A) \[{{\sin }^{-1}}(0.8)\]
B) \[60{}^\circ \]
C) \[45{}^\circ \]
D) \[30{}^\circ \]
Correct Answer: C
Solution :
For minimum deviation through a prism, the refractive index of material of prism is given by \[\mu =\frac{\sin \left( \frac{A+{{\delta }_{m}}}{2} \right)}{\sin \left( \frac{A}{2} \right)}\] Given, \[A={{60}^{o}},\mu =\sqrt{2}\] \[\therefore \] \[\sqrt{2}=\frac{\sin \left( \frac{A+{{\delta }_{m}}}{2} \right)}{\sin \left( \frac{{{60}^{o}}}{2} \right)}\] or \[\sin \left( \frac{A+{{\delta }_{m}}}{2} \right)=\sqrt{2}\sin {{30}^{o}}\] or \[\sin \left( \frac{A+{{\delta }_{m}}}{2} \right)=\sqrt{2}\times \frac{1}{2}=\frac{1}{\sqrt{2}}\] or \[\sin \left( \frac{A+{{\delta }_{m}}}{2} \right)=\sin {{45}^{o}}\] or \[\frac{A+{{\delta }_{m}}}{2}={{45}^{o}}\] but we know angle of incidence \[i=\frac{A+{{\delta }_{m}}}{2}={{45}^{o}}\]
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