Total Internal Reflection in Coated Equilateral Prism

To find the required refractive index $n_2$ of the transparent coating for total internal reflection (TIR) to occur at the bottom face of the prism, we analyze the path of the light ray as shown in the diagram.

1. Analysis at the First Surface

From the diagram, the light ray is incident normally on the first surface of the equilateral prism.

• Angle of incidence at the first face: $i = 0^\circ$
• Prism angle for an equilateral prism: $A = 60^\circ$
• Refractive index of the prism: $n_1 = 1.6 = \frac{8}{5}$

Applying Snell's Law at the first surface: $\sin i = n_1 \sin r_1$ $\sin(0^\circ) = 1.6 \sin r_1 \implies r_1 = 0^\circ$

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2. Angle of Incidence at the Painted Surface

The relationship between the angle of the prism $A$ and the internal angles $r_1$ and $r_2$ is: $r_1 + r_2 = A$

Substituting $r_1 = 0^\circ$ and $A = 60^\circ$: $0^\circ + r_2 = 60^\circ \implies r_2 = 60^\circ$

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3. Condition for Total Internal Reflection (TIR)

For total internal reflection to occur at the interface between the prism (refractive index $n_1$) and the painted coating (refractive index $n_2$), the angle of incidence $r_2$ must be greater than or equal to the critical angle $\theta_c$:

$r_2 \ge \theta_c$

Taking the sine on both sides: $\sin r_2 \ge \sin \theta_c$

The critical angle $\theta_c$ between two media is given by $\sin \theta_c = \frac{n_2}{n_1}$. Thus: $\sin(60^\circ) \ge \frac{n_2}{n_1}$

Substitute $n_1 = 1.6$ and $\sin(60^\circ) = \frac{\sqrt{3}}{2}$: $\frac{\sqrt{3}}{2} \ge \frac{n_2}{1.6}$

$n_2 \le 1.6 \times \frac{\sqrt{3}}{2}$

$n_2 \le \frac{8}{5} \times \frac{\sqrt{3}}{2} = \frac{4\sqrt{3}}{5}$

Thus, the limiting value of $n_2$ required for total internal reflection to occur is: $n_2 = \frac{4\sqrt{3}}{5}$

Correct Option: D