Question

A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. By rotating the prism, the angle of minimum deviation is measured to be ${40}^0$. What is the refractive index of the material of the prism? If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light. The refracting angle of the prism is ${60}^0$.

Answer 1

Given,

$A={60}^0$

${\delta }_m={40}^0$

So from formula,

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

$\mu =\frac{sin\left({50}^0\right)}{sin{30}^0}=2sin{50}^0$

$\mu =1.532$

if it is dipped in water,

So, ${\mu }_1=\frac{1.532}{1.33}=1.1519$

So $1.1519=\frac{sin\left(\frac{60+{\delta }_m}{2}\right)}{sin30}$

$\Rightarrow \left(\frac{1.1512}{2}\right)=sin\left(\frac{60+{\delta }_m}{2}\right)$

$\left(35.1678\right)=\frac{60+{\delta }_m}{2}$

${\delta }_m=10.3356^0$