Question

The angle of minimum deviation produced by a ${60}^0$ prism is ${40}^0$. Calculate the refractive index of the material of the prism.

• A. 1.33
• B. 1.53
• C. 1.44
• D. 1.63

Answer 1

Answer: (B)

$A={60}^0,\mu ={40}^0,\delta =?$

The refractive index of the material of the prism,

$\mu =\frac{sin\left(\frac{A+\delta }{2}\right)}{sin\frac{A}{2}}=\frac{sin\left(\frac{{60}^0+{40}^0}{2}\right)}{sin\left(\frac{{60}^0}{2}\right)}$

$\frac{sin{50}^0}{sin{30}^0}=\frac{0.7660}{1}\times 2=1.532$