Refraction is the bending or change of direction of light as it moves from one medium

to another. e.g
When a ruler is placed in a glass of water, it appears bent at the boundary of air and
water. Similarly, a coin placed in water appears to be nearer the surface than it actu-
ally is.

Ruler Bent Coin appear

nearer

Water
Coin

In a uniform medium, light travels in a straight line, but as it moves from one medi-
um to another of different optical density (eg air to water) its velocity changes and
hence it bends.
When light travels from a less dense medium (air) to a denser medium (glass), it is
refracted towards the normal.
Incident ray
i
Air

Glass
r
Refracted ray

The incident angle i is greater than the refracted angle r. i.e i > r
When light travels from a denser medium (glass) to a less dense medium (air), it is
refracted away from the normal.
Incident ray

i
Glass

Air
r Refracted ray

The incident angle i is less than the refracted angle r. i.e i <r
However, if ray is travelling perpendicular (900) to the boundary it is not refracted.

Air

Water

F O R M 3 N O T E S: REFRACTION OF LIGHT Pg 1
Laws of refraction of light
1. The incident ray, the refracted ray and the normal, at the point of
incidence, they all lie on the same plane.
2. The ratio of sine of angle of incidence to the sine of angle of refrac-
tion, is a constant for a given pair of a media. (also called the Snell’s
law) i.e

i
𝑺𝒊𝒏 𝒊
= 𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕
r 𝑺𝒊𝒏 𝒓

Emergent ray
e

This constant is called the refractive index (n)

The ray that emerges out of the block is called the emergent and is parallel to the
incident ray hence the emergent angle (e) is equal to the incident angle (i)

REFRACTIVE INDEX (n)

Consider a ray of light traveling from medium 1 to medium 2.

Medium 1 i

Medium 2
r

The refractive index for medium 2 with respect to medium 1 is written as 1n2
If the light moves from medium 1 to medium 2 of refractive index n1 and n2 and
with angles i = θ1 and r = θ2 respectively as shown below, then:
n1sin θ1= n2 sin θ2
Note: The refractive index of air =1.0, hence if the first media is air, then
1xsin θ1=n2 sin θ2

𝑠𝑖𝑛 𝜽1 𝒔𝒊𝒏 𝒊
𝒏2 = 𝒏 =
sin 𝜽2 𝟏 𝟐 𝒔𝒊𝒏 𝒓
Hence,

F O R M 3 N O T E S: REFRACTION OF LIGHT Pg 2
 EXAMPLES
1. Find the refractive index of the glass medium below.
750
Air
Glass
400

2. Find the refractive index of water from the information below.

Air 400
Water
350

3. Given that anw= 1.33 and ang= 1.65. Find angle θ below

θ
Water
Glass
300

4. The refractive index of ethanol is 1.36. Find the refractive index of glycerol.

530
Glycerol
Ethanol
600

Note: If the ray of light reverses and moves from medium 2 to 1, it

travels back thru the same path making the same angles by
the principal of reversibility of light.

Medium 1

Medium 2

𝟏
Then the refractive index of medium 1 with respect to 2 is 𝒏 =
𝟐 𝟏 𝒏
𝟏 𝟐
EXAMPLES
1. The refractive index for light traveling from air to glass is 1.75. Find the re-
fractive index of glass to air.
2. A ray of light is refracted thru an angle of 350. Find the angle of incidence
given that gna= 0.67.

F O R M 3 N O T E S: REFRACTION OF LIGHT Pg 3
 REFRACTIVE INDEX IN TERMS OF VELOCITY OF LIGHT
The speed of light depends with the media in which it is traveling in. In vacuum
(air) it travels at a velocity of 3 x 108 m/s.
Absolute refractive index of a material is the ratio of velocity of light in vacuum to
the velocity of light in the material. i.e
𝐯𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝒐𝒇 𝒍𝒊𝒈𝒉𝒕 𝒊𝒏 𝒗𝒂𝒄𝒖𝒖𝒎
𝐧=
𝐯𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝒐𝒇 𝒍𝒊𝒈𝒉𝒕 𝒊𝒏 𝒎𝒂𝒕𝒆𝒓𝒊𝒂𝒍
EXAMPLES
1. Calculate the refractive index of glass given that the velocity of light in air is
3x 108 ms-1 and velocity of light in glass is 2.4 x 108ms-1.
2. Calculate the speed of light in water given that nw = 4/3 and C= 3 x 108m/s

