A prism of refractive index n1 and another prism of refractive index n2 are stuck together (as shown

Q: Why did the problem give so many angles in the diagram if we don't use them?

This is a classic distractor technique used in competitive exams! The angles are there to test your conceptual clarity. If you know that 'no bending' means equal refractive indices, you can confidently ignore the geometry.

Q: What are these equations for $n_1$ and $n_2$?

These equations represent Cauchy's dispersion formula, which shows how the refractive index of a material decreases as the wavelength of light increases.

Q: Why do we ignore the negative square root when solving for $\lambda$?

Wavelength represents a physical distance (the length of one wave cycle), so it must always be a positive value.

Q: How did $9.0 / 0.25$ become $36$ so quickly?

Remember that $0.25$ is the same as the fraction $1/4$. Dividing by a fraction is the same as multiplying by its reciprocal. So, $9.0 / (1/4) = 9.0 \times 4 = 36$.

Q: Would the ray bend if it hit the interface exactly at $90^\circ$ (normal incidence) even if $n_1 \neq n_2$?

No, a ray hitting exactly along the normal never bends, regardless of the refractive indices. However, the problem specifies that it passes without bending for *any* angle, which strictly requires $n_1 = n_2$.

Question

A prism of refractive index $n_1$ and another prism of refractive index $n_2$ are stuck together (as shown in the figure). $n_1$ and $n_2$ depend on $\lambda$, the wavelength of light, according to the relation
$n_1 = 1.2 + \frac{10.8 	imes 10^{-14}}{\lambda^2}$
and $n_2 = 1.45 + \frac{1.8 	imes 10^{-14}}{\lambda^2}$
The wavelength for which rays incident at any angle on the interface BC pass through without bending at that interface will be ............... nm.

Sigma Percentile · Accepted Answer

### The Illusion of Complexity

When you first look at this problem, your eyes are immediately drawn to the intricate geometry of the two prisms. You see angles of $60^\circ$, $70^\circ$, $90^\circ$, and $40^\circ$. It is incredibly tempting to start drawing normals, applying geometry, and setting up complex trigonometric relations.

However, this is a classic psychological trap set by the examiners! The key to unlocking this problem lies not in the geometry, but in a single, powerful phrase hidden in the text: **"rays incident at any angle on the interface BC pass through without bending."**

### The Core Principle: Index Matching

Let's think about the physics of refraction. According to Snell's Law, $n_1 \sin i = n_2 \sin r$. Bending occurs because light changes its speed when it transitions between media with different optical densities.

If a light ray passes through a boundary without any deviation (meaning the angle of incidence $i$ equals the angle of refraction $r$) for *every possible angle*, it implies that the light doesn't even "feel" the boundary. Optically speaking, the two media are identical. Therefore, the fundamental condition we must satisfy is simply:

$$n_1 = n_2$$

When two materials share the exact same refractive index, the interface between them becomes optically invisible. This phenomenon is known as **index matching**.

### The Algebra of Light

The problem provides us with expressions for $n_1$ and $n_2$ that depend on the wavelength $\lambda$. These expressions are forms of Cauchy's dispersion formula. By equating them, we set up our master equation:

$$1.2 + \frac{10.8 	imes 10^{-14}}{\lambda^2} = 1.45 + \frac{1.8 	imes 10^{-14}}{\lambda^2}$$

Now, it's just a matter of careful algebra. We group the constant terms on the right side and the terms containing $\lambda^2$ on the left side:

$$\frac{10.8 	imes 10^{-14}}{\lambda^2} - \frac{1.8 	imes 10^{-14}}{\lambda^2} = 1.45 - 1.2$$

Subtracting the numerators on the left and the constants on the right yields a much cleaner equation:

$$\frac{9.0 	imes 10^{-14}}{\lambda^2} = 0.25$$

### The Final Reveal

To isolate $\lambda^2$, we rearrange the equation. A great mental math trick here is to recognize that dividing by $0.25$ is mathematically identical to multiplying by $4$:

$$\lambda^2 = \frac{9.0 	imes 10^{-14}}{0.25} = 9.0 	imes 10^{-14} 	imes 4$$

$$\lambda^2 = 36 	imes 10^{-14} 	ext{ m}^2$$

Taking the square root of both sides gives us the wavelength in meters. Since wavelength is a physical distance, we only consider the positive root:

$$\lambda = \sqrt{36 	imes 10^{-14}} = 6 	imes 10^{-7} 	ext{ m}$$

Finally, we convert this result into nanometers, the standard unit for the wavelength of visible light. Since $1 	ext{ nm} = 10^{-9} 	ext{ m}$, we multiply our result by $10^9$:

$$\lambda = 600 	imes 10^{-9} 	ext{ m} = 600 	ext{ nm}$$

And there we have it! By focusing on the core physical principle and ignoring the geometric noise, a seemingly complex optics problem collapses into a straightforward algebraic calculation.

Enter Numerical Value:

Visualized Solution

Analyzing the Interface

Condition for No Refraction

Equating Refractive Indices

Rearranging the Equation

Solving for Wavelength Squared

Final Calculation

The Way Forward

The Sigma Insight: Refraction and Dispersion through Prism

The Illusion of Complexity

The Core Principle

The Algebra of Light

The Final Reveal

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