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 $B C$ ) to $A$ is ______$\times 10^{-10} s .$ (Given, speed of light in vacuum $=3 \times 10^{8} m / s$ and $\left.\cos 30^{\circ}=\frac{\sqrt{3}}{2}\right)$

Answer 1

$ i = A =60^{\circ}$
$\underline{\underline{\delta}}_{\min }=2 i - 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$=\frac{5 \times 10^{-2}}{3 \times 10^{8}} \times \sqrt{3} \times \sqrt{3}$
$=5 \times 10^{-10} sec$
$= 5$