REFRACTION THRU’ SUCCESSIVE MEDIA

If light passes thru a multiple transparent media whose boundaries are parallel to
each other, as shown below:
i Air 1

r1 Medium 2

Medium 3

Medium 4

Air
Then the combined refractive index of the multiple media is the product of all the re-
fractive indices of each media ie 1 n4=1n2 x 2n3X 3n4
EXAMPLES
1. A ray of light travels thru the media below. Calculate the refractive index 1n3.
400 AIR 1

240 Medium 2

260 Medium 3

2. If anw= 𝟒⁄𝟑 and a ng =𝟑⁄𝟐 . Find the refractive index of water with respect to
glass.

F O R M 3 N O T E S: REFRACTION OF LIGHT Pg 4
 3. The refractive index of ethanol is 𝟑𝟒⁄𝟐𝟓 and that of ruby is 𝟒𝟒⁄𝟐𝟓. Calculate
the refractive index of ruby with respect to ethanol.

REFRACTIVE INDEX IN TERMS OF REAL AND APPARENT DEPTH

When an object is placed under a transparent liquid or glass, it appears to be nearer
the surface than it actually is.
N1

S N2
Water Apparent depth

Real depth I 𝛉 Vertical displacement

O
Coin
The actual position of the coin is at O but it appears to be at I. The depth from ob-
ject to the surface (OS) is called the real depth while the depth from image to
the surface (IS) is called the apparent depth. The depth OI in which the ob-
ject is displaced from the bottom is called the vertical displacement.
The apparent depth depends with the angle of view from the normal N1. As the
angle reduces, the apparent depth increases such that it is maximum when 𝛉= 00,
i.e viewed perpendicularly. The refractive index of the medium is given as;
𝑹𝒆𝒂𝒍𝒅𝒆𝒑𝒕𝒉
𝒏=
𝑨𝒑𝒑𝒂𝒓𝒆𝒏𝒕𝒅𝒆𝒑𝒕𝒉

Note; While using this formula, the correct apparent depth is obtained when
the object is viewed perpendicularly thru N1

EXAMPLES
1. A coin placed at the bottom of a basin full of water to a depth of 32cm appears
to be 24cm from the surface. Find the refractive index of water.
2. A pin is placed at the bottom of a beaker containing a transparent liquid to a
depth of 24cm. If the pin is apparently displaced by 9cm from the bottom; cal-
culate the refractive index of the liquid.
3. A microscope is focused on a mark on a surface. A glass block 30mm thick is
placed on the mark. The microscope is then adjusted 10mm upwards to bring
the mark back to focus. Determine the refractive index of the glass.
4. A traveling microscope (M) is focused on a coin placed at the bottom of an emp-
ty beaker as shown below.
M
M 3cm

Coin Coin image

F O R M 3 N O T E S: REFRACTION OF LIGHT Pg 5
 When water of refractive index 𝟒⁄𝟑 is poured into the beaker, the microscope has to
be raised thru 3cm to focus the image of the coin. Calculate the height of water
poured into the beaker.
5. In an experiment to determine refractive index of water, the following results
were recorded from the experiment
Real depth (cm) 8.1 12 16 20
Vertical Displacement (cm) 2.2 2.9 4 4.9
Apparent depth(cm)
(i) Complete the table for apparent depth row.
(ii) Plot a graph of real depth against apparent depth.
(iii) Determine the refractive index for the water.

TOTAL INTERNAL REFLECTION

When light passes from an optically denser medium to a less dense medium
(eg glass to air), the refracted ray bends away from the normal ie i < r

Glass

Air
r
As the angle of incidence i increases, the angle of refraction r increases also.
Since the angle of refraction is always bigger than the angle of incidence, it reaches
a point where the refracted angle is 900 ie the refracted ray travels along the
boundary.
Glass

Where C is the critical angle

C

Air

Critical angle (C)– This is the angle of incidence in the denser medium
when the angle of refraction in the less dense medium is 900.
However, when the angle of incidence is increased beyond the critical angle, all the
light is internally reflected inside the denser medium in a process called total in-
ternal reflection.

i r

In this case the laws of reflection are obeyed. i.e ∠i = ∠r

F O R M 3 N O T E S: REFRACTION OF LIGHT Pg 6
 Conditions necessary for total internal reflection to occur
(i) Light must travel from an optically denser medium to a less
dense medium.
(ii) The incident angle must be greater than the critical angle.
REFRACTIVE INDEX IN TERMS OF CRITICAL ANGLE
𝒔𝒊𝒏 𝑪
n =
𝒈 𝒂 𝒔𝒊𝒏 𝟗𝟎𝟎
C Glass

