Question

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 to 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 another surface $CD$ is given by

• A. ${\sin }^{-1}\left[\frac{n_1}{n_2}\cos \left({\sin }^{-1}\left(\frac{n_2}{n_1}\right)\right)\right]$
• B. ${\sin }^{-1}\left[n_1\cos \left({\sin }^{-1}\left(\frac{1}{2}\right)\right)\right]$
• C. ${\sin }^{-1}\left(\frac{n_1}{n_2}\right)$
• D. ${\sin }^{-1}\left(\frac{n_2}{n_1}\right)$

Answer 1

The correct option is A ${\sin }^{-1}\left[\frac{n_1}{n_2}\cos \left({\sin }^{-1}\left(\frac{n_2}{n_1}\right)\right)\right]$

Rays comes out only from $CD$, means rays after refraction from $AB$ get totally internally reflected at $AD$.
From the figure,
$r_1+r_2=90$
$r_1=90-r_2$
${\left(r_1\right)}_{max}={90}^o-{\left(r_2\right)}_{max}$ and ${\left(r_2\right)}_{min}={\theta }_C$ (fro total internal reflection at $AD$)
where, $\sin {\theta }_C=\frac{n_2}{n_1}$ or ${\theta }_C={\sin }^{-1}\left(\frac{n_2}{n_1}\right)$
$\therefore {\left(r_1\right)}_{max}={90}^o-{\theta }_C$
Now, applying Snell's law at face $AB$, we get
$\frac{n_1}{n_2}=\frac{\sin {\alpha }_{max}}{\sin {\left(r_1\right)}_{max}}=\frac{\sin {\alpha }_{max}}{\sin (90-{\theta }_C)}=\frac{\sin {\alpha }_{max}}{\cos {\theta }_C}$
$\sin {\alpha }_{max}=\frac{n_1}{n_2}\cos {\theta }_C$
$\therefore {\alpha }_{max}={\sin }^{-1}\left[\frac{n_1}{n_2}\cos {\theta }_C\right]$
$={\sin }^{-1}\left[\frac{n_1}{n_2}\cos \left({\sin }^{-1}\left(\frac{n_2}{n_1}\right)\right)\right]$