question_answer

A thin equilateral triangle has a refractive index of \[\sqrt{3}\]. Find the angle of minimum deviation :

A) \[30{}^\circ \]                                   

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

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

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

Correct Answer: B

Solution :

                For a prism of angle A and\[{{\delta }_{M}}\]the minimum angle of deviation, refractive index is given by \[\mu =\frac{\sin \left( \frac{A+{{\delta }_{m}}}{2} \right)}{\sin \frac{A}{2}}\] Given,                   \[\mu =\sqrt{3},\text{ }A={{60}^{o}}\] \[\therefore \]  \[\sqrt{3}=\sin \left( \frac{\frac{{{60}^{o}}+{{\delta }_{m}}}{2}}{\sin \frac{{{60}^{o}}}{2}} \right)\] \[\Rightarrow \]               \[\frac{{{60}^{o}}+{{\delta }_{m}}}{2}={{60}^{o}}\] \[\Rightarrow \]               \[{{\delta }_{m}}={{60}^{o}}\]