Question 59 of 295

A ray of light entering from air into a denser medium of refractive index $\dfrac{4}{3}$, 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${\sin }^{-1}\dfrac{\sqrt{7}}{3}$B${\sin }^{-1}\dfrac{\sqrt{5}}{4}$C${\sin }^{-1}\dfrac{\sqrt{7}}{4}$D${\sin }^{-1}\dfrac{\sqrt{5}}{3}$

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Answer: (A) ${\sin }^{-1}\dfrac{\sqrt{7}}{3}$

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

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