Question

For a prism, angle of prism is ${60}^{\circ }$ and its refractive index is $1.5$. Find angle of incidence corresponding to the angle of minimum deviation.

Answer 1

Given,

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

Refractive index, $\mu =1.5$

We know,

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

$1.5=\frac{\sin \left(\frac{60+D}{2}\right)}{\sin 30}$

$\Rightarrow 1.5\times \frac{1}{2}=\sin \frac{(60+D)}{2}$

$\Rightarrow {\sin }^{-1}\left(0.75\right)=\frac{60+D}{2}$

$\Rightarrow 2\times 48.6=60+D$

$\Rightarrow D={37.2}^o$

$i=\frac{D+A}{2}$

$\Rightarrow =\frac{37.2+60}{2}$

$i={48.6}^0$