A ray of light is incident at an angle of 60° on one face of a prism which has an angle of 30°. The

Q: Why is the angle of deviation exactly $30^\circ$?

The problem states that the emergent ray makes an angle of $30^\circ$ with the incident ray. By definition, the angle between the incident ray's original path and the emergent ray is the angle of deviation, $\delta$.

Q: How do we know $r_2 = 0^\circ$ when $i_2 = 0^\circ$?

According to Snell's Law at the second interface, $\mu \sin r_2 = 1 \cdot \sin i_2$. If $i_2 = 0^\circ$, then $\sin i_2 = 0$, which means $\sin r_2$ must also be zero. Therefore, $r_2 = 0^\circ$.

Q: What does it mean physically when $i_2 = 0^\circ$?

It means the light ray is traveling exactly along the normal to the surface. It hits the surface head-on at a $90^\circ$ angle to the boundary, so it doesn't bend as it exits.

Q: Why did the question say 'material of the lens' instead of prism?

That is likely a slight typographical error in the original exam paper! The physical principles and the object described are clearly that of a prism.

Q: Can we use the minimum deviation formula here?

No! The minimum deviation formula $\mu = \frac{\sin((A+\delta_m)/2)}{\sin(A/2)}$ only applies when $i_1 = i_2$. Here, $i_1 = 60^\circ$ and $i_2 = 0^\circ$, so the prism is not in the position of minimum deviation.

Question

A ray of light is incident at an angle of 60° on one face of a prism which has an angle of 30°. The ray emerging out of the prism makes an angle of 30° with the incident ray. Show that the emergent ray is perpendicular to the face through which it emerges and calculate the refractive index of the material of the lens.

Sigma Percentile · Accepted Answer

### The Setup: Visualizing the Prism

Imagine a prism with an apex angle of $30^\circ$. A light ray strikes one of its faces at an angle of incidence of $60^\circ$. The question states that the angle between the emergent ray and the incident ray is $30^\circ$. By definition, the angle between the original path of the incident ray and the final path of the emergent ray is the **angle of deviation**, denoted by $\delta$. Therefore, we are given that the total deviation $\delta = 30^\circ$.

### The Master Equation: Angle of Deviation

To unlock this problem, we need to recall the master formula for the angle of deviation in a prism. The total deviation $\delta$ is equal to the sum of the angle of incidence ($i_1$) and the angle of emergence ($i_2$), minus the prism angle ($A$). Mathematically, this is written as:

$$\delta = (i_1 + i_2) - A$$

Let's substitute the values we know into this equation. We have $\delta = 30^\circ$, $i_1 = 60^\circ$, and $A = 30^\circ$. Plugging these in, we get:

$$30^\circ = (60^\circ + i_2) - 30^\circ$$

### The Revelation: Angle of Emergence

Look closely at the equation we just formed. If we simplify the right side, $60^\circ - 30^\circ$ leaves us with $30^\circ$. So, we have $30^\circ = 30^\circ + i_2$. Solving this simple relation gives us the value of $i_2$:

$$i_2 = 0^\circ$$

What does an angle of emergence of $0^\circ$ physically mean? It means the emergent ray travels exactly along the normal to the second face. In other words, the ray exits the prism perfectly **perpendicular** to the face through which it emerges! This beautifully proves the first part of our question.

### Inside the Prism: The Geometry of Refraction

Now, let's look inside the prism to understand the path of the refracted ray. The geometry of a prism dictates that the sum of the two internal angles of refraction, $r_1$ (at the first face) and $r_2$ (at the second face), must equal the prism angle $A$:

$$r_1 + r_2 = A$$

Because the ray exits along the normal ($i_2 = 0^\circ$), Snell's Law at the second interface tells us that the internal angle $r_2$ must also be zero. Substituting $r_2 = 0^\circ$ and $A = 30^\circ$ into our geometric relation, we immediately find the first angle of refraction:

$$r_1 + 0^\circ = 30^\circ \implies r_1 = 30^\circ$$

### The Final Stroke: Snell's Law

We now have everything we need to find the refractive index of the material. We know the angle of incidence at the first face ($i_1 = 60^\circ$) and the corresponding angle of refraction ($r_1 = 30^\circ$). It's time to bring in Snell's Law:

$$\mu = \frac{\sin i_1}{\sin r_1}$$

Substituting our known angles into Snell's Law, we get:

$$\mu = \frac{\sin 60^\circ}{\sin 30^\circ}$$

We know the standard trigonometric values: $\sin 60^\circ = \frac{\sqrt{3}}{2}$ and $\sin 30^\circ = \frac{1}{2}$. Plugging these in:

$$\mu = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}$$

The denominators cancel out perfectly, leaving us with a clean, elegant final answer:

$$\mu = \sqrt{3}$$

And there we have it! By systematically applying the prism deviation formula, internal geometry, and Snell's Law, we've completely unraveled the physics of this optical system.

Visualized Solution

\text{Visualizing the Prism}

\text{Angle of Deviation}

\text{Substituting Values}

\text{Angle of Emergence}

\text{Internal Angles Relation}

\text{Finding } r_1

\text{Snell's Law}

\text{Refractive Index}

The Sigma Insight: Refraction and Dispersion through Prism

The Setup

The Master Equation

The Revelation

Inside the Prism

The Final Stroke

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