A light ray falls on a square glass slab as shown in the diagram. The index of refraction of the glass, if total internal reflection is to occur at the vertical face, is equal to :

$$1 \cdot \sin(45^\circ) = n \cdot \sin(r)$$

$$\frac{1}{\sqrt{2}} = n \sin(r) \implies \sin(r) = \frac{1}{n\sqrt{2}}$$

For total internal reflection to occur, the angle of incidence must be greater than or equal to the critical angle ($$\theta_c$$), where $$\sin(\theta_c) = \frac{1}{n}$$

$$\sin(i') \geq \sin(\theta_c)$$

$$\sin(90^\circ - r) \geq \frac{1}{n} \implies \cos(r) \geq \frac{1}{n}$$

$$\cos(r) = \sqrt{1 - \sin^2(r)}$$,

$$\sqrt{1 - \left(\frac{1}{n\sqrt{2}}\right)^2} \geq \frac{1}{n}$$

$$\sqrt{1 - \frac{1}{2n^2}} \geq \frac{1}{n}$$

$$1 - \frac{1}{2n^2} \geq \frac{1}{n^2}$$

$$1 \geq \frac{3}{2n^2} \implies 2n^2 \geq 3$$

$$n^2 \geq \frac{3}{2} \implies n \geq \sqrt{\frac{3}{2}}$$