Question

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

Answer 1

The angle of the prism, ${60}^o$.

Minimum deviation $\delta ={40}^o$

${\mu }^{{}^{\prime }}$ is the refractive index of the material of the prism.

${\mu }^{{}^{\prime }}=\frac{sin\left(\frac{A+\delta m}{2}\right)}{sin\left(\frac{A}{2}\right)}=1.5321$

Let ${\delta }_1$ be new minimum deviation, now it is put under water $\mu =1.33$

$\Rightarrow \frac{{\mu }^{{}^{\prime }}}{\mu }=\frac{sin\left(\frac{A+\delta m}{2}\right)}{sin\left(\frac{A}{2}\right)}$

$\Rightarrow {\delta }_1={10.32}^o$