question_answer

A prism of refractive index\[\sqrt{2}\]has refracting angle of\[{{60}^{o}}\]. At what angle a ray must be incident one it so that it suffers a minimum deviation?

A) \[45{}^\circ \]                                   

B) \[60{}^\circ \]

C) \[90{}^\circ \]

D)        \[180{}^\circ \]

Correct Answer: A

Solution :

In the position of minimum deviation angle of incidence is equal to angle of emergence. Let a ray of monochromatic light \[PQ\] be incident on face\[AB\]. \[PQRS\] is path of light ray, where \[i\] is angle of incidence \[r\] angle of refraction, \[r\]angle of incidence and\[i\]angle of emergence. In position of minimum deviation                   \[i=i,\,\,r=\delta =\delta m\]                 \[2r=A\]or\[r=\frac{A}{2}\] Given    \[A={{60}^{o}}\],                  \[r=\frac{60}{2}={{30}^{o}}\] Also from Snells law                 \[\mu =\frac{\sin i}{\sin r}=\frac{\sin i}{\sin {{30}^{o}}}\]             \[\sqrt{2}=\frac{\sin i}{\sin {{30}^{o}}}\] \[\Rightarrow \]     \[\sin i=\sqrt{2}\times \frac{1}{2}=\frac{1}{\sqrt{2}}\] \[\Rightarrow \]          \[i={{45}^{o}}\]