A transparent block A having refractive index $\mu=1.25$ is surrounded by another medium of refractive index $\mu=1.0$ as shown in figure. A light ray is incident on the flat face of the block with incident angle $\theta$ as shown in figure. What is the maximum value of $\theta$ for which light suffers total internal reflection at the top surface of the block?

• A

  $\tan ^{-1}(4 / 3)$
• B

  $\tan ^{-1}(3 / 4)$
• Correct answer
  C

  $\sin ^{-1}(3 / 4)$
• D

  $\cos ^{-1}(3 / 4)$

Solution

$\begin{aligned} & \mathrm{r}+\theta_{\mathrm{C}}=90^{\circ} \\ & \mu_1 \sin \theta=\mu_2 \sin \mathrm{r} \\ & \sin \theta=\frac{\mu_2}{\mu_1} \sin \left(90-\theta_{\mathrm{C}}\right) \\ & \sin \theta=\frac{\mu_2}{\mu_1} \cos \theta_{\mathrm{C}} \\ & \sin \theta_{\mathrm{C}}=\frac{\mu_1}{\mu_2}\end{aligned}$
$\begin{aligned} & \sin \theta=\frac{\mu_2}{\mu_1} \sqrt{1-\frac{\mu_1^2}{\mu_2^2}} \\ & \sin \theta=\sqrt{\frac{\mu_2^2-\mu_1^2}{\mu_1^2}}=\sqrt{\frac{\frac{25}{16}-1}{1}} \\ & \sin \theta=\frac{3}{4} \\ & \theta=\sin ^{-1}\left(\frac{3}{4}\right)\end{aligned}$

Hence, the correct answer is option: 'C'.