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.