A monochromatic light is incident from air on a refracting surface of a prism of angle 75 degrees an

Q: Why do we assume the ray hits the second surface exactly at the critical angle?

The problem states TIR occurs for $\theta \le 60^\circ$. This means $60^\circ$ is the boundary condition. At this maximum incident angle, the ray is on the verge of escaping, which corresponds exactly to the critical angle.

Q: What happens if the incident angle is less than 60 degrees?

If $\theta$ decreases, the refracted angle $r_1$ also decreases according to Snell's Law. Since $r_1 + r_2 = A$, a smaller $r_1$ means a larger $r_2$. A larger $r_2$ exceeds the critical angle, ensuring TIR still occurs safely!

Q: Why is the prism angle formula A = r1 + r2 valid here?

This is a fundamental geometric property of any triangular prism. The angle between the two normals inside the prism is $180^\circ - A$. In the triangle formed by the normals and the ray path, the sum of angles gives $r_1 + r_2 = A$.

Q: Can we use the minimum deviation formula here?

No, the minimum deviation formula ($\delta_{min}$) assumes symmetric ray paths where $i = e$ and $r_1 = r_2$. Here, the ray is undergoing TIR at the second surface, so the path is highly asymmetric.

Q: Why is the refractive index of air taken as 1?

In optics, the refractive index of a perfect vacuum is exactly 1, and for air, it is approximately 1.0003. For all practical JEE calculations, we take it as exactly 1 to simplify the math unless specified otherwise.

Question

A monochromatic light is incident from air on a refracting surface of a prism of angle $75^\circ$ and refractive index $n_0 = \sqrt{3}$. The other refracting surface of a prism is coated by a thin film of material of refractive index $n$ as shown in figure. The light suffers total internal reflection at the coated prism surface for an incidence angle of $	heta \le 60^\circ$. The value of $n^2$ is_________.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are a photon of light embarking on a journey through a beautifully crafted optical system. We are presented with a prism possessing an apex angle of $A = 75^\circ$ and a refractive index of $n_0 = \sqrt{3}$. A monochromatic light ray enters the left face of this prism from the air.

The right face of the prism is coated with a thin film of an unknown refractive index $n$. Our mission is to uncover the value of $n^2$, given a very specific and fascinating constraint: the light suffers Total Internal Reflection (TIR) at the coated surface for any incidence angle $	heta \le 60^\circ$.

### The Master Equation: Snell's Law

To trace the path of our photon, we must first understand how it bends upon entering the prism. This is governed by Snell's Law, the master equation of refraction. We apply it at the first air-prism interface:

$$n_{	ext{air}} \sin 	heta = n_0 \sin r_1$$

The problem provides a boundary condition: the maximum angle of incidence is $	heta = 60^\circ$. Let's substitute the known values into our equation. The refractive index of air is $1$, and the prism's refractive index is $\sqrt{3}$:

$$1 \cdot \sin 60^\circ = \sqrt{3} \sin r_1$$

Knowing that $\sin 60^\circ = \frac{\sqrt{3}}{2}$, we can easily solve for the angle of refraction $r_1$:

$$\frac{\sqrt{3}}{2} = \sqrt{3} \sin r_1 \implies \sin r_1 = \frac{1}{2}$$

This beautifully simplifies to $r_1 = 30^\circ$. The ray bends sharply towards the normal as it enters the denser medium of the prism.

### The Geometry of the Prism

Now, the ray travels through the prism and strikes the second, coated surface. To find out what happens next, we need the angle of incidence at this second surface, which we will call $r_2$.

Here, we rely on a fundamental geometric property of all triangular prisms. The angle of the prism $A$ is always equal to the sum of the two internal angles made with the normals:

$$A = r_1 + r_2$$

We know the prism angle is $75^\circ$, and we just calculated $r_1$ to be $30^\circ$. Substituting these values gives us:

$$75^\circ = 30^\circ + r_2 \implies r_2 = 45^\circ$$

Our photon strikes the coated boundary at exactly $45^\circ$.

### The Critical Condition for TIR

Here is where the physics gets truly elegant. The problem states that TIR occurs for all angles $	heta \le 60^\circ$. This implies that $	heta = 60^\circ$ is the extreme boundary case. At this exact angle, the ray is on the absolute verge of escaping the prism.

In optical terms, the ray hits the second surface exactly at the **critical angle**. At the critical angle, the refracted ray grazes the boundary, meaning the angle of refraction is $90^\circ$. We apply Snell's Law once more at this second interface:

$$n_0 \sin r_2 = n \sin 90^\circ$$

### Final Calculation

Let's plug in the values we've gathered. The prism's refractive index is $\sqrt{3}$, the internal angle $r_2$ is $45^\circ$, and $\sin 90^\circ$ is simply $1$:

$$\sqrt{3} \sin 45^\circ = n \cdot 1$$

Since $\sin 45^\circ = \frac{1}{\sqrt{2}}$, we find the refractive index of the coating:

$$n = \sqrt{3} \cdot \frac{1}{\sqrt{2}} = \sqrt{\frac{3}{2}}$$

The question asks for the value of $n^2$. By squaring our result, we arrive at the final, elegant answer:

$$n^2 = \frac{3}{2} = 1.50$$

**Conceptual Check:** What if the incident angle $	heta$ was less than $60^\circ$? A smaller $	heta$ leads to a smaller $r_1$. Because $r_1 + r_2 = 75^\circ$, a smaller $r_1$ forces $r_2$ to be larger than $45^\circ$. Since $45^\circ$ is the critical angle, any angle larger than it guarantees Total Internal Reflection. The physics perfectly aligns with the mathematical constraints!

Enter Numerical Value:

Visualized Solution

$A na l y z in g t h e O pt i c a l S y s t e m$

$S n e l l^{'} s L a w a t F i r s t I n t er f a ce$

$S u b s t i t u t in g K n o w n V a l u es$

$C a l c u l a t in g r_{1}$

$G eo m e t r y o f t h e P r i s m$

$P r i s m A n g l e R e l a t i o n$

$C a l c u l a t in g r_{2}$

$T o t a l I n t er na l R e f l ec t i o n$

$C r i t i c a l A n g l e C o n d i t i o n$

$S u b s t i t u t in g a t S eco n d I n t er f a ce$

$S o l v in g f or n$

$F ina l C a l c u l a t i o n$

$C o n ce pt u a l R e f l ec t i o n$

The Sigma Insight: Refraction and Total Internal Reflection

Analyzing the Setup

The Master Equation

The Geometry of the Prism

The Critical Condition for TIR

Final Calculation

Similar Questions

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the Optical System

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Snell′s Law at First Interface

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Known Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating r1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Geometry of the Prism

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Prism Angle Relation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating r2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Internal Reflection

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Critical Angle Condition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting at Second Interface

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for n

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conceptual Reflection