# Cross-section view of a prism is the equilateral triangle

Question:

Cross-section view of a prism is the equilateral triangle $\mathrm{ABC}$ in the figure. The minimum deviation is observed using this prism when the angle of incidence is equal to the prism angle. The time taken by light to travel from $\mathrm{P}$ (midpoint of $\mathrm{BC}$ ) to $\mathrm{A}$ is $\times 10^{-10} \mathrm{~s}$. (Given, speed of light in vacuum $=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ and

$\left.\cos 30^{\circ}=\frac{\sqrt{3}}{2}\right)$

Solution:

$\mathrm{i}=\mathrm{A}=60^{\circ}$

$\underline{\underline{\delta}}_{\min }=2 \mathrm{i}-\mathrm{A}$

$=2 \times 60^{\circ}-60^{\circ}=60^{\circ}$

$\mu=\frac{\sin ^{-1}\left(\frac{\delta_{\min }+\mathrm{A}}{2}\right)}{\sin ^{-1}\left(\frac{\mathrm{A}}{2}\right)}$

$=\sqrt{3}$

$\mathrm{V}_{\text {prism }}=\frac{3 \times 10^{8}}{\sqrt{3}}$

$\mathrm{AP}=10 \times 10^{-2} \times \frac{\sqrt{3}}{2}$

time $=\frac{5 \times 10^{-2}}{3 \times 10^{8}} \times \sqrt{3} \times \sqrt{3}$

$=5 \times 10^{-10} \mathrm{sec}$

$\mathrm{Ans}=5$