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 $\mu ,$ the material will be :

• A. 1.531
• B. 1.414
• C. 1.593
• D. 1.825

Answer 1

Answer: (B)

The refractive index [a] of prism is given by
$\mu =\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \frac{A}{2}}$
where $A$ is angle of prism and ${\delta }_m$ is angle of minimum deviation. Given, $A={60}^o,{\delta }_m={30}^o$
$\therefore \mu =\frac{\sin \left(\frac{{60}^o+{30}^o}{2}\right)}{\sin \frac{{60}^o}{2}}=\frac{\sin {45}^o}{\sin {30}^o}$
$=\frac{0.707}{0.5}=1.414$
Note: For a specific prism, there is one and only one angle of minimum deviation.