Question

The refractive index of the material of a prism is $1.5$. When the prism is placed in minimum deviation position, the angle of incidence is $51$. Calculate the angle of prism and the angle of minimum deviation.

• A. ${28}^0$
• B. ${17}^0$
• C. ${22}^0$
• D. ${39.6}^0$

Answer 1

The correct option is D ${39.6}^0$

Given:- $i={51}^o$ and $\mu =1.5$

For minimum deviation, $r_1=r_2=r\left(say\right)$ and $r_1+r_2=2r=A$

Now, at surface AB:-

$\frac{\sin r}{\sin i}=\frac{1}{\mu }$

$\sin r=\frac{\sin {51}^o}{1.5}=0.518$

$⟹r={\sin }^{-1}0.518={31.2}^o$

Hence, angle of prism=$A=2r={62.4}^o$

Angle of deviation=$\delta =2i-A={102}^o-{62.4}^o={39.6}^o$