Refraction of Light

Light travels in straight line when it travels in a transparent medium. But as light pass through one transparent object to another different transparent object, e.g. air to water, it changes direction at the boundary of the two objects thus making it appear to be bending. This bending of light from one medium to another is called as the refraction of light.

Refraction depends on the medium where the light travels. The statements below describes the bending of light through different medium.

  1. The ray of light bends toward the normal as it enters an optically denser medium. Thus, the angle of refraction is less than the angle of incidence. The term optically dense is used to refer to material in which the speed of light decreases.
  2. As it enters a less optically dense medium, the ray of light is bent away from the normal. Thus, the angle of refraction is greater than the angle of incidence.

Light is bent due to the changing speed of light as it travels from one medium to another. As light travels from air (vacuum) to another material, the ratio of the speed of light in a vacuum to the speed of light in the other matherial is given as 

\({\text{speed of light in a vacuum} \over \text{speed of light in other material}}={c \over v}\).

This ratio is called as the index of refraction, symbolized as n, and is expressed as

\(n={c \over v}\).

The speed of the wave varies inversely to the index of refraction. The greater the index of refraction in the material, the slower the light travels in it. Light always travels more slowly in a vacuum than in any other material. The index of refraction in a vacuum is 1 and more than 1 for all the other materials.

The table below contains the indices of refraction of some materials.


Example 1.

If the speed of light in water is 2.25 x 10m/s, what is its index of refraction in water?

Given:

speed of light in water, v=2.25 x 10m/s

speed of light in air=3.0 x 10m/s

Find: index of refraction, n

Solution:

\(n={c \over v}={3.0 \times 10^8\;m/s \over 2.25 \times 10^8\;m/s}=1.33\)

Thus, the index of refraction in water is 1.33.


Example 2.

The index of refraction of diamond is 2.42. How fast does light travel in diamond?

Given:

n=2.42

Solution:

Derive the formula for speed from the equation: \(n={c \over v} \implies v={c \over n}\).

Substituting the values to c and n, we have

\(v={3.0 \times 10^8\;m/s \over 2.42}=1.24 \times 10^8 \;m/s\)


When a monochromatic light pass through one material to another, say, material a and b, the ratio of the sines of the angles \(\theta_a\) and \(\theta_b\), both angles are measured from the normal to the surface, is equal to the inverse ratio of the two indexes of refraction. Mathematically, it is written as

\({\sin \theta_a \over \sin \theta_b} = {n_b \over n_a}\)

or

\(n_a\sin \theta_a = n_b\sin \theta_b\)

The equation \(n_a\sin \theta_a = n_b\sin \theta_b\) and the observation that the incident ray, refracted ray, and the normla all lie on the same plane is salled as the Law of Refraction also called as Snell's Law which is named after the Dutch scientist Willebrord Snell.


Example 3.

Light travels from air into glass. If the angle of incidence is 35º, and the index of refraction of the glass is 1.52, what is the angle of refraction?

Given:

nair=1

nglass=1.52

\(\theta\)air=35º

Find: \(\theta\)glass

Solution:

Using the Snell's law equation \(n_a\sin \theta_a = n_b\sin \theta_b\), we have

\(n_{air}\sin \theta_{air} = n_{glass}\sin \theta_{glass}\)\(\implies\)\(\sin \theta_{glass}={n_{air}\sin \theta_{air} \over n_{glass}}\)

Substituting all the given values,

\(\sin \theta_{glass}={1(\sin 30^\circ) \over 1.52}={0.5 \over 1.52}=0.329\).

Take the inverse function of \(\sin \theta_{glass}\),

\(\theta_{glass}=\sin^{-1} 0.329=70.792^\circ \).

The angle of refraction is 70.792º.