Question

A rectangular glass slab $A B C D$, 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 $A B$ 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 $C D$ 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 $C D$, means rays after refraction from $A B$ get, total internally reflected at $A D$

$\frac{n_{1}}{n_{2}}=\frac{\sin a_{\max }}{\sin r_{1}} $
$\Rightarrow a_{\max }=\sin ^{-1}\left[\frac{n_{1}}{n_{2}} \sin r_{1}\right] \quad$...(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) $
$\Rightarrow r_{1}=90-\sin ^{-1}\left(\frac{n_{2}}{n_{1}}\right) \ldots( ii )$
Hence from equations (i) and (ii)
$a_{\max }=\sin ^{-1}\left[\frac{n_{1}}{n_{2}} \sin \left\{90-\sin ^{-1} \frac{n_{2}}{n_{1}}\right\}\right]$
$=\sin ^{-1}\left[\frac{n_{1}}{n_{2}} \cos \left(\sin ^{-1} \frac{n_{2}}{n_{1}}\right)\right]$