of light in vacuum, which would need n=1. Options A and D are larger than c, which is impossible, because the speed of light in a material medium is always less than in vacuum.

FBISE SSC-II 2026 Annual I, Set D, Q.2 (iv) OR1+2 = 3 marks

On what principle do optical fibres work? Give any two advantages of using optical fibres.

Solution

Principle (1 mark). Optical fibres work on total internal reflection. Light enters the core at an angle greater than the critical angle of the core and cladding boundary, so it is totally reflected again and again and travels along the fibre.

Advantages (any two, 2 marks).

High bandwidth: a fibre carries much more data per second than a copper wire.
Resistance to electrical interference: the signal is light, so electrical noise does not affect it.
Also: faster speed (data travel as light pulses), long distance without much loss of signal, and low power use.

FBISE SSC-II 2026 Annual I, Set D, Q.2 (ix) OR1.5+1.5 = 3 marks

If M=0.5 and a 2\,\mathrm{cm} tall object is placed at 5\,\mathrm{cm} distance, calculate the image distance and the image height.

Solution

Given

M = 0.5
O = 2\,\mathrm{cm} height of object
p = 5\,\mathrm{cm} object distance

To find

Image distance q
Image height I

The image distance follows from M=\dfrac{q}{p} (1.5 marks).
q=Mp=0.5\times5\,\mathrm{cm}=2.5\,\mathrm{cm}
The image height follows from M=\dfrac{I}{O} (1.5 marks).
I=MO=0.5\times2\,\mathrm{cm}=1\,\mathrm{cm}
Since M<1, the image is smaller than the object: it is half the size and half as far from the lens.

q=2.5\,\mathrm{cm},\qquad I=1\,\mathrm{cm}

Note. The paper does not say whether the image is real or virtual, so M is used as a number, as the formula list on the paper does.

FBISE SSC-II 2026 Annual I, Set D, Q.2 (x)3 marks

Differentiate between real and virtual images formed by a lens or mirror. (Three differences)

Solution

Real image	Virtual image
Can be caught on a screen	Cannot be caught on a screen
Formed where the rays of light actually meet	Formed where the rays only appear to meet (their backward extensions meet)
Always inverted	Always erect

A convex lens or a concave mirror can form a real image. A plane mirror, a concave lens, or a convex lens with the object inside its focal length form a virtual image.

FBISE SSC-II 2026 Annual I, Set D, Q.4 OR2+3 = 5 marks

a. What is critical angle? Give the mathematical formula of critical angle. b. State the conditions necessary for total internal reflection to occur.

Solution

(a) What is critical angle? Give the mathematical formula of critical angle.

Critical angle. The critical angle is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is 90^\circ, so that the refracted ray travels along the boundary.

Formula. Applying Snell's law with \theta_r=90^\circ gives, for light going from a medium of index n into air:

\sin\theta_c=\dfrac{1}{n}\qquad\text{or}\qquad n=\dfrac{1}{\sin\theta_c}

For two media, \sin\theta_c=\dfrac{n_2}{n_1}, where n_1>n_2.

(b) State the conditions necessary for total internal reflection to occur.

Two conditions must both be met:

The light must travel from a denser medium to a rarer medium (for example from glass or water into air).
The angle of incidence in the denser medium must be greater than the critical angle of the two media.

If light travels from a rarer to a denser medium, or if the angle of incidence is less than \theta_c, part of the light is refracted and total internal reflection does not occur.

Note. The marks for this question are not clearly printed; a 2+3 split is assumed.

FBISE SSC-II 2026 Annual I, Set M, MCQ 21 mark

If the angle of incidence is zero, the refracted ray will:

AContinue along the normalAnswer
BBend away from the normal
CBend towards the normal
DBe totally reflected

Why

If the angle of incidence is zero, the ray travels along the normal. Snell's law gives n_1\sin0=n_2\sin\theta_r, so \sin\theta_r=0 and \theta_r=0: the ray is not bent, whatever the refractive indices are.

Options B and C describe bending, which happens only when the angle of incidence is more than zero and the speed of light changes. Option D needs light going from a denser to a rarer medium at an angle greater than the critical angle.

FBISE SSC-II 2026 Annual I, Set M, Q.2 (iv) OR2+1 = 3 marks

State Snell's law. Give its mathematical expression.

Solution

Snell's law (2 marks). The laws of refraction are known as Snell's law:

First law: the incident ray, the normal and the refracted ray all lie in the same plane.
Second law: the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for a given pair of media. It relates the angles to the refractive indices of the two media.

Mathematical expression (1 mark).

n_1\sin\theta_i=n_2\sin\theta_r

For light entering a medium of index n from air (n_1=1) this becomes n=\dfrac{\sin\theta_i}{\sin\theta_r}, which is the form on the paper's formula list.

FBISE SSC-II 2026 Annual I, Set M, Q.2 (ix) OR2+1 = 3 marks

An image formed by a lens has a height of 10\,\mathrm{cm} while the object height is 5\,\mathrm{cm}. Find the magnification. Also state the condition when linear magnification is greater than one.

