A rectangular glass slab ABCD, of refractive index $n_1$, is immersed in water of refractive index $n_2 (n_1 > n_2)$. A ray of light is incident on the surface AB of the slab as shown. The maximum value of the angle of incidence $α_{max}$, such that the ray emerges only from the surface CD is given by
Answer & explanation
Correct answer: option 1
Here, total internal reflection must take place at AD Applying Snell’s law at P we obtain,
$\frac{\sin α_m}{\sin r}=\frac{n_1}{n_2}$
Geometrically $r + θ_c = π/2$
where critical angle $θ_c =\sin^{−1} (n_2/n_1)$
$⇒ α_m = \sin^{−1} [ (\sin r) (n_1/n_2)]$
Putting $\sin r = \sin (π/2 - θ_c) = \cos θ_c = \cos [\sin^{−1}(n_2/n_1) ]$ in the above equation we obtain
$α_m = \sin^{−1}[\frac{n_1}{n_2}\cos(\sin^{-1}\frac{n_2}{n_1})]$