Question

A ray of light passing through an equilateral triangular glass prism from air undergoes minimum deviation when angle of incidence is 3/4 of the angle of prism. Calcute the speed of light in prism. $(2.1\times {10}^{3}m/s)$

Answer 1

Given that, on incidence :-

When $i=\frac{3}{4}A$, minimum deviation occurs

In minimum deviation ease, $i=e$

using.

$A+\delta =i+e$

$\Rightarrow A+\delta min=2i=2\left(\frac{3}{4}A\right)$

$\Rightarrow \delta min=\frac{A}{2}$. [angle of minimum deviation ]

using minimum deviation formula,

$\mu =\frac{\sin \left(\frac{A+\delta min}{2}\right)}{\sin \left(A/2\right)}=\frac{\sin \left(\frac{3A}{4}\right)}{\sin \left(A/2\right)}$

AS prism is equilateral, $A={60}^o$

Thus,

$\mu =\frac{c}{v}=\frac{\sin \left({45}^o\right)}{\sin \left({30}^o\right)}$ [$v=$ velocity of eight in prism ]

$\Rightarrow v=\frac{c\sin \left({30}^o\right)}{\sin \left({45}^o\right)}=\frac{3\times {10}^8\times 2}{\sqrt{2}}=2.1\times {10}^{8}m/s$.