Question

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

• A. $si{n}^{-1}\left[\frac{n_1}{n_2}cos\right(si{n}^{-1}\frac{n_2}{n_1}\left)\right]$
• B. $si{n}^{-1}\left[n_{1}cos\right(si{n}^{-1}\frac{n_1}{n_2}\left)\right]$
• C. $si{n}^{-1}\left(\frac{n_1}{n_2}\right)$
• D. $si{n}^{-1}\left(\frac{n_2}{n_1}\right)$

Answer 1

Answer: (A)

Rays come out only from CD, means rays after refraction
from A B get total internally reflected at AD.
From the figure
$r_1+r_2=90^{\circ }$
$(r_1{)}_{max}=901^{\circ }-(r_2{)}_{min}and(r_2{)}_{min}={\theta }_C$( for total internal reflection at AD) where, $sin{\theta }_C=\frac{n_2}{n_1}$
or ${\theta }_C=si{n}^{-1}\left(\frac{n_2}{n_1}\right)$
$\therefore (r_1{)}_{max}=90^{\circ }-{\theta }_C$
Now, applying Snell's law at face A B
$\frac{n_1}{n_2}=\frac{sin{\alpha }_{max}}{sin(r_1{)}_{max}}$
$=\frac{sin{\alpha }_{max}}{sin(90^{\circ }-{\theta }_C}$
$=\frac{sin{\alpha }_{max}}{cos{\theta }_C}$
$orsin{\alpha }_{max}=frac{n}_1{n}_{2}cos{\theta }_C$
$\therefore {\alpha }_{max}=si{n}^{-1}\left[\frac{n_1}{n_2}cos{\theta }_C\right]$
$=si{n}^{-1}\left[\frac{n_1}{n_2}cossi{n}^{-1}\right(\frac{n_2}{n_1}\left)\right]$