Question

Cross-section view of a prism is the equilateral triangle $ABC$ 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 $P$ (midpoint of $BC$ ) to $A$ is ______$\times 10^{-10}s.$ (Given, speed of light in vacuum $=3\times 10^{8}m/s$ and $\cos 30^{\circ }=\frac{\sqrt{3}}{2})$

Answer 1

Answer: 5

$i=A=60^{\circ }$
${\underline{\underline{\delta }}}_{\min }=2i-A$
$=2\times 60^{\circ }-60^{\circ }=60^{\circ }$
$\mu =\frac{{\sin }^{-1}\left(\frac{{\delta }_{\min }+A}{2}\right)}{{\sin }^{-1}\left(\frac{A}{2}\right)}$
$=\sqrt{3}$
$V_{\text{prism }}=\frac{3\times 10^8}{\sqrt{3}}$
AP $=10\times 10^{-2}\times \frac{\sqrt{3}}{2}$
time$=\frac{5\times 10^{-2}}{3\times 10^8}\times \sqrt{3}\times \sqrt{3}$
$=5\times 10^{-10}sec$
$=5$