A) \[\sqrt{2}\]
B) \[\sqrt{3}\]
C) \[\sqrt{\frac{3}{2}}\]
D) \[\frac{3}{2}\]
Correct Answer: A
Solution :
For minimum angle of deviation for a prism \[A=2r\], \[\therefore \] \[A={{60}^{o}}\] Now \[\mu =\frac{\sin \frac{60+30}{2}}{\sin \frac{60}{2}}=\frac{\sin {{45}^{o}}}{\sin {{30}^{o}}}=\frac{1}{\sqrt{2}}\times \frac{2}{1}=\sqrt{2}\]
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