Question

A rectangular glass slab $ABCD$, of refractive index $n_1$, is immersed in water of refractive index $n_2\left(n_1>n_2\right).$ A ray of light in 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. ${\sin }^{-1}\left[\frac{n_1}{n_2}\cos \left({\sin }^{-1}\frac{n_2}{n_1}\right)\right]$
• B. ${\sin }^{-1}\left[n_1\cos \left({\sin }^{-1}\frac{1}{n_2}\right)\right]$
• C. ${\sin }^{-1}\left(\frac{n_1}{n_2}\right)$
• D. ${\sin }^{-1}\left(\frac{n_2}{n_1}\right)$

Answer 1

Answer: (A)

Ray comes out from $CD$, means rays after refraction from $AB$ get, total internally reflected at $AD$

$\frac{n_1}{n_2}=\frac{\sin a_{\max }}{\sin r_1}$
$\Rightarrow a_{\max }={\sin }^{-1}\left[\frac{n_1}{n_2}\sin r_1\right]$...(i)
Also $r_1+r_2=90^{\circ }$
$\Rightarrow r_1=90-r_2=90-C$
$r_1=90-{\sin }^{-1}\left(\frac{1}{{}_2{\mu }_2}\right)$
$\Rightarrow r_1=90-{\sin }^{-1}\left(\frac{n_2}{n_1}\right)\dots (ii)$
Hence from equations (i) and (ii)
$a_{\max }={\sin }^{-1}\left[\frac{n_1}{n_2}\sin \left\{90-{\sin }^{-1}\frac{n_2}{n_1}\right\}\right]$
$={\sin }^{-1}\left[\frac{n_1}{n_2}\cos \left({\sin }^{-1}\frac{n_2}{n_1}\right)\right]$