question_answer

A prism of refractive index \[\sqrt{2}\] has refracting angle of \[{{60}^{\circ }}\]. At what angle a ray must be incident on it that it suffers a minimum deviation?                                                                                                                             [BHU M-2003]

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

B)  \[{{60}^{\circ }}\]

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

D)  \[{{180}^{\circ }}\]

Correct Answer: A

Solution :

                 Key Idea: 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                                 \[i'=i,\,\,r'=r,\,\delta ={{\delta }_{m}}\]                 \[\therefore \]                  \[2r=A\,\,or\,\,r=\frac{A}{2}\] Given, \[A={{60}^{\circ }}\], r\[r=\frac{60}{2}={{30}^{\circ }}\] Also from Snell?s law                                                 \[\mu =\frac{\sin \,i}{\sin \,r}=\frac{\sin \,i}{\sin \,{{30}^{\circ }}}\]                                                 \[\sqrt{2}=\frac{\sin \,i}{\sin \,{{30}^{\circ }}}\]                 \[\Rightarrow \]                               \[\sin \ i=\sqrt{2}\times \frac{1}{2}=\frac{1}{\sqrt{2}}\]                 \[\Rightarrow \]                                               \[i={{45}^{\circ }}\]