Two beams, A and B, of plane polarised light with mutually perpendicular planes of polarisation are

Q: Why is the initial angle for beam B considered $90^\circ$?

The problem states that initially, beam A has maximum intensity and beam B has zero intensity. For a polaroid to completely block a plane-polarised beam, its transmission axis must be perpendicular ($90^\circ$) to the beam's plane of polarisation.

Q: Does it matter which direction we rotate the polaroid?

No, it doesn't. Whether you rotate it $+30^\circ$ or $-30^\circ$, the angle with beam A will be $30^\circ$, and the angle with beam B will be $90^\circ - 30^\circ = 60^\circ$ or $90^\circ + 30^\circ = 120^\circ$. Since $\cos^2(60^\circ) = \cos^2(120^\circ) = 1/4$, the result remains exactly the same.

Q: What is Malus's Law?

Malus's Law states that when completely plane-polarised light is incident on an analyzer, the intensity of the light transmitted is directly proportional to the square of the cosine of the angle between the transmission axes of the polarizer and analyzer.

Q: Why do we square the cosine term?

The amplitude of the transmitted electric field is $E_0 \cos\theta$. Since intensity is proportional to the square of the amplitude ($I \propto E^2$), the transmitted intensity becomes proportional to $\cos^2\theta$.

Q: Can this principle be used to determine the initial intensities if they are unknown?

Yes! By rotating the polaroid and finding the angle where the transmitted intensities are equal, you can easily calculate the ratio of their initial intensities, just like we did in this problem.

Question

Two beams, $A$ and $B$, of plane polarised light with mutually perpendicular planes of polarisation are seen through a polaroid. From the position, when the beam $A$ has maximum intensity (and beam $B$ has zero intensity), a rotation of polaroid through $30^\circ$ makes the two beams appear equally bright. If the initial intensities of the two beams are $I_A$ and $I_B$ respectively, then $I_A / I_B$ equals

Options:

(A) 3

(B) 3/2

(C) 1

(D) 1/3

Sigma Percentile · Accepted Answer

The phenomenon of polarization is one of the most fascinating aspects of wave optics. It reveals the transverse nature of light. In this problem, we are dealing with two mutually perpendicular plane-polarized beams and a rotating polaroid. Let's break down the physics step-by-step.

### The Initial Setup: A Tale of Two Beams

Imagine you are looking through a polaroid. We are given two beams, $A$ and $B$, which have mutually perpendicular planes of polarization.

Initially, the problem states that beam $A$ shines through with **maximum intensity**, while beam $B$ has **zero intensity**. What does this physical observation tell us about the geometry of the setup?

It implies that the polaroid's transmission axis is perfectly aligned (parallel) with the plane of polarization of beam $A$. Consequently, since beam $B$ is perpendicular to beam $A$, the transmission axis is exactly at $90^\circ$ to the plane of polarization of beam $B$, completely blocking it.

### The Rotation: Malus's Law in Action

To analyze what happens when we rotate the polaroid, we rely on **Malus's Law**. This fundamental law states that when completely plane-polarized light is incident on an analyzer, the transmitted intensity $I$ is given by:

$$I = I_0 \cos^2 	heta$$

where $I_0$ is the initial intensity and $	heta$ is the angle between the light's plane of polarization and the polaroid's transmission axis.

Now, we rotate the polaroid by $30^\circ$. Let's determine the new angles for both beams.

Since the transmission axis was initially parallel to beam $A$, the new angle for beam $A$ is simply:

$$	heta_A = 30^\circ$$

For beam $B$, the transmission axis was initially at $90^\circ$. After a $30^\circ$ rotation, the new angle becomes:

$$	heta_B = 90^\circ - 30^\circ = 60^\circ$$

### The Balancing Act: Equating Intensities

Using Malus's Law, we can express the new transmitted intensities for both beams. For beam $A$, the transmitted intensity is:

$$I'_A = I_A \cos^2 30^\circ$$

For beam $B$, the transmitted intensity is:

$$I'_B = I_B \cos^2 60^\circ$$

The problem provides a crucial condition: after the rotation, the two beams appear **equally bright**. This means their transmitted intensities are equal. We can set up our master equation:

$$I'_A = I'_B$$

$$I_A \cos^2 30^\circ = I_B \cos^2 60^\circ$$

### Final Calculation

Now, we simply substitute the standard trigonometric values. We know that $\cos 30^\circ = \frac{\sqrt{3}}{2}$ and $\cos 60^\circ = \frac{1}{2}$. Substituting these into our equation yields:

$$I_A \left(\frac{\sqrt{3}}{2}ight)^2 = I_B \left(\frac{1}{2}ight)^2$$

Squaring the terms, we get:

$$I_A \left(\frac{3}{4}ight) = I_B \left(\frac{1}{4}ight)$$

The denominators cancel out beautifully. Rearranging the equation to find the ratio of the initial intensities, we arrive at our final answer:

$$\frac{I_A}{I_B} = \frac{1}{3}$$

This elegant result shows how a simple rotation can balance the intensities of two perpendicular beams, provided their initial intensities are in a specific ratio!

Select Answer:

Visualized Solution

$I ni t ia l S e t u p$

$M a l u s^{'} s L a w$

$R o t a t i o n b y 3 0^{\circ}$

$E q u a t in g I n t e n s i t i es$

$S u b s t i t u t i o n$

$F ina l A n s w er$

$T h e W a y F or w a r d$

The Sigma Insight: Polarization

The Initial Setup

The Rotation

The Balancing Act

Final Calculation

Similar Questions

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When an unpolarised light of intensity $I_{0}$ is incident on a polarising sheet, the intensity of the light which does not get transmitted is

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Malus′s Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rotation by 30∘

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating Intensities

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward