Consider a configuration of n identical units, each consisting of three layers. The first layer is a

Q: Why do we use $\tan \theta$ for the lateral shift?

In a right triangle formed by the thickness of the layer and the ray's path, the horizontal side (lateral shift) is opposite to the angle of refraction, and the thickness is adjacent. So, $\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$, which means $\text{shift} = \text{thickness} \times \tan \theta$.

Q: Can we use a generalized Snell's Law from the first to the last layer?

Yes! $n_1 \sin \theta_1 = n_k \sin \theta_k$ holds true across any number of parallel layers. This means you can directly relate the angle in the air to the angle in the final layer without calculating the intermediate steps, though here we needed all angles to find the total shift.

Q: What if the ray undergoes Total Internal Reflection (TIR)?

We must always check if the ray moves from a denser to a rarer medium and if the angle exceeds the critical angle. Here, the indices are $1 \to \sqrt{1.5} \to \sqrt{3}$, so it's moving into progressively denser media, meaning no TIR can occur.

Q: Why is the total shift simply $n$ times the shift of one unit?

Because the units are identical and the interfaces are perfectly parallel, the ray emerges from each unit at the exact same angle it entered ($60^\circ$). This makes the process perfectly periodic, so the shifts just add up linearly.

Q: Is the lateral shift dependent on the order of the layers?

Interestingly, yes! While the final emergent angle remains the same regardless of the order, the total lateral shift depends on the specific path taken, which changes if the layer order is swapped.

Question

Consider a configuration of n identical units, each consisting of three layers. The first layer is a column of air of height $h = \frac{1}{3}	ext{ cm}$, and the second and third layers are of equal thickness $d = \frac{\sqrt{3}-1}{2}	ext{ cm}$, and refractive indices $\mu_1 = \sqrt{\frac{3}{2}}$ and $\mu_2 = \sqrt{3}$, respectively. A light source O is placed on the top of the first unit, as shown in the figure. A ray of light from O is incident on the second layer of the first unit at an angle of $	heta = 60^\circ$ to the normal. For a specific value of n, the ray of light emerges from the bottom of the configuration at a distance $l = \frac{8}{\sqrt{3}}	ext{ cm}$, as shown in the figure. The value of n is ________.

Sigma Percentile · Accepted Answer

### The Beauty of Multiple Refractions

Imagine a beam of light traversing through a meticulously arranged stack of transparent layers. Every time the light crosses a boundary, it bends, obeying the fundamental laws of optics. This problem from JEE Advanced takes us on a journey through such a configuration, challenging us to track the light's path and calculate its total lateral displacement.

We are given a system composed of $n$ identical units. Each unit is a sandwich of three distinct layers: a column of air with height $h = \frac{1}{3}	ext{ cm}$, followed by two denser media with equal thickness $d = \frac{\sqrt{3}-1}{2}	ext{ cm}$ and refractive indices $\mu_1 = \sqrt{\frac{3}{2}}$ and $\mu_2 = \sqrt{3}$. A light ray enters the first air layer at an angle of $	heta = 60^\circ$ to the normal. Our goal is to find the number of units $n$ given that the total lateral shift is $l = \frac{8}{\sqrt{3}}	ext{ cm}$.

### Snell's Law in Action

To trace the ray's path, we must apply Snell's Law at each interface. Snell's Law states that $n_1 \sin 	heta_1 = n_2 \sin 	heta_2$. Let's analyze the refraction step-by-step as the ray travels through a single unit.

First, the light travels from air ($n = 1$) into the first medium ($\mu_1 = \sqrt{\frac{3}{2}}$). Applying Snell's Law:

$$1 \cdot \sin 60^\circ = \sqrt{\frac{3}{2}} \sin 	heta_1$$

Substituting $\sin 60^\circ = \frac{\sqrt{3}}{2}$, we get:

$$\frac{\sqrt{3}}{2} = \sqrt{\frac{3}{2}} \sin 	heta_1 \Rightarrow \sin 	heta_1 = \frac{1}{\sqrt{2}}$$

This gives us the angle of refraction in the first medium: $	heta_1 = 45^\circ$.

Next, the light travels from the first medium into the second medium ($\mu_2 = \sqrt{3}$). Applying Snell's Law again:

$$\sqrt{\frac{3}{2}} \sin 45^\circ = \sqrt{3} \sin 	heta_2$$

Substituting $\sin 45^\circ = \frac{1}{\sqrt{2}}$, we get:

$$\sqrt{\frac{3}{2}} \cdot \frac{1}{\sqrt{2}} = \sqrt{3} \sin 	heta_2 \Rightarrow \frac{\sqrt{3}}{2} = \sqrt{3} \sin 	heta_2 \Rightarrow \sin 	heta_2 = \frac{1}{2}$$

This gives us the angle of refraction in the second medium: $	heta_2 = 30^\circ$.

### The Geometry of Lateral Shift

As the ray travels through each layer, it shifts horizontally. The lateral shift in a single layer of thickness $t$ and angle of refraction $\phi$ is given by simple trigonometry: $\Delta x = t 	an \phi$.

Therefore, the total lateral shift $\Delta x$ for one complete unit is the sum of the shifts in its three layers:

$$\Delta x = h 	an 60^\circ + d 	an 45^\circ + d 	an 30^\circ$$

Now, we carefully substitute the given values for $h$ and $d$:

$$\Delta x = \left(\frac{1}{3}ight)\sqrt{3} + \left(\frac{\sqrt{3}-1}{2}ight)(1) + \left(\frac{\sqrt{3}-1}{2}ight)\frac{1}{\sqrt{3}}$$

### The Grand Summation

This expression might look intimidating, but it simplifies beautifully. Let's find a common denominator, which is $6$:

$$\Delta x = \frac{2\sqrt{3}}{6} + \frac{3\sqrt{3}-3}{6} + \frac{3-\sqrt{3}}{6}$$

Combining the numerators:

$$\Delta x = \frac{2\sqrt{3} + 3\sqrt{3} - 3 + 3 - \sqrt{3}}{6} = \frac{4\sqrt{3}}{6} = \frac{2}{\sqrt{3}}	ext{ cm}$$

So, the ray shifts horizontally by $\frac{2}{\sqrt{3}}	ext{ cm}$ every time it passes through one unit. Since the configuration consists of $n$ identical units, the total lateral shift $l$ is simply $n$ times the shift of a single unit:

$$l = n \cdot \Delta x$$

We are given that the total shift is $l = \frac{8}{\sqrt{3}}	ext{ cm}$. Equating the two expressions:

$$\frac{8}{\sqrt{3}} = n \cdot \frac{2}{\sqrt{3}}$$

Solving for $n$, we find:

**$$n = 4$$**

The elegance of this problem lies in how the complex-looking algebraic terms perfectly cancel out, leaving a clean, integer answer. It's a testament to the beautiful symmetry often found in physics!

Enter Numerical Value:

Visualized Solution

\text{Analyzing the Setup}

\text{Snell's Law}

\text{Refraction at 1st Interface}

\text{Refraction at 2nd Interface}

\text{Lateral Shift in One Unit}

\text{Substituting Values}

\text{Calculating } \Delta x

\text{Total Shift and Final Answer}

\text{The Way Forward}

The Sigma Insight: Refraction and Total Internal Reflection

The Beauty of Multiple Refractions

Snell's Law in Action

The Geometry of Lateral Shift

The Grand Summation

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