But since sin 900 =1

hence
r Air
𝒔𝒊𝒏 𝑪
From Snell’s law
𝒔𝒊𝒏 𝒊 n =
n = 𝒈 𝒂 𝟏
𝒈 𝒂 𝒔𝒊𝒏 𝒓
Then
But at critical angle C =i and r = 900
hence 𝟏
𝒏 =
𝒂 𝒈 𝒔𝒊𝒏 𝑪
EXAMPLES
1. The refractive index for air-water boundary is 1.33. Calculate the critical angle
for water–air interface.
2. Critical angle of a material is 420, determine the angle of refraction of light in the
material when the incidence angle is 300.
3. Figure below shows the path of light thru’ a transparent material placed in air.
Calculate the refractive index of the transparent material.
Air

1380
Ray of light
Transparent
material

EFFECTS OF TOTAL INTERNAL REFLECTION OF LIGHT

(i) Mirages–This is a phenomenon observed during a hot day on a road
when the sky is clear which appears like a pool of water.
On a hot day the layers of air next to the ground are warmer and less dense than
the layers above. Since denser air has a higher refractive index, the refractive index
increases gradually from the ground upwards.
Rays of light from the sky undergoes multiple refraction and eventually they are
totally internally reflected. The observer sees the image of the sky on the ground
ahead of him which looks like pool of water.
Mirages are also observed in very cold regions but the light curves in opposite
direction such that a polar bear seems to be upside down in the sky

F O R M 3 N O T E S: REFRACTION OF LIGHT Pg 7
(ii) Atmospheric refraction – in the morning the sun is seen b4 it rises and
in the evening it is seen even after it has set due to refraction in the earth’s
atmosphere.
(iii) Glittering of precious stones. The critical angle of some precious stones
such as diamond is very small and that is why they glitter due to total internal
reflection

TOTAL INTERNAL REFLECTION PRISMS

These are right angled isosceles triangle of 900, 450, 450.

Deviation thru 900

The ray is turned thru 900
A
450

450
450
450
B C
C
The refractive index of glass is 1.5 hence its critical angle is
𝟏 𝟏 C =𝐬𝐢𝐧−𝟏 𝟎. 𝟔𝟔𝟕
𝒏 = 𝒏 = = 𝟎. 𝟔𝟔𝟕
𝒂 𝒈 𝒔𝒊𝒏 𝑪 𝒂 𝒈 𝟏. 𝟓 Thus C= 420

Therefore the ray enters side AB at a right angle hence it is not refracted. On
reaching side AC it strikes the prism at an incident angle of 450.Which is bigger
than the critical angle hence it is totally internally reflected. The ray meets side BC
normally and passes thru unrefracted.

Inversion with Deviation thru 1800

The ray is turned thru 1800

450

The image produced is always inverted.

450

F O R M 3 N O T E S: REFRACTION OF LIGHT Pg 8
 Inversion without Deviation
The image formed is inverted but not deviated.

This setup is used to view inverted

objects which make them upright.
45 45
0 0

EXAMPLES
1. Complete the path of the ray shown 2. Complete the path of the ray
until it leaves the glass prism, given shown until when it leaves the glass
that refractive index for glass is 1.6 prism, given that the refractive index
(Show all the angles). for glass is n= 1.3 (Show all the an-
gles).

450
300

APPLICATION OF TOTAL INTERNAL REFLECTION

1. Prism Periscope.
It is used to observed objects obstructed by obstacles. It uses two prisms of 900,
450,450

Prism

Obstacle

Observer

The image obtained is erect and virtual.

Prisms are preferred to plane mirrors in periscopes because
(i) Mirrors absorb some of the incident light.
(ii) Thick mirrors produce images that are blurred
(iii) The silvering on the mirror get tarnished and peel off.

F O R M 3 N O T E S: REFRACTION OF LIGHT Pg 9
 2. Optical fibre
These are thin glass fibres equal to the size of a hair strand. The inner core is made
with a glass of higher refractive index than the outer one.
A ray of light enters thru one end and undergoes repeated total internal reflection
until it emerges out of the other end of the tube.

