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)].

Answer 1

according to Snell's law,

$\mu_1sini=\mu_2sinr$

where , $\mu_1$ is refractive index of first medium , i is angle of incidence , r is angle of refraction and $\mu_2$ is refractive index of 2nd medium.

(a) applying Snell's law for the refraction from air to glass.

i.e., $\mu_a sini=\mu_g sinr$

or, $\frac{\mu_g}{\mu_a}=\frac{sini}{sinr}=\frac{sin60}{sin35}$

= 0.866/0.5736 = 1.51

hence, refractive index of glass with respect to air is 1.51

(b) now Snell's law for the refraction from air to water.

$\mu_asini=\mu_wsinr$

or, $\frac{\mu_w}{\mu_a}=\frac{sini}{sinr}=\frac{sin60}{sin47}$

= 0.866/0.6560 = 1.32

hence, refractive index of water with respect to air is 1.32

(c) Now the beam is incident at an angle of 45° from water to glass.

so, $\mu_w sini=\mu_gsinr$

or, $\frac{\mu_g}{\mu_w}=\frac{sini}{sinr}=\frac{sin45}{sinr}$

or, 1.51/1.32 = (0.7071)/sinr

or, sinr = 0.7071 × 1.32/1.51 = 0.6181

or, r ≈ 38.2°