Step-by-Step Solution

1. Understand the Geometry and Physical Principles

• A monochromatic light ray enters a prism from air (refractive index $n_{air} = 1$).
• The prism has an angle $A = 75^\circ$ and refractive index $n_0 = \sqrt{3}$.
• The first surface has an angle of incidence $i_1 = \theta$ and an angle of refraction $r_1$.
• The ray travels through the prism and strikes the second surface at an angle of incidence $r_2$.
• The second surface is coated with a material of refractive index $n$.
• The light undergoes Total Internal Reflection (TIR) at this second surface.
• The condition for the problem is that TIR occurs for all angles of incidence $\theta \le 60^\circ$.

2. Apply Snell's Law and Prism Equations

• At the first refracting surface (Air to Prism): Applying Snell's Law: $n_{air} \sin(i_1) = n_0 \sin(r_1)$ $1 \cdot \sin(\theta) = \sqrt{3} \sin(r_1)$

• Inside the prism: The relationship between the angles in a prism is given by: $r_1 + r_2 = A$ $r_1 + r_2 = 75^\circ$ Therefore, $r_2 = 75^\circ - r_1$.

• At the second refracting surface (Prism to Film): For Total Internal Reflection (TIR) to occur, the angle of incidence $r_2$ must be greater than or equal to the critical angle $C$. $r_2 \ge C$ The critical angle $C$ is defined by: $\sin(C) = \frac{n}{n_0}$ So, the condition for TIR is $\sin(r_2) \ge \frac{n}{n_0}$.

3. Analyze the Critical Condition

• The problem states that TIR happens for all $\theta \le 60^\circ$.
• Let's analyze the relationship between $\theta$ and $r_2$. From Snell's law at the first surface, as $\theta$ increases, $\sin(\theta)$ increases, which means $r_1$ also increases. Since $r_2 = 75^\circ - r_1$, as $r_1$ increases, $r_2$ decreases.
• This means the angle of incidence on the second surface, $r_2$, is at its minimum value when the angle of incidence on the first surface, $\theta$, is at its maximum value.
• The given range is $\theta \le 60^\circ$, so the maximum value of $\theta$ is $60^\circ$.
• For TIR to occur for the entire range of $\theta$, the condition $r_2 \ge C$ must hold even for the minimum value of $r_2$. The minimum value of $r_2$ occurs at $\theta = 60^\circ$. Therefore, the limiting condition for TIR is when the angle of incidence on the second face is equal to the critical angle at $\theta = 60^\circ$. $r_{2, min} = C \quad \text{at} \quad \theta = 60^\circ$

4. Calculate the Angles at the Limiting Condition ($\ heta = 60^\circ$)

• Step 4.1: Find $r_1$ Using Snell's Law at the first surface with $\theta = 60^\circ$: $1 \cdot \sin(60^\circ) = \sqrt{3} \sin(r_1)$ $\frac{\sqrt{3}}{2} = \sqrt{3} \sin(r_1)$ $\sin(r_1) = \frac{1}{2}$ $r_1 = 30^\circ$

• Step 4.2: Find $r_2$ Using the prism angle equation: $r_2 = A - r_1 = 75^\circ - 30^\circ = 45^\circ$

• Step 4.3: Relate $r_2$ to the critical angle $C$ At this limiting condition, $r_2$ is equal to the critical angle $C$. $C = r_2 = 45^\circ$

5. Calculate the Refractive Index $n$ and $n^2$

• Using the formula for the critical angle: $\sin(C) = \frac{n}{n_0}$ $\sin(45^\circ) = \frac{n}{\sqrt{3}}$ $\frac{1}{\sqrt{2}} = \frac{n}{\sqrt{3}}$ $n = \frac{\sqrt{3}}{\sqrt{2}}$

• The question asks for the value of $n^2$. $n^2 = \left( \frac{\sqrt{3}}{\sqrt{2}} \right)^2 = \frac{3}{2}$ $n^2 = 1.5$

Thus, the value of $n^2$ is 1.5.