A) \[60{}^\circ \]
B) \[45{}^\circ \]
C) \[30{}^\circ \]
D) None of these
Correct Answer: A
Solution :
Given \[AQ=AR\]and \[\angle A=60{}^\circ \] |
\[\therefore \,\,\angle AQR=\angle ARQ=60{}^\circ \] |
\[\therefore \,\,\,{{r}_{1}}={{r}_{2}}=30{}^\circ \] |
Applying Snell's law on face AB. |
1. \[\sin \,{{i}_{1}}=\mu \,\sin \,{{r}_{1}}\] |
\[\Rightarrow \,\,\sin {{i}_{1}}=\sqrt[{}]{3}\sin 30{}^\circ =\sqrt{3}\times \frac{1}{2}=\frac{\sqrt{3}}{2}\] |
\[\therefore \,\,{{i}_{1}}=60{}^\circ \] |
Similarly, \[{{i}_{2}}=60{}^\circ \] |
In a prism, deviation \[\delta ={{i}_{1}}+{{i}_{2}}-A=60{}^\circ +60{}^\circ -60{}^\circ =60{}^\circ \] |
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