Question

The angle of incidence for a ray of light at a refracting surface of a prism is ${45}^o$. The angle of prism is ${60}^o$. If the ray suffers minimum deviation through the prism, the angle of minimum deviation and refractive index of the material of the prism respectively, are

• A. ${45}^o;\frac{1}{\sqrt{2}}$
• B. ${30}^o;\sqrt{2}$
• C. ${30}^o;\frac{1}{\sqrt{2}}$
• D. ${45}^o;\sqrt{2}$

Answer 1

Answer: (B)

Angle of incidence, $i={45}^0$

Angle of prism, $A={60}^0$

Angle of minimum deviation, ${\delta }_m=2i-A={30}^0$

$\Rightarrow \mu =\frac{sin\left(\frac{A+{\delta }_m}{2}\right)}{sin\frac{A}{2}}=\frac{sin{45}^0}{sin{30}^0}=\frac{2}{\sqrt{2}}=\sqrt{2}$