A) \[{{25}^{o}}\]
B) \[{{60}^{o}}\]
C) \[{{45}^{o}}\]
D) \[{{30}^{o}}\]
Correct Answer: B
Solution :
Refractive index of the material\[\mu =\sqrt{3}\] The relation for refractive index is \[\mu =\frac{\sin i}{\sin r}\] \[\sqrt{3}=\frac{\sin i}{\sin r}\] \[\sqrt{3}=\frac{\sin \frac{A+{{\delta }_{m}}}{2}}{\sin \frac{A}{2}}=\frac{\sin \frac{A+{{\delta }_{m}}}{2}}{\sin \frac{{{60}^{o}}}{2}}\] \[{{v}_{3}}=\frac{\sin \frac{A+{{\delta }_{m}}}{2}}{\frac{1}{2}}\] or \[\sin \frac{A+{{\delta }_{m}}}{2}=\frac{\sqrt{3}}{2}\] \[=\sin {{60}^{o}}\] (where \[A\] is the angle of equilateral prism) Hence, \[\frac{A+{{\delta }_{m}}}{2}={{60}^{o}}\] Thus, \[A+{{\delta }_{m}}=120\] \[{{\delta }_{m}}={{120}^{o}}-{{60}^{o}}\] \[{{\delta }_{m}}={{60}^{o}}\]
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