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

$$\sin \theta_C = \frac{\mu_{\text{rarer}}}{\mu_{\text{denser}}} = \frac{\mu_1}{\mu_2}$$

$$\sin \theta_C = \frac{1.0}{1.25} = \frac{4}{5}$$

$$\cos \theta_C = \sqrt{1 - \sin^2 \theta_C} = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \frac{3}{5}$$

Using the geometric relation $$r = 90^\circ - \theta_C$$: $$\sin r = \sin(90^\circ - \theta_C) = \cos \theta_C = \frac{3}{5}$$

$$\mu_1 \sin \theta = \mu_2 \sin r$$

$$1.0 \times \sin \theta = 1.25 \times \frac{3}{5}$$

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