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Physics Question on Refraction of Light

A rectangular glass slab ABCD of refractive index n1n_1 is A rectangular glass slab ABCD of refractive index n2(n1>n2)n_2(n_1 > n_2) A ray of light is incident at the surface AB of the slab as shown. The maximum value of the angle of incidence αmax\alpha_{max} such that the ray comes out only from the other surface CD, is given by

A

sin−1[n1n2cos(sin−1n2n1)]sin^{-1}\Big[\frac{n_1}{n_2}cos\Big(sin^{-1}\frac{n_2}{n_1}\Big)\Big]

B

sin−1[n1cos(sin−1n1n2)]sin^{-1}\Big[n_1cos\Big(sin^{-1}\frac{n_1}{n_2}\Big)\Big]

C

sin−1(n1n2)sin^{-1}\Big(\frac{n_1}{n_2}\Big)

D

sin−1(n2n1)sin^{-1}\Big(\frac{n_2}{n_1}\Big)

Answer

sin−1[n1n2cos(sin−1n2n1)]sin^{-1}\Big[\frac{n_1}{n_2}cos\Big(sin^{-1}\frac{n_2}{n_1}\Big)\Big]

Explanation

Solution

Rays come out only from CD, means rays after refraction
from A B get total internally reflected at AD.
From the figure
r1+r2=90∘r_1+r_2=90^\circ
(r1)max=901∘−(r2)minand(r2)min=θC(r_1)_{max}=901^\circ-(r_2)_{min}and (r_2)_{min}=\theta_C( for total internal reflection at AD)
where, \hspace15mm sin\theta_C=\frac{n_2}{n_1}
or \hspace15mm \theta_C=sin^{-1}\Big(\frac{n_2}{n_1}\Big)
\therefore \hspace15mm (r_1)_{max}=90^\circ-\theta_C
Now, applying Snell's law at face A B
n1n2=sinαmaxsin(r1)max\frac{n_1}{n_2}=\frac{sin\alpha_{max}}{sin(r_1)_{max}}
=sinαmaxsin(90∘−θC=\frac{sin \alpha_{max}}{sin(90^\circ-\theta_C}
=sinαmaxcosθC=\frac{sin\alpha_{max}}{cos\theta_C}
or \hspace15mm sin\alpha_{max}=frac{n_1}{n_2}cos \theta_C
\therefore\hspace15mm \alpha_{max}=sin^{-1}\Bigg[\frac{n_1}{n_2}cos \theta_C\Bigg]
=sin−1[n1n2cossin−1(n2n1)]=sin^{-1}\Bigg[\frac{n_1}{n_2}cos sin^{-1}\Big(\frac{n_2}{n_1}\Big)\Bigg]