Question

The angle of a prism is ${60}^o$ and its refractive index is $\sqrt{2}$. The angle of minimum deviation suffered by a ray of light in passing through it is:

• A. About ${20}^o$
• B. ${30}^o$
• C. ${60}^o$
• D. ${45}^o$

Answer 1

The correct option is B ${30}^o$

Angle of minimum deviation is

$\mu =\frac{\sin \left(\frac{A+{\delta }_{min}}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

$\sqrt{2}=\frac{\sin \left(\frac{60+{\delta }_{min}}{2}\right)}{\sin \left(\frac{60}{2}\right)}$

$\frac{\sqrt{2}}{2}=\sin \left(\frac{60+{\delta }_{min}}{2}\right)$

${\sin }^{-}\left(\frac{1}{\sqrt{2}}\right)=\frac{60+{\delta }_{min}}{2}$

$90=60+{\delta }_{min}$

${\delta }_{min}={30}^0$