question_answer

The light ray is incidence at angle of 60° on a prism of angle 45°. When the light ray falls on the other surface at 90°, the refractive index of the material of prism  \[\mu \] and the angle of deviation \[\delta \] are given by       [DPMT 2001]

A)            \[\mu =\sqrt{2}\,,\,\delta ={{30}^{o}}\]                      

B)            \[\mu =1.5\,,\,\delta ={{15}^{o}}\]

C)            \[\mu =\frac{\sqrt{3}}{2},\delta ={{30}^{o}}\]          

D)            \[\mu =\sqrt{\frac{3}{2}},\delta ={{15}^{o}}\]

Correct Answer: D

Solution :

                   From figure it is clear that \[\angle \,e=\angle \,{{r}_{2}}=0\] From \[A={{r}_{1}}+{{r}_{2}}\] \[\Rightarrow {{r}_{1}}=A={{45}^{o}}\] \[\therefore \mu =\frac{\sin i}{\sin {{r}_{1}}}=\frac{\sin 60}{\sin 45}=\sqrt{\frac{3}{2}}\] Also from\[i+e=A+\delta \]\[\Rightarrow 60+0=45+\delta \]\[\Rightarrow \delta ={{15}^{o}}\]