For a prism of prism angle θ = 60°, the refractive indices of the left half and the right half are,

Q: Why does the ray travel horizontally inside the prism initially?

In a symmetric prism at minimum deviation, the ray path is symmetric. This means the angle of refraction at both faces is equal ($r_1 = r_2 = \theta/2$). Geometrically, this forces the ray to be parallel to the base of the prism.

Q: Why doesn't the ray bend at the vertical interface when $n_2$ changes?

Because the ray is traveling horizontally, it strikes the vertical interface at exactly $90^\circ$ to the surface (an angle of incidence of $0^\circ$ with respect to the normal). According to Snell's law, a ray incident at $0^\circ$ passes through undeviated, regardless of the change in refractive index.

Q: Why do we differentiate Snell's law instead of just solving for $e$?

Solving for $e$ directly would involve inverse sine functions of small increments, which requires complex Taylor series expansions. Differentiating the equation implicitly is a much faster and more elegant way to relate small changes (differentials) like $\Delta n$ and $\Delta e$.

Q: How did we know $\cos e = \sqrt{7}/4$?

We evaluated the derivative at the initial state where $\Delta n = 0$. In this state, the emergent angle $e$ is equal to the incident angle $i$. Since we already calculated $\sin i = 3/4$, we know $\sin e = 3/4$. Using the identity $\cos e = \sqrt{1 - \sin^2 e}$, we get $\sqrt{1 - 9/16} = \sqrt{7}/4$.

Q: What if the prism wasn't equilateral?

The problem states the prism angle is $60^\circ$ and the diagram implies symmetry (isosceles). If it weren't symmetric, the vertical interface wouldn't be exactly perpendicular to the horizontal ray, and the ray would refract at the middle boundary, making the calculation significantly more complex.

Question

For a prism of prism angle $	heta = 60^\circ$, the refractive indices of the left half and the right half are, respectively, $n_1$ and $n_2$ ($n_2 \ge n_1$) as shown in the figure. The angle of incidence $i$ is chosen such that the incident light rays will have minimum deviation if $n_1 = n_2 = n = 1.5$. For the case of unequal refractive indices, $n_1 = n$ and $n_2 = n + \Delta n$ (where $\Delta n \ll n$), the angle of emergence $e = i + \Delta e$. Which of the following statement(s) is (are) correct?

Options:

(A) The value of $\Delta e$ (in radians) is greater than that of $\Delta n$

(B) $\Delta e$ is proportional to $\Delta n$

(C) $\Delta e$ lies between 2.0 and 3.0 milliradians, if $\Delta n = 2.8 	imes 10^{-3}$

(D) $\Delta e$ lies between 1.0 and 1.6 milliradians, if $\Delta n = 2.8 	imes 10^{-3}$

Sigma Percentile · Accepted Answer

### The Anatomy of the Prism

Imagine a beautiful equilateral glass prism, but with a secret: it's split right down the middle.

The left half has a refractive index of $n_1$, and the right half has $n_2$.

Initially, the problem sets a beautifully symmetric stage. Both halves have the exact same refractive index, $n_1 = n_2 = 1.5$.

In this state, the prism is perfectly tuned to produce **minimum deviation**.

### The Magic of Minimum Deviation

When a prism is at minimum deviation, the light ray passing through it travels perfectly parallel to the base.

Because of this symmetry, the angle of refraction at the first surface, $r_1$, and the angle of incidence at the second surface, $r_2$, are exactly equal.

Geometrically, they are exactly half of the prism angle $	heta$.

Since $	heta = 60^\circ$, we immediately know that $r_1 = r_2 = 30^\circ$.

Let's lock down the angle of incidence $i$ using Snell's Law at the first surface:

$$1 \cdot \sin i = n_1 \sin r_1$$

Substituting our known values:

$$\sin i = 1.5 \cdot \sin 30^\circ = 1.5 \cdot \frac{1}{2} = \frac{3}{4}$$

This incident angle $i$ is now a fixed parameter for the rest of our journey.

### The Twist: A Slight Perturbation

Now, the problem throws a curveball. The right half of the prism gets a tiny boost in its refractive index: $n_2 = n + \Delta n$.

But what happens to the ray in the left half? Absolutely nothing!

The incident angle $i$ hasn't changed, and $n_1$ is still $1.5$. Therefore, the ray still refracts at $r_1 = 30^\circ$ and travels perfectly horizontally.

Here is the crucial insight: because the ray is horizontal, it strikes the vertical boundary between the two halves at exactly $90^\circ$ to the surface.

In the language of optics, its angle of incidence at this middle interface is $0^\circ$.

According to Snell's Law, a ray hitting a boundary head-on passes through completely undeviated!

It enters the right half still traveling horizontally, meaning it strikes the final exit surface at the exact same angle, $r_2 = 30^\circ$.

### The Power of Calculus in Optics

Now we must find the new emergent angle, $e = i + \Delta e$. Let's write Snell's Law for the final surface:

$$n_2 \sin r_2 = 1 \cdot \sin e$$

Substituting our perturbed values:

$$(n + \Delta n) \sin 30^\circ = \sin(i + \Delta e)$$

We could try to solve this using inverse trigonometry, but that leads to a messy algebraic nightmare.

Instead, let's use the elegance of calculus! Since $\Delta n$ and $\Delta e$ are infinitesimally small, we can simply differentiate our Snell's Law equation with respect to $n_2$:

$$\Delta n \cdot \sin 30^\circ = \Delta e \cdot \cos e$$

Rearranging this gives us a beautiful, direct relationship:

$$\Delta e = \frac{\sin 30^\circ}{\cos e} \Delta n$$

### The Final Calculation

To evaluate this, we need $\cos e$. Since the perturbation is tiny, we evaluate the derivative at the initial state where $e = i$.

We already know $\sin e = \sin i = \frac{3}{4}$. Using the Pythagorean identity:

$$\cos e = \sqrt{1 - \sin^2 e} = \sqrt{1 - \left(\frac{3}{4}ight)^2} = \frac{\sqrt{7}}{4}$$

Now, plug everything back into our differential equation:

$$\Delta e = \frac{1/2}{\sqrt{7}/4} \Delta n = \frac{2}{\sqrt{7}} \Delta n$$

Since $\sqrt{7} \approx 2.64$, the fraction $\frac{2}{\sqrt{7}}$ is approximately $0.75$.

This tells us two things immediately: $\Delta e$ is directly proportional to $\Delta n$, and $\Delta e$ is strictly less than $\Delta n$.

Finally, let's crunch the numbers for the given $\Delta n = 2.8 	imes 10^{-3}$:

$$\Delta e = \frac{2}{\sqrt{7}} 	imes 2.8 	imes 10^{-3} = \frac{5.6}{2.645} 	imes 10^{-3} \approx 2.11 	imes 10^{-3} 	ext{ rad}$$

This is $2.11 	ext{ mrad}$, which sits perfectly between $2.0$ and $3.0 	ext{ mrad}$.

A stunning problem that beautifully marries geometric optics with differential calculus!

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Visualized Solution

$Analyzing the Prism Setup$

$Minimum Deviation Condition$

$Calculating Angle of Incidence$

$Changing the Refractive Index$

$Snell’s Law at Second Surface$

$Differentiating the Equation$

$Evaluating cos e$

$Relation between Δ e and Δ n$

$Calculating the Numerical Value$

The Sigma Insight: Refraction and Dispersion through Prism

The Anatomy of the Prism

The Magic of Minimum Deviation

The Twist

The Power of Calculus in Optics

The Final Calculation

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Visualized Solution

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Changing the Refractive Index

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Relation between Δe and Δn

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