Refraction and Critical Angle
CPD Resource for First Teaching
Refraction and Critical Angle
In a homogeneous medium a beam of light will propagate in a straight line. When it encounters a
boundary between two different transmission media, such as the boundary between water and air,
the beam will change in direction. This phenomenon is called refraction and it can be experienced by
any wave, including electromagnetic waves (such as light) and water waves. In this document, we
will only look at beams of light.
The “speed of light” in common usage often refers to the speed in vacuum, which is the highest
possible speed light can travel at. Light will slow down in other media, especially in liquids and solids.
As the speed of light even in a “slow” medium is still extremely high compared to the size of
everyday objects, the additional time necessary to travel through the medium can only be observed
with specialised equipment. Refraction of the beam, however, is easily observable by eye.
The ratio between the speed of light v in a substance and
vacuum c is called the refractive index n
𝑛=
𝑐
𝑣
The higher the refractive index of a material, the slower the
light will travel through it. In table 1 the refractive index for a
number of common substances is given.
Table 1: Refractive indices of common
materials
Material
Vacuum
Air
Water
Crown glass
Diamond
n
1
1.000293
1.33
1.52
2.42
To investigate how refraction changes the
direction of travel of a beam, we will now
consider a ray of light travelling through a
block of glass with 𝑛 = 1.52 and air with
𝑛 = 1.00 as shown in figure 1. When the
ray travels from glass to air, the light will
speed up and refract away from the
normal. If travelling from air to glass, the
light will slow down and refract towards
the normal. A small amount of light will
be reflected within the glass as shown.
Note that the change in angle only occurs
on the boundary between the two
materials! Occasionally the
Figure 1: The beam refracts when it travels from glass to air. A
misconception arises that the angle
small amount of light is reflected internally.
changes continuously while in the
material with the higher refractive index. This is not the case: the ray will travel in a straight line
through the glass.
Refraction and Critical Angle
CPD Resource for First Teaching
The angles made by a ray of light can be calculated using Snell’s law. One expression of this law
states that
𝑛 sin 𝜃 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
with 𝑛 the refractive index of the medium and 𝜃 the angle between the normal and the incoming or
refracted ray. Note that this is not a universal constant! It relates the angle of incidence to an angle
of refraction for a certain substance.
Snell’s law can also be expressed as
𝑛1 sin 𝜃𝑖 = 𝑛2 sin 𝜃𝑟
where 𝑛1 is the index of refraction of the medium the light is travelling from, 𝜃𝑖 the angle of
incidence, 𝑛2 the index of refraction of the medium the light is travelling into and 𝜃𝑟 the angle of
refraction.
Applying Snell’s law to the example in figure 1, we can calculate the angle of incidence. For the
transition from glass to air, Snell’s Law can be rewritten to find the angle of refraction 𝜃𝑟
𝑛1 sin 𝜃𝑖 = 𝑛2 sin 𝜃𝑟
sin 𝜃𝑟 =
n𝑔𝑙𝑎𝑠𝑠
sin 𝜃𝑖
n𝑎𝑖𝑟
For 𝜃𝑖 = 30° this becomes
𝑛𝑔𝑙𝑎𝑠𝑠
sin 𝜃𝑖 = 1.52 sin 30°
𝑛𝑎𝑖𝑟
1.52
𝜃𝑟 = sin−1
= 49°
2.0
sin 𝜃𝑟 =
When looking closely at Snell’s law it is clear that if the
angle of incidence becomes large there is a problem: the
RHS of the equation can become >1. Recall that sin 𝜃𝑟 ≤
1. The critical angle C where sin 𝜃𝑟 = 1 can be found by
substituting sin 𝜃𝑟 = 1 into Snell’s equation and
rearranging
sin 𝐶 =
𝑛2
𝑛1
Figure 2: Total internal reflection
where 𝑛2 < 𝑛1 .
If 𝑛2 > 𝑛1 , there is no critical angle. The refractive index of air is very close to 1, we can simplify to
sin 𝐶 =
1
𝑛
Refraction and Critical Angle
CPD Resource for First Teaching
The critical angle of incidence from glass to air is then given by
sin 𝐶 =
1
=
1
1.52
𝑛𝑔𝑙𝑎𝑠𝑠
1
𝐶 = sin−1
= 41°
1.52
When the angle of incidence becomes greater than the critical angle, total internal reflection occurs,
as shown in figure 2. All of the light will now be reflected back into the glass, with no light crossing
the boundary into the air.
Application
The principle of total internal reflection is used in fibre optic cables for high-bandwidth digital
communications. The fibre is a very thin and very long glass (or plastic) tube with a light source
(usually a laser) on one end and a detector on the other. Due to the high refractive index of the glass
the beam of light undergoes total internal reflection and is transmitted with a low loss over a large
distance. A video showing this principle is available on Youtube:
https://www.youtube.com/watch?v=0MwMkBET_5I
Summary
Refraction occurs when light encounters a boundary between two transmission media with a
different speed of light, also called a different refractive index. Snell’s Law can be used to determine
the angle of refraction. When travelling from a medium with a higher refractive index to a lower
refractive index, total internal reflection can occur.
Refraction has many practical applications within optics, especially when constructing lenses where
carefully shaped materials with high refractive indices are used to magnify objects. This is however
beyond the current Physics A specification.
Refraction and Critical Angle
CPD Resource for First Teaching
Exercises

A beam of light passes from air into water at 𝜃𝑖 = 25°.
o Calculate the angle of refraction.

A beam passes from diamond into air
o Calculate the angle of refraction when 𝜃𝑖 = 10°
o Calculate the critical angle of diamond

A beam hits a block of glass at 𝜃𝑖 = 0°
o Calculate 𝜃𝑟

A ray of light is shone at a block of an unknown material. Measurements show that 𝜃𝑖 = 15°
and 𝜃𝑟 = 9.0°
o Calculate the refractive index of the medium (answer 1.65).

A beam of light enters a fish tank of 50cm
wide at 𝜃𝑖 = 25° as shown. Assume the walls
are thin and have no influence on the result
o Calculate the angle of refraction going
in to the water.
o Calculate the angle of refraction
coming out of the water.

If an unknown material is made into a semi-circular
block, it can be used to determine the refractive
index by measuring the critical angle
o Why does the ray not refract when it first hits
the surface of the block?
o A student performs the experiment and finds
𝐶 = 48°, what is the refractive index of the
block? (answer 1.35)