A transparent cube of side d, made of a material of refractive index $$\mu_2$$, is immersed in a liquid of refractive index $$\mu_1$$ ($$\mu_1 < \mu_2$$). A ray is incident on the face AB at an angle $$\theta$$ (shown in the figure). Total internal reflection takes place at point E on the face BC.

Then $$\theta$$ must satisfy

Snell's Law at face AB: $$\mu_1 \sin\theta = \mu_2 \sin r \implies \sin r = \frac{\mu_1}{\mu_2}\sin\theta$$

Angle of incidence at face BC: $$\phi = 90^\circ - r$$

Condition for TIR at point E on face BC:

$$\sin\phi > \frac{\mu_1}{\mu_2}$$

$$\sin(90^\circ - r) > \frac{\mu_1}{\mu_2} \implies \cos r > \frac{\mu_1}{\mu_2}$$

$$1 - \sin^2 r > \frac{\mu_1^2}{\mu_2^2}$$

$$1 - \left(\frac{\mu_1}{\mu_2}\sin\theta\right)^2 > \frac{\mu_1^2}{\mu_2^2}$$

$$\left(\frac{\mu_1}{\mu_2}\sin\theta\right)^2 < 1 - \frac{\mu_1^2}{\mu_2^2}$$

$$\sin^2\theta < \frac{\mu_2^2}{\mu_1^2} - 1$$

$$\theta < \sin^{-1} \sqrt{\frac{\mu_2^2}{\mu_1^2} - 1}$$