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) \[{{\alpha }_{\max }}={{\sin }^{-1}}\left[ \frac{{{n}_{1}}}{{{n}_{2}}}\sin \left\{ 90-{{\sin }^{-1}}\frac{{{n}_{2}}}{{{n}_{1}}} \right\} \right]\] = \[{{\sin }^{-1}}\left[ \frac{{{n}_{1}}}{{{n}_{2}}}\cos \left( {{\sin }^{-1}}\frac{{{n}_{2}}}{{{n}_{1}}} \right) \right]\]You need to login to perform this action.
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