question_answer

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

Correct Answer: A

Solution :

[a] Ray comes out from CD, means rays after refraction form 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 \[{{r}_{1}}+{{r}_{2}}=90{}^\circ \Rightarrow {{r}_{1}}=90-{{r}_{2}}=90-c\] \[{{r}_{1}}=90-{{\sin }^{-1}}\left( \frac{1}{2{{\mu }_{2}}} \right)\Rightarrow {{r}_{1}}=90-{{\sin }^{-1}}\left( \frac{{{n}_{2}}}{{{n}_{1}}} \right)\]                                                    ...(ii) Hence form equations (i) and (ii) \[{{a}_{\max }}={{\sin }^{-1}}\left[ \frac{{{n}_{1}}}{{{n}_{2}}}\sin \left\{ 90-{{\sin }^{-1}}\frac{{{n}_{2}}}{{{n}_{1}}} \right\} \right]\] \[={{\sin }^{-1}}\left[ \frac{{{n}_{1}}}{{{n}_{2}}}\cos \left( {{\sin }^{-1}}\frac{{{n}_{2}}}{{{n}_{1}}} \right) \right]\]