question_answer

For the angle of minimum deviation of a prism to be equal to its refracting angle, the prism must be made of a material whose refractive index                                                                               

A)  lies between \[\sqrt{2}\] and 1

B)  lies between 2 and \[\sqrt{2}\]

C)  is less than 1

D)  is greater than 2

Correct Answer: B

Solution :

\[\mu =\frac{\sin \left( \frac{A+\delta m}{2} \right)}{\sin A/2}\]                \[=\frac{\sin \left( \frac{A+A}{2} \right)}{\sin A/2}\] In given situation value of A varies from 0 to \[90{}^\circ \]. So, \[{{\mu }_{\min }}=2\cos \frac{{{90}^{o}}}{2}=\sqrt{2}\]and\[{{\mu }_{\max }}=2\cos {{0}^{o}}=2\]