Question

A prism of refracting angle $60{}^{\circ }$ is made with a material of refractive index $\mu$ .For a certain wavelength of light, the angle of minimum deviation is $30{}^{\circ }$ . For this wavelength, the value of u. of material is:

• A. 1.820
• B. 1.503
• C. 1.414
• D. 1.731

Answer 1

Answer: (C)

The value of refractive index of material of prism is $\mu =\frac{\sin \frac{A+{\delta }_m}{2}}{\sin \frac{A}{2}}$ $\left(\begin{array}{rl}& \text{Given:}{\delta }_m={30}^o\\ & A={60}^o\end{array}\right)$ $\mu =\frac{\sin \left(\frac{{60}^o+{30}^o}{2}\right)}{\sin \frac{{60}^o}{2}}$ $=\frac{\sin {45}^o}{\sin {30}^o}$ $=\frac{1}{\sqrt{2}}\times \frac{2}{1}$ $=\sqrt{2}=1.414$