Q.A ray of light travelling in glass (μ = 3/2) is incident on a horizontal glass-air surface at the critical angle θc. if a thin layer of water (μ = 4/3) is now poured on the glass-air surface, the angle at which the ray emerges into air the water-air surface is
(a) 60°
(b) 45°
(c) 90°
(d) 180°
Ans: (c)
Sol: For glass – air interface
$ \mu_g sin\theta_c = \mu_a sin90 $ ..(i)
For glass -water interface
$ \mu_g sin\theta_c = \mu_w sin\theta_1 $ …(ii)
For water -air interface
$\mu_w sin\theta_1 = \mu_a sin\theta_2 $ …(iii)
Solving (i) , (ii) & (iii)
$\theta_2 = 90°$