Question

A ray of light passing through an equilateral glass prism from air undergoes minimum duration when angle of incidence is ${\frac{3}{4}}^{th}$ the angle of prism, calculate the speed of light in prism.

Answer 1

$\begin{array}{c}\Rightarrow GivenA={60}^{\circ }\\ i=\frac{3}{4}A\\ =\frac{3}{4}\times 60\\ ={45}^{\circ }\end{array}$

For minimum Deviation $i_1=i_2=iand{r}_1=r_2=\frac{A}{2}=\frac{60}{2}={30}^{\circ }$d

Reflective of prism n $=\frac{\sin i}{\sin v}$

$\begin{array}{c}=\frac{\sin 45}{\sin 30}\\ =\frac{1}{\frac{\sqrt{2}}{\frac{1}{2}}}\\ =\frac{2}{\sqrt{2}}=\sqrt{2}=1.414\end{array}$

Speed of the light in prism $=v=\frac{c}{n}$

$\begin{array}{c}=\frac{3\times {10}^8}{1.414}\\ =2.13\times {10}^{8}m/s\end{array}$