Two equilateral-triangular prisms P1 and P2 are kept with their sides parallel to each other, in vac

Q: Why is the angle of refraction inside P2 exactly 30°?

For minimum deviation in a prism, the ray passes symmetrically, meaning $r_1 = r_2 = A/2$. Since it's an equilateral prism, $A=60^\circ$, making the internal angles 30°.

Q: How do we know the angle of emergence from P1 is the same as the angle of incidence on P2?

The problem states the adjacent sides of the prisms are parallel. A ray passing through the vacuum gap between two parallel surfaces will have alternate interior angles equal.

Q: Why did we convert 15° to $\pi/12$?

The final expression in the question is given in radians ($\pi/\beta$). Converting $15^\circ$ to radians ($15 \times \pi/180$) gives $\pi/12$, allowing us to directly compare and find $\beta$.

Q: Does the distance between the prisms affect the angles?

No, as long as the faces remain parallel and the medium between them is uniform (vacuum), the angles of incidence and emergence at the parallel interfaces remain equal regardless of the gap distance.

Q: What if the prisms had different angles, say 45° and 60°?

The logic remains the same, but you would use $A_2/2$ for the minimum deviation in the second prism, and $r_1 + r_2 = A_1$ for the first prism. The parallel face condition would still hold.

Question

Two equilateral-triangular prisms $P_1$ and $P_2$ are kept with their sides parallel to each other, in vacuum, as shown in the figure. A light ray enters prism $P_1$ at an angle of incidence $	heta$ such that the outgoing ray undergoes minimum deviation in prism $P_2$. If the respective refractive indices of $P_1$ and $P_2$ are $\sqrt{\frac{3}{2}}$ and $\sqrt{3}$ , then $	heta = \sin^{-1} \left[ \sqrt{\frac{3}{2}} \sin \left( \frac{\pi}{\beta} ight) ight]$ , where the value of $\beta$ is

Sigma Percentile · Accepted Answer

The problem of the double prism might look intimidating at first glance, but it holds a beautiful secret. The key to unlocking this puzzle isn't to start from the beginning, but to start from the end! Let's embark on a fascinating journey backwards through the light ray's path.

### The Anchor: Minimum Deviation in the Second Prism

The most crucial piece of information given to us is that the light ray undergoes **minimum deviation** in the second prism, $P_2$. What does this physically mean?

When a ray passes through a prism with minimum deviation, it travels perfectly symmetrically. Inside the prism, the ray becomes parallel to the base. Because of this symmetry, the angle of refraction at the first face is exactly equal to the angle of incidence at the second face, and both are exactly half of the prism's apex angle.

Since $P_2$ is an equilateral prism, its apex angle is $A = 60^\circ$. Therefore, the internal angle of refraction, let's call it $r_1'$, is:
$$r_1' = \frac{A}{2} = \frac{60^\circ}{2} = 30^\circ$$

### Bridging the Gap: Snell's Law and Parallel Faces

Now that we know the ray's angle inside $P_2$, we can find out how it entered. We apply **Snell's Law** at the first surface of $P_2$. Let the angle of incidence from the vacuum gap be $\alpha$. The refractive index of $P_2$ is given as $n_2 = \sqrt{3}$.

$$1 \cdot \sin \alpha = n_2 \sin r_1'$$
$$\sin \alpha = \sqrt{3} \sin 30^\circ$$
$$\sin \alpha = \sqrt{3} \left(\frac{1}{2}ight) = \frac{\sqrt{3}}{2}$$

This is a standard trigonometric value! It tells us that the angle $\alpha$ is exactly $60^\circ$.

Here is where the geometry of the setup comes to our rescue. The problem states that the adjacent faces of the two prisms are **parallel** to each other. When a transversal line (our light ray) intersects two parallel lines (the prism faces), the alternate interior angles are equal.

This means the angle at which the ray emerges from the first prism $P_1$ must be exactly equal to the angle at which it strikes the second prism $P_2$. Thus, the angle of emergence from $P_1$ is also $60^\circ$.

### Tracing Backwards: Inside the First Prism

Armed with the emergence angle, we can now trace the ray backwards into the first prism, $P_1$. We apply Snell's Law at its exiting face. Let the internal angle of incidence at this face be $r_2$. The refractive index of $P_1$ is $n_1 = \sqrt{\frac{3}{2}}$.

$$n_1 \sin r_2 = 1 \cdot \sin \alpha$$
$$\sqrt{\frac{3}{2}} \sin r_2 = \sin 60^\circ$$
$$\sqrt{\frac{3}{2}} \sin r_2 = \frac{\sqrt{3}}{2}$$

By canceling $\sqrt{3}$ from both sides, we get:
$$\frac{1}{\sqrt{2}} \sin r_2 = \frac{1}{2}$$
$$\sin r_2 = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$$

This reveals that the internal angle $r_2$ is exactly $45^\circ$.

Now, we use a fundamental property of any prism: the sum of the two internal angles of refraction, $r_1$ and $r_2$, is always equal to the apex angle of the prism. Since $P_1$ is also an equilateral prism, its angle is $60^\circ$.

$$r_1 + r_2 = A$$
$$r_1 + 45^\circ = 60^\circ$$
$$r_1 = 15^\circ$$

We have successfully traced the ray all the way back to the very first surface where the light entered!

### The Final Strike: Finding the Angle of Incidence

Finally, we apply Snell's Law one last time at the entry point of $P_1$ to find the initial angle of incidence, $	heta$.

$$1 \cdot \sin 	heta = n_1 \sin r_1$$
$$\sin 	heta = \sqrt{\frac{3}{2}} \sin 15^\circ$$

Taking the inverse sine, we get:
$$	heta = \sin^{-1} \left( \sqrt{\frac{3}{2}} \sin 15^\circ ight)$$

To match the format given in the question, we convert $15^\circ$ into radians. Since $180^\circ = \pi$ radians, $15^\circ$ is exactly $\frac{\pi}{12}$ radians.

$$	heta = \sin^{-1} \left[ \sqrt{\frac{3}{2}} \sin \left( \frac{\pi}{12} ight) ight]$$

Comparing this with the given expression $	heta = \sin^{-1} \left[ \sqrt{\frac{3}{2}} \sin \left( \frac{\pi}{\beta} ight) ight]$, we can clearly see that the value of $\beta$ perfectly matches **12**.

By working backwards step-by-step, what seemed like a complex optical maze unraveled into a beautiful sequence of logical deductions!

Enter Numerical Value:

Visualized Solution

$P_{2}$ Minimum Deviation$

$r_{1}^{'} = \frac{A}{2}$

$1 sin α = n_{2} sin r_{1}^{'}$

$α = 6 0^{\circ}$

$Parallel Faces$

$n_{1} sin r_{2} = 1 sin α$

$r_{2} = 4 5^{\circ}$

$r_{1} + r_{2} = A$

$r_{1} = 1 5^{\circ}$

$1 sin θ = n_{1} sin r_{1}$

$β = 12$

The Sigma Insight: Refraction and Dispersion through Prism

The Anchor

Bridging the Gap

Tracing Backwards

The Final Strike

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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; } P2​ Minimum Deviation$

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r1′​=2A​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1sinα=n2​sinr1′​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α=60∘

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n1​sinr2​=1sinα

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r2​=45∘

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r1​+r2​=A

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r1​=15∘

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1sinθ=n1​sinr1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β=12