A) \[30{}^\circ \]
B) \[45{}^\circ \]
C) \[60{}^\circ \]
D) \[90{}^\circ \]
Correct Answer: D
Solution :
Angle of incidence at B is same as angle at A because angle opposite to equal side in triangle will be equal. Hence angle at B becomes critical and hence angle of incidence at A is 90 degree. METHOD DETAIL TIR will take place if \[{{r}_{2}}>{{\theta }_{C}}\] Snell's law at B, \[\mu \,{{\operatorname{sinr}}_{2}}=1\times \sin \,{{i}_{2}}\] ?(1) \[\mu ={{\sin }^{-1}}\left[ \left( \frac{1}{\mu } \right)\sin \,{{i}_{2}} \right]\] ?(2) \[{{r}_{1}}={{r}_{2}}\](since OA = OB) Snell?s law at A: \[1\times \sin {{i}_{1}}=\mu \,\sin \,{{r}_{1}}=\mu \,\sin \,{{r}_{2}}=\sin \,{{i}_{2}}\][from eq. (1) So \[{{i}_{1}}={{i}_{2}}\] From eq. (2) \[{{r}_{2}}={{\sin }^{-1}}\left[ \left( \frac{1}{\mu } \right)\sin \,{{i}_{1}} \right]\] \[={{q}_{C}}\,if\,\sin \,i=1\,and\,\sin \,{{q}_{C}}=\frac{1}{\mu }\] or \[i=90{}^\circ \]
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