A plane polarized blue light ray is incident on a prism such that there is no reflection from the su

Q: Why does 'no reflection' mean it's Brewster's angle?

According to Brewster's law, when unpolarized light is incident at a specific angle (Brewster's angle), the reflected light is completely s-polarized. If the incident light is purely p-polarized (polarized in the plane of incidence), there is zero reflection. The light is 100% transmitted.

Q: How did we know the light was p-polarized?

The problem states there is 'no reflection'. This is only physically possible if the light is incident at Brewster's angle AND is polarized entirely in the plane of incidence. If it had any s-polarized component, that component would reflect.

Q: Why is the deviation formula $\delta = i + e - A$?

The total deviation is the sum of the deviations at each surface. At the first surface, the ray bends by $(i - r_1)$. At the second surface, it bends by $(e - r_2)$. Total deviation $\delta = (i - r_1) + (e - r_2) = i + e - (r_1 + r_2)$. Since $r_1 + r_2 = A$, we get $\delta = i + e - A$.

Q: How did we solve the equation $\sqrt{3} \sin(A - 30^\circ) = \sin A$?

We used the compound angle formula $\sin(A - B) = \sin A \cos B - \cos A \sin B$. Substituting $B = 30^\circ$, we get $\sqrt{3}(\sin A \cdot \frac{\sqrt{3}}{2} - \cos A \cdot \frac{1}{2}) = \sin A$. Expanding and grouping the $\sin A$ and $\cos A$ terms leads directly to $\tan A = \sqrt{3}$, which means $A = 60^\circ$.

Q: Why doesn't the prism angle $A$ change for the red light?

The prism angle $A$ is a physical, geometric property of the glass prism itself. It is the angle between the two refracting faces. Changing the color (wavelength) of the light changes the refractive index and how the light bends, but it cannot change the physical shape of the glass.

Question

A plane polarized blue light ray is incident on a prism such that there is no reflection from the surface of the prism. The angle of deviation of the emergent ray is $\delta = 60^\circ$ (see Figure-1). The angle of minimum deviation for red light from the same prism is $\delta_{	ext{min}} = 30^\circ$ (see Figure-2). The refractive index of the prism material for blue light is $\sqrt{3}$. Which of the following statement(s) is(are) correct?

Options:

(A) The blue light is polarized in the plane of incidence.

(B) The angle of the prism is $45^\circ$.

(C) The refractive index of the material of the prism for red light is $\sqrt{2}$.

(D) The angle of refraction for blue light in air at the exit plane of the prism is $60^\circ$.

Sigma Percentile · Accepted Answer

### Decoding the "No Reflection" Mystery

Imagine you are shining a blue laser at a glass prism. Normally, you'd expect some light to bounce off the surface while the rest enters the glass. But the problem states something extraordinary: **there is absolutely no reflection from the surface of the prism**.

When does a surface completely refuse to reflect light? This phenomenon occurs exclusively at **Brewster's angle**, provided the incident light is purely p-polarized (polarized parallel to the plane of incidence). If the light had any s-polarized component, that component would reflect. Therefore, the blue light must be polarized in the plane of incidence, making **Option (A) correct** right out of the gate.

### The Master Equation: Tracing the Blue Light

Since we are at Brewster's angle, we can immediately find the angle of incidence $i$ using Brewster's Law:

$$ 	an i = \mu_B $$

Given the refractive index for blue light $\mu_B = \sqrt{3}$, we find:

$$ 	an i = \sqrt{3} \implies i = 60^\circ $$

Now, let's trace the light inside the prism. Applying Snell's Law at the first interface:

$$ 1 \cdot \sin 60^\circ = \sqrt{3} \cdot \sin r_1 $$

$$ \frac{\sqrt{3}}{2} = \sqrt{3} \sin r_1 \implies \sin r_1 = \frac{1}{2} \implies r_1 = 30^\circ $$

Next, we look at the overall bending of the ray. The total angle of deviation $\delta$ is given by the classic prism formula:

$$ \delta = i + e - A $$

The problem gives us a deviation $\delta = 60^\circ$. Plugging in our known values:

$$ 60^\circ = 60^\circ + e - A \implies e = A $$

This is a beautiful, elegant relation! The angle of emergence $e$ is exactly equal to the prism angle $A$.

### Unlocking the Prism Angle

Inside the prism, geometry dictates that the sum of the two internal angles of refraction must equal the prism angle:

$$ r_1 + r_2 = A $$

Substituting $r_1 = 30^\circ$, we can express $r_2$ purely in terms of $A$:

$$ r_2 = A - 30^\circ $$

We are now ready to tackle the second surface where the light exits. Applying Snell's Law again:

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

Substituting everything we know:

$$ \sqrt{3} \sin(A - 30^\circ) = \sin A $$

I know this trigonometric equation looks a bit intimidating, but let's take a breath and expand it using the compound angle formula $\sin(A - B) = \sin A \cos B - \cos A \sin B$:

$$ \sqrt{3} \left(\sin A \cdot \frac{\sqrt{3}}{2} - \cos A \cdot \frac{1}{2}ight) = \sin A $$

$$ \frac{3}{2} \sin A - \frac{\sqrt{3}}{2} \cos A = \sin A $$

Rearranging the terms, we find:

$$ \frac{1}{2} \sin A = \frac{\sqrt{3}}{2} \cos A \implies 	an A = \sqrt{3} $$

Therefore, the prism angle **$A = 60^\circ$**. This immediately tells us that **Option (B) is incorrect**. Furthermore, since $e = A$, the angle of emergence is $60^\circ$, making **Option (D) correct**.

### The Red Light Shift

Let's shift our focus to the red light. The problem states that the red light experiences a minimum deviation of $\delta_{	ext{min}} = 30^\circ$. Remember, the physical prism hasn't changed, so its angle $A$ remains $60^\circ$.

To find the refractive index for red light, we use the minimum deviation formula:

$$ \mu_R = \frac{\sin\left(\frac{A + \delta_{	ext{min}}}{2}ight)}{\sin\left(\frac{A}{2}ight)} $$

Let's plug in the numbers:

$$ \mu_R = \frac{\sin\left(\frac{60^\circ + 30^\circ}{2}ight)}{\sin\left(\frac{60^\circ}{2}ight)} = \frac{\sin 45^\circ}{\sin 30^\circ} $$

$$ \mu_R = \frac{1/\sqrt{2}}{1/2} = \sqrt{2} $$

So, the refractive index for red light is $\sqrt{2}$, making **Option (C) correct** as well. This problem is a masterclass in combining polarization, Brewster's angle, and prism kinematics!

Select Answer:

Visualized Solution

\text{The "No Reflection" Condition}

\text{Brewster's Angle Calculation}

\text{Snell's Law at First Surface}

\text{The Deviation Equation}

\text{Internal Prism Geometry}

\text{Snell's Law at Second Surface}

\text{Solving for Prism Angle } A

\text{Evaluating Options (B) and (D)}

\text{Red Light: Minimum Deviation}

\text{Calculating } \mu_R

\text{The Way Forward}

The Sigma Insight: Refraction and Dispersion through Prism

Decoding the "No Reflection" Mystery

The Master Equation

Unlocking the Prism Angle

The Red Light Shift

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