A monochromatic light is travelling in a medium of refractive index n = 1.6. It enters a stack of gl

Q: Why can we directly equate the first and the $m$th layer using Snell's law?

Because the interfaces are parallel, the angle of refraction in one layer becomes the angle of incidence for the next. This creates a chain: $n_1 \sin\theta_1 = n_2 \sin\theta_2 = n_3 \sin\theta_3$, allowing us to equate any two layers directly.

Q: What does it mean when the ray is 'refracted out parallel to the interface'?

It means the ray grazes the boundary between the two media. In this case, the angle of refraction in the $m$th slab is exactly $90^\circ$ with respect to the normal.

Q: Why is the answer mathematically correct but physically impossible?

Solving the equation gives $m = 8$, which means the refractive index of the 8th slab is $1.6 - 8(0.1) = 0.8$. Since the refractive index of a medium cannot be less than 1 (the refractive index of a vacuum), this specific setup cannot exist in reality.

Q: Does the thickness of the slabs affect the final angle?

No, the thickness of the parallel slabs only affects the lateral displacement of the ray, not the angle at which it travels in any given medium.

Q: What would happen if the refractive index was increasing instead of decreasing?

If the refractive index was increasing, the ray would bend towards the normal at each interface, and it would never graze the interface or undergo total internal reflection.

Question

A monochromatic light is travelling in a medium of refractive index $n = 1.6$. It enters a stack of glass layers from the bottom side at an angle $	heta = 30^\circ$. The interfaces of the glass layers are parallel to each other. The refractive indices of different glass layers are monotonically decreasing as $n_m = n - m\Delta n$, where $n_m$ is the refractive index of the $m$th slab and $\Delta n = 0.1$ (see the figure). The ray is refracted out parallel to the interface between the $(m-1)$th and $m$th slabs from the right side of the stack. What is the value of $m$?

Sigma Percentile · Accepted Answer

Have you ever wondered what happens to a light ray as it travels through a medium where the refractive index keeps changing? Imagine a stack of glass slabs, each slightly less dense than the one below it. As the light ray moves upward, it bends further and further away from the normal. Eventually, it might just bend so much that it travels perfectly parallel to the interface! Let's dive into this fascinating optical journey.

### The Magic of Parallel Interfaces

When light travels across multiple parallel boundaries, we don't need to calculate the angle of refraction at every single step. Thanks to the geometry of parallel lines, the angle of refraction in one medium becomes the exact angle of incidence for the next.

This allows us to use the **Generalized Snell's Law**, which states that the product of the refractive index and the sine of the angle with the normal remains constant across all layers:
$$n \sin	heta = n_1 \sin	heta_1 = n_2 \sin	heta_2 = \dots = 	ext{constant}$$

This powerful principle means we can directly connect the very first medium to the final medium, completely bypassing the intermediate layers!

### The Grazing Condition

In our problem, the ray enters the first medium (with $n = 1.6$) at an angle of $	heta = 30^\circ$. As it moves up the stack, the refractive index decreases according to the formula $n_m = n - m\Delta n$, where $\Delta n = 0.1$.

The problem states that the ray is refracted out parallel to the interface between the $(m-1)$th and $m$th slabs. This is a classic description of **grazing emergence**. It means that in the $m$th slab, the angle of refraction is exactly $90^\circ$.

### The Master Equation

Let's apply our generalized Snell's law between the initial medium and the $m$th slab:
$$n \sin	heta = n_m \sin 90^\circ$$

We know that $\sin 90^\circ = 1$. Substituting the given values, we get:
$$1.6 \sin 30^\circ = (1.6 - m 	imes 0.1) 	imes 1$$

Since $\sin 30^\circ = 0.5$, the left side simplifies beautifully:
$$1.6 	imes 0.5 = 1.6 - 0.1m$$
$$0.8 = 1.6 - 0.1m$$

### The Final Calculation

Now, it's just a matter of simple algebra to find $m$:
$$0.1m = 1.6 - 0.8$$
$$0.1m = 0.8$$
$$m = 8$$

So, the ray grazes the interface at the 8th slab!

### A Physical Catch

While our math is flawless, there's a hidden twist in this problem. Let's calculate the actual refractive index of that 8th slab:
$$n_8 = 1.6 - 8(0.1) = 0.8$$

Wait a minute! The refractive index of a vacuum is 1, and no normal material can have a refractive index less than 1. This means our calculated medium is **physically impossible**!

Even though the setup cannot exist in the real world, it serves as a brilliant mathematical exercise to test your understanding of Snell's law and total internal reflection. Always trust the math, but never forget to view the results through the lens of physical reality!

Enter Numerical Value:

Visualized Solution

\text{Understanding the Setup}

\text{Generalized Snell's Law}

\text{Applying Snell's Law}

\text{Substituting Values}

\text{Solving for } m

\text{Final Calculation}

\text{Physical Validity Check}

The Sigma Insight: Refraction and Total Internal Reflection

The Magic of Parallel Interfaces

The Grazing Condition

The Master Equation

The Final Calculation

A Physical Catch

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