question_answer

A rectangular glass slab ABCD, of refractive index n1, is immersed in water of refractive index n2 (n1>n2).  A ray of light in incident at the surface AB of the slab as shown. The maximum value of the angle of incidence amax, such that the ray comes out only from the other surface CD is given by [IIT-JEE (Screening) 2000]

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)            \[{{\sin }^{-1}}\left( \frac{{{n}_{2}}}{{{n}_{1}}} \right)\]

Correct Answer: A

Solution :

                   Ray comes out from CD, means rays after refraction from AB get, total internally reflected at AD                    \[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{\sin {{\alpha }_{\max }}}{\sin {{r}_{1}}}\Rightarrow {{\alpha }_{\max }}={{\sin }^{-1}}\left[ \frac{{{n}_{1}}}{{{n}_{2}}}\sin {{r}_{1}} \right]\]             ?(i)                    Also \[{{r}_{1}}+{{r}_{2}}={{90}^{o}}\Rightarrow {{r}_{1}}=90-{{r}_{2}}=90-C\]                    \[\Rightarrow \]\[{{r}_{1}}=90-{{\sin }^{-1}}\left( \frac{1}{_{2}{{\mu }_{1}}} \right)\Rightarrow {{r}_{1}}=90-{{\sin }^{-1}}\left( \frac{{{n}_{2}}}{{{n}_{1}}} \right)\]  ...(ii)                    Hence from equation (i) and (ii)                    \[{{\alpha }_{\max }}={{\sin }^{-1}}\left[ \frac{{{n}_{1}}}{{{n}_{2}}}\sin \left\{ 90-{{\sin }^{-1}}\frac{{{n}_{2}}}{{{n}_{1}}} \right\} \right]\]                     = \[{{\sin }^{-1}}\left[ \frac{{{n}_{1}}}{{{n}_{2}}}\cos \left( {{\sin }^{-1}}\frac{{{n}_{2}}}{{{n}_{1}}} \right) \right]\]