An unpolarised light beam is incident on the polariser of a polarisation experiment and the intensit

Q: Why do we use the initial emerging intensity as $I_0$?

Because the 100 lumens represents light that is already polarized by the analyser in its initial position. Malus's law applies to the angle between the light's current polarization direction and the analyser's new pass axis.

Q: Does it matter if the rotation is clockwise or counter-clockwise?

No, it doesn't matter. Mathematically, $\cos(-\theta) = \cos(\theta)$, and since we are squaring the cosine value anyway, the intensity drop is perfectly symmetric in either direction.

Q: What if the incident light was completely unpolarized?

If unpolarized light of intensity $I_