Cross-section view of a prism is the equilateral triangle $ABC$ shown 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 $P$ (midpoint of $BC$) to $A$ is ______ $\times10^{-10}$ s. (Given, speed of light in vacuum $= 3\times10^8\ \text{m s}^{-1}$ and $\cos30^\circ = \dfrac{\sqrt3}{2}$)
Numerical value type. Enter your answer.
Answer: 5
At minimum deviation with $i = A = 60^\circ$: $r = \dfrac A2 = 30^\circ$, so $\mu = \dfrac{\sin60^\circ}{\sin30^\circ} = \sqrt3$.
$PA = 10\cos30^\circ = 5\sqrt3$ cm; speed in the prism $= \dfrac{c}{\sqrt3}$.
$$t = \frac{5\sqrt3\times10^{-2}\times\sqrt3}{3\times10^8} = \frac{0.15}{3\times10^8} = 5\times10^{-10}\ \text{s}$$
Solution by Sreeraj P, M.Sc Physics