Optical fibres are used in;

(i) Medicine to view internal organs
(ii) Telecommunication to transmit information such as data
and telephone networks. The message is converted in to a coded
light pulse. Optical fibres are better than ordinary copper wires cables be-
cause;
- They carry more info simultaneously,
- They are lighter, thinner and cheaper.

DISPERSION OF WHITE LIGHT

This is the separation of the white light in to its constituent seven colors .
When a narrow beam of light is directed to an equilateral prism, a patch of seven
colours of the rainbow called the spectrum is obtained ie Red, orange, Yel-
low, Green, blue, Indigo, and Violet in that order form top to bottom.
Screen

Red

Light
Violet

In vacuum, all colours of travel with the same velocity of 3 x 108m/s. However
their velocities changes when they enter into another media.
When white light enters the prism, it undergoes refraction. Each colour is refracted
at its own angle. The colour red with the longest wavelength and highest speed is
refracted (deviated) the least while violet with the shortest wavelength and lowest
speed is deviated the most.

RECOMBINIG THE COLOURS OF THE SPECTRUM

The seven colours of the spectrum can be recombined into the original white light
using:
(i) Using two prisms
(ii) Using a concave mirror
(iii) Using Newton’s disc

F O R M 3 N O T E S: REFRACTION OF LIGHT Pg 10
1. Using two prisms
Two equilateral triangle are placed as shown below.

Light

White light

2. Using a concave mirror

The mirror focuses the spectrum at one point on the screen.

Concave mirror

Light

White sport
Screen

3. Using Newton’s disc

This is a card with all the colours of the spectrum painted on it in equal areas

Yellow
Orange

Green
Red
Blue
Violet Indigo

When the card is rotated very fast, all the seven colours appear like one white col-
our.
THE RAINBOW
This is a natural phenomenon where light from the sun is dispersed and totally in-
ternally reflected
R

V
Total internal
V Reflection

NOTE; The observer of the rainbow looks away from the sun in order to see the
rainbow.

F O R M 3 N O T E S: REFRACTION OF LIGHT Pg 11
EXAMPLES
The figure shows the path of a yellow light through a glass prism. The speed of yel-
low light in the prism is 1.88 x 108 m/s.

600
Ө
r
Yellow light

a) Determine the refractive index of the prism material for the light. (Speed of
light in vacuum = 3.0 x 108 ms-1)
b) Given that r= 21.20, determine angle Ө

F O R M 3 N O T E S: REFRACTION OF LIGHT Pg 12
Exp to verify Snell’s law
Apparatus
Drawing pins, white sheet of paper, soft board and rectangular glass block.

P1

P2
450

D O C
r

A B
E
P3

Emergent ray
P4

- Fix the plain white paper on soft board using drawing pins.
- Place the rectangular block on the paper and trace its outline ABCD
- Remove the glass block and draw a normal at point O.
- Draw an incident ray making an angle of incident i= 450with the normal.
- Place the glass block to the outline again.
- Fix two pins P1and P2on the line.
- View pins P1and P2from AB and fix pins P3 and P4 such that all the pins appear
to be in a straight line.
- Draw a line joining P3and P4and produce the line to meet AB at E
- Draw a line joining point E to O
- Measure the angle of refraction r
𝒔𝒊𝒏𝒊
- Find the refractive index 𝒏=
𝒔𝒊𝒏𝒓

- Repeat the procedure with an angle of incident i = 600.

F O R M 3 N O T E S: REFRACTION OF LIGHT Pg 13
 Drawing pins, white sheet of paper, soft board and rectangular glass block.

P1
P2
i
D C

A B

- Fix the plain white paper on soft board.

- Place the rectangular block on the paper and trace its outline ABCD
- Remove the glass block and draw a normal at point O.
- Draw an incident ray making an angle of 200with the normal.
- Place the glass block to the outline.
- Fix two pins P1and P2on the line.
- View pins P1and P2 from AB and fix pins P3and P4 such that all the pins appear to
be in a straight line.
- Draw a line joining P3 and P4 and produce the line to meet AB
- Measure the angle of refraction r
- Repeat the procedure for other values of i =300, 400, 500, 600, and 700.

Angle of Incidence Angle of refraction Sin i Sin r

(i) (r)
200
300
400
500
600
700

(i) Plot A graph of sin r against sin r

(ii) From the graph determine the refractive index.

F O R M 3 N O T E S: REFRACTION OF LIGHT Pg 14