Question

A plane light wave propagating from a rarer into a denser medium, is incident at an angle i on the surface separating two media. Using Huygen’s principle, draw the refracted wave and hence verify Snell’s law of refraction.

Answer 1

In order to determine the shape of the refracted wavefront, we draw a sphere of radius \(v_2\tau\) from the point A in the second medium. Let CE represent a tangent plane drawn from the point ‘C’ on to the sphere. Then \(AE = v_2 \tau\) and CE would represent the refracted plane wavefront.

At ‘A’ draw a normal N₁AN₂. In the \(\Delta\)ABC, AB is perpendicular to AA′ and AC is the perpendicular to AN1. We know that the angles between two lines and their perpendiculars are same. So, the angle between AB and AC

= angle between AA′ and AN₁ = i

In triangle AEC,

\(\angle A + \angle B +\angle C = 180^{\circ}\)

\((90^\circ-r)+ 90^\circ + \angle180^\circ, \text{so} \ \angle C = r\) 

If we now consider the triangles ABC and AEC, we obtain

\(sin i = \frac{BC}{AC} = \frac{v_1 \tau}{AC} \ \text{and}\ sin r = \frac{AE}{AC} = \frac{v_2\tau}{AC}\)

Here i and r are the angles of incidence and refraction respectively.

Thus we obtain \(\frac{sin\ i }{sin\ r} = \frac{V_1}{V_2}\)

If c represents the speed of light in vacuum, then

\(\mu_1 = \frac{c}{v_1}\ \text{and}\ \mu_2 = \frac{c}{v_2}\) are known as the refractive indices of medium 1 and medium 2 respectively

\(\therefore \frac{\mu_2}{\mu_1} = \frac{v_1}{v_2}; \ \text{Hence} \ \frac{sin \ i}{sin \ r} = \frac{\mu_2}{\mu_1}\)

\(\mu_1 sin\ i = \mu_2sin\ r\)

This is the Snell’s law for refraction