# Figures 9.34(a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass-air and water-air interface, respectively.

Question:

Figures 9.34(a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water is 45º with the normal to a water-glass interface [Fig. 9.34(c)]. Solution:

As per the given figure, for the glass − air interface:

Angle of incidence, = 60°

Angle of refraction, r = 35°

The relative refractive index of glass with respect to air is given by Snell’s law as:

$\mu_{\mathrm{g}}^{\mathrm{a}}=\frac{\sin i}{\sin r}$

$=\frac{\sin 60^{\circ}}{\sin 35^{\circ}}=\frac{0.8660}{0.5736}=1.51$    ..(1)

As per the given figure, for the air − water interface:

Angle of incidence, i = 60°

Angle of refraction, r = 47°

The relative refractive index of water with respect to air is given by Snell’s law as:

$\mu_{\mathrm{w}}^{\mathrm{a}}=\frac{\sin i}{\sin r}$

$=\frac{\sin 60}{\sin 47}=\frac{0.8660}{0.7314}=1.184$     ...(2)

Using (1) and (2), the relative refractive index of glass with respect to water can be obtained as:

$\mu_{\mathrm{g}}^{\mathrm{w}}=\frac{\mu_{\mathrm{g}}^{\mathrm{a}}}{\mu_{\mathrm{w}}^{\mathrm{a}}}$

$=\frac{1.51}{1.184}=1.275$

The following figure shows the situation involving the glass − water interface. Angle of incidence, i = 45°

Angle of refraction = r

From Snell’s law, r can be calculated as:

$\frac{\sin i}{\sin r}=\mu_{\mathrm{g}}^{\mathrm{w}}$

$\frac{\sin 45^{\circ}}{\sin r}=1.275$

$\sin r=\frac{\frac{1}{\sqrt{2}}}{1.275}=0.5546$

$\therefore r=\sin ^{-1}(0.5546)=38.68^{\circ}$

Hence, the angle of refraction at the water − glass interface is 38.68°.