Question

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

Answer 1

The correct option is A ${\sin }^{-1}\frac{\sqrt{7}}{3}$


At maximum angle $\theta$ ray at point B, goes in gazing emergence,

Applying Snell's law at point B,

$\frac{4}{3}\times \sin \theta "=1\times \sin {90}^{\circ }$

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

From the figure,

${\theta }^{\prime }=(\frac{\pi }{2}-\theta ")$

Now applying snell's law at point A,

$1\times \sin \theta =\frac{4}{3}\times \sin {\theta }^{\prime }$

$\sin \theta =\frac{4}{3}\times \sin {\theta }^{\prime }$

$\sin \theta =\frac{4}{3}\times \sin (\frac{\pi }{2}-\theta ")$

$\sin \theta =\frac{4}{3}\times \cos {\theta }^{\prime \prime }$

From the above triangle,

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

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

$\Rightarrow \sin \theta =\frac{4}{3}\times \frac{\sqrt{7}}{4}$

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

Hence, option $\left(A\right)$ is correct.