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 45s with the normal to a water- glass interface [Fig. 9.34(c)].

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

Angle of incidence, i = 60°

Angle of refraction, r = 35°

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

μ_g^a= sin i/sin r

= sin 60°/sin 35°= 0.8660/0.5736 = 1.51 ————–Eq-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 :

μ_w^a= sin i/sin r

= sin 60°/sin 47°= 0.8660/0.7314 = 1.184————–Eq-2

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

μ_g^w = μ_g^a/μ_w^a

=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:

sin i/sin r= μ_g^w

=>sin 45°/sin r = 1.275

sin r = (1/√2)/1.275 = 0.5546

Therefore ,r = sin⁻¹ (0.5546)

= 38.68°

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