As shown in the figure, a ray AB of unpolarized light enters from water of refractive index nw = 4/3

Q: Why did we ignore the glass layer completely?

Because the interfaces are perfectly parallel. According to Snell's Law, the quantity $n \sin \theta$ is conserved across all parallel layers. The intermediate glass layer shifts the ray laterally but does not change the angle it makes when it returns to a medium of the same refractive index (water).

Q: How do we know the reflection happens at the top interface?

The diagram visually traces the ray passing through the glass and the second water layer before reflecting. The reflection is explicitly shown at the boundary between the upper water layer ($n_w$) and the top medium ($n_p$).

Q: What do the dots on the reflected ray mean?

The dots represent the electric field vector oscillating perpendicular to the plane of the screen. This indicates that the light is completely plane-polarized, which is a direct signature of reflection occurring exactly at Brewster's angle.

Q: Why is $\tan \theta_B = n_p / n_w$ and not the other way around?

Brewster's angle is defined as $\tan \theta_B = n_2 / n_1$, where $n_1$ is the medium the light is coming *from*, and $n_2$ is the medium the light is reflecting *off*. Here, light travels from water ($n_w$) and reflects off the top medium ($n_p$).

Q: What if the initial medium was air instead of water?

If the initial medium was air ($n=1$), then by Snell's Law, $1 \cdot \sin i = n_w \sin \theta_B$. You would first find $\theta_B = 60^\circ$, and then calculate $\sin i = (4/3) \sin 60^\circ$ to find the new initial angle.

Question

As shown in the figure, a ray $AB$ of unpolarized light enters from water of refractive index $n_w = 4/3$ into a medium of refractive index $n_p = 4/\sqrt{3}$ after passing through a glass plate of refractive index $n_g=1.5$ and a layer of water. At a particular incident angle $i$ the reflected ray $CD$ is polarized in the direction as shown in the figure. The value of $i$ (in degrees) is:

Sigma Percentile · Accepted Answer

### The Setup: A Multi-Layer Optical Path

Imagine a beam of unpolarized light embarking on a journey through a multi-layered optical sandwich. It starts in water, passes through a glass plate, re-enters water, and finally strikes a new medium at the top.

The problem asks us to find the initial angle of incidence, $i$, at the very first boundary. At first glance, tracing the ray through all these refractions seems like a mathematical nightmare. However, the problem gives us a massive clue hidden in plain sight.

### The Clue: Polarization by Reflection

Look closely at the reflected ray $CD$ in the diagram. It is adorned with dots, which is the universal physics symbol for light that is completely plane-polarized perpendicular to the plane of incidence.

This is the signature of **Brewster's Law**. Light becomes completely polarized upon reflection only when it strikes the interface at a very specific angle, known as Brewster's angle ($	heta_B$). The law states that the tangent of this angle is equal to the ratio of the refractive indices of the two media:

$$	an 	heta_B = \frac{n_{	ext{ref}}}{n_{	ext{inc}}}$$

In our setup, the reflection occurs at the top interface. The light is traveling from water ($n_w = 4/3$) and reflecting off the top medium ($n_p = 4/\sqrt{3}$). Let's plug these values into Brewster's equation:

$$	an 	heta_B = \frac{4/\sqrt{3}}{4/3} = \sqrt{3}$$

Since $	an 60^\circ = \sqrt{3}$, we immediately know that the angle of incidence at this top interface is exactly $60^\circ$.

### The Magic of Parallel Interfaces

Now we know the angle at the top, but how does that help us find the initial angle $i$ at the bottom? This is where the elegance of parallel interfaces comes into play.

According to Snell's Law, when light passes through a series of parallel boundaries, the product of the refractive index and the sine of the angle remains constant across every single layer:

$$n_1 \sin 	heta_1 = n_2 \sin 	heta_2 = \dots = n_k \sin 	heta_k$$

This means we can completely ignore the intermediate glass layer! We can directly equate the conditions at the very first interface to the conditions at the interface just before the reflection.

### The Final Calculation

Let's apply this generalized Snell's Law. The initial medium is water ($n_w$), and the medium just before the reflection is also water ($n_w$).

$$n_w \sin i = n_w \sin 	heta_B$$

Because the refractive indices are identical, they cancel out perfectly:

$$\sin i = \sin 	heta_B$$

Therefore, the initial angle of incidence $i$ must be equal to Brewster's angle $	heta_B$.

$$i = 60^\circ$$

And just like that, a seemingly complex multi-layer problem collapses into a beautiful, single-line realization. The intermediate layers shifted the ray laterally, but they could not alter its fundamental angular relationship with the parallel boundaries.

Enter Numerical Value:

Visualized Solution

Visualizing the Optical Path

Brewster's Law

Setting up Brewster's Equation

Calculating Brewster's Angle

Generalized Snell's Law

Applying Snell's Law

Final Calculation

The Way Forward

The Sigma Insight: Polarization

The Setup

The Clue

The Magic of Parallel Interfaces

The Final Calculation

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