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 $B$
$\frac{4}{3}\times \sin {\theta }^{\prime \prime }=1\times \sin 90^{\circ }$
${\theta }^{\prime \prime }={\sin }^{-1}\left(\frac{3}{4}\right)$
${\theta }^{\prime }=\left(\frac{\pi }{2}-{\theta }^{\prime \prime }\right)$
At point $A$
$1\times \sin \theta =\frac{4}{3}\times \sin {\theta }^{\prime }$
$\sin \theta =\frac{4}{3}\times \sin \left(\frac{\pi }{2}-{\theta }^{\prime \prime }\right)$
$\sin \theta =\frac{4}{3}\cos \left[{\cos }^{-1}\frac{\sqrt{7}}{4}\right]$

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