Question

Using Huygen's wave theory, derive Snell's law of refraction.

Answer 1

Snell's law of refraction using Huygen's Principle:

Consider a plane wavefront $AB$ incident on a surface $PQ$ separating two media 1 and 2.

The medium 1 is a rarer medium of refractive index $n_1$ in which light travels with a velocity $C_1$. The medium 2 is denser medium of refractive index $n_2$ in which light travels with a velocity $C_2$.

The angle between the incident ray $FA$ and the normal $NA$ at the point of incidence $A$ is equal to $i$.

The angle is also equal to the angle between the incident plane wavefront $BA$ and the surface of separation $PQ$.

So $\angle BAD$ is the angle of incidence of the incident plane wavefront $AB$.

Similarly, the angle between the refracted wavefront and the surface of separation $PQ$ is equal to the angle of refraction $r$.

i.e. $\angle ADC=r$.

Consider the triangles $BAD$ $ACD$ figure above.

$\sin i=\sin \angle BAD=\frac{BD}{AD}=\frac{C_{1}r}{AD}$

$\sin r=\sin \angle ADC=\frac{AC}{AD}=\frac{C_{2}r}{AD}$

$\frac{\sin i}{\sin r}=\frac{C_{1}t}{C_{2}t}=\frac{C_1}{C_2}=$ constant

This constant is called the refractive index of the second medium (2) with respect to the first medium (1).

$\frac{C_1}{C_2}=\frac{n_2}{n_1}=1n_2$

This equation proves the Snell's law.