Question

A ray of light entering from air into a denser medium of refractive index $\frac{4}{3}$, as shown in figure. The light ray suffers total internal reflection at the adjacent surface as shown. The maximum value of angle theta should be equal to :

• A. $\sin ^{-1} \frac{\sqrt{5}}{4}$
• B. $\sin ^{-1} \frac{\sqrt{7}}{3}$
• C. $\sin ^{-1} \frac{\sqrt{5}}{3}$
• D. $\sin ^{-1} \frac{\sqrt{7}}{4}$

Answer 1

Answer: (A)

At maximum angle $\theta$ ray at point $B$ goes in gazing emergence, at all less values of $\theta$, TIR occurs.
At point $B$
$\frac{4}{3} \times \sin \theta''=1 \times \sin 90^{\circ}$
$\theta''=\sin ^{-1}\left(\frac{3}{4}\right)$
$\theta'=\left(\frac{\pi}{2}-\theta''\right)$
At point $A$
$1 \times \sin \theta=\frac{4}{3} \times \sin \theta'$
$\sin \theta=\frac{4}{3} \times \sin \left(\frac{\pi}{2}-\theta''\right)$
$\sin \theta=\frac{4}{3} \cos \left[\cos ^{-1} \frac{\sqrt{7}}{4}\right]$

$\sin \theta=\frac{4}{3} \times \frac{\sqrt{7}}{4}$
$\theta=\sin ^{-1}\left(\frac{\sqrt{7}}{3}\right)$