Solution

Given

I = 10\,\mathrm{cm} height of image
O = 5\,\mathrm{cm} height of object

To find

Magnification M
Condition for M>1

Use M=\dfrac{h_i}{h_o}=\dfrac{I}{O} (2 marks).
M=\dfrac{10\,\mathrm{cm}}{5\,\mathrm{cm}}=2
Condition (1 mark). Linear magnification is greater than one, |M|>1, when the image is larger than the object, that is when the image distance is greater than the object distance (q>p). For a convex lens this happens when the object is placed between F and 2F (real, inverted image) or inside F (virtual, erect image, as in a magnifying glass).

M=2

FBISE SSC-II 2026 Annual I, Set M, Q.2 (x)1x3 = 3 marks

Write the characteristics of the image formed by a magnifying glass when the object is placed within its focal length.

Solution

A magnifying glass is a convex lens. When the object is placed between the lens and its focal point (p<f), the image is:

Virtual: it cannot be caught on a screen, and it is on the same side of the lens as the object.
Erect: it has the same orientation as the object.
Enlarged: it is larger than the object (|M|>1).

FBISE SSC-II 2026 Annual I, Set M, Q.4 OR2.5+2.5 = 5 marks

Describe the use of lenses in the correction of long-sightedness and short-sightedness. Also draw diagrams.

Solution

Long-sightedness Long-sightedness (hypermetropia): correction with a lens, with diagram.

Defect. A long-sighted person can see distant objects clearly but near objects look blurred. The eye lens is too weak, or the eyeball is too short, so the rays from a near object would meet behind the retina.

Correction. A convex (converging) lens is placed in front of the eye, in spectacles or as a contact lens. It converges the diverging rays from the near object before they enter the eye, so the eye lens focuses them exactly on the retina.

Long-sightedness corrected: a convex lens converges the rays from a near object so that the eye lens focuses them on the retina.

Short-sightedness Short-sightedness (myopia): correction with a lens, with diagram.

Defect. A short-sighted person can see near objects clearly but distant objects look blurred. The eye lens is too strong, or the eyeball is too long, so the rays from a distant object meet in front of the retina.

Correction. A concave (diverging) lens is placed in front of the eye. It spreads out the rays from the distant object, so the eye lens converges them exactly on the retina.

Short-sightedness corrected: a concave lens spreads the rays from a distant object so that the eye lens focuses them on the retina.

Defect	Image forms	Corrected with
Long-sightedness	Behind the retina	Convex lens
Short-sightedness	In front of the retina	Concave lens

Exam focus

Total internal reflection: define it, draw the ray diagram, and derive n=1/\sin\theta_c; state the conditions for it. (long)
Set B Q.4 OR (a and b, 5 marks) and Set D Q.4 OR (critical angle formula and conditions, 5 marks) both asked it, and Set B MCQ 2 used \sin\theta_c=1/n for polystyrene. Same as comprehensive question 8.
Go to answer
Defects of vision: long-sightedness and short-sightedness, their causes, correcting lenses and diagrams. (long)
Set M Q.4 OR (5 marks, 2.5 + 2.5); same as comprehensive question 18. It is a diagram question, so practise both corrected-eye figures.
Go to answer
Linear magnification numericals using M=I/O=q/p and the lens formula 1/f=1/p+1/q. (numerical)
Q.2 (ix) OR appeared in all three sets: Set B (find f and M from p=2\,\mathrm{cm}, q=6\,\mathrm{cm}), Set D (find q and I from M=0.5) and Set M (M from heights and the condition |M|>1). Book numerical 5 is the same type.
Go to answer
Differentiate between real and virtual images (three differences). (short)
Q.2 (x) in Set B and Set D (3 marks each); same as comprehensive question 12.
Go to answer
Refractive index and the speed of light in a medium: n=c/v. (numerical)
Set B Q.2 (iv) OR (define n, find n when v=2\times10^{8}\,\mathrm{m/s}) and Set D MCQ 2 (v in glass with n=1.5). Book numericals 2 and MCQ 8 use the same formula.
Go to answer
State Snell's law and its expression n_1\sin\theta_i=n_2\sin\theta_r; the ray along the normal does not bend. (short)
Set M Q.2 (iv) OR (State Snell's law and its expression) and Set M MCQ 2 (angle of incidence zero). Book numericals 1 and 3 apply the law.
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Optical fibres: the working principle and two advantages. (short)
Set D Q.2 (iv) OR (principle and two advantages); book exercise MCQs 10 and 17 and comprehensive question 9 ask about the same principle.
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Characteristics of the image formed by a magnifying glass (object within the focal length). (short)
Set M Q.2 (x) (3 marks, one for each of virtual, erect, enlarged).
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Ray diagrams for a converging lens and the nature of the image for each object position. (long)
Not asked directly in the 2026 papers, but the real and virtual image and magnification questions above all depend on it; comprehensive question 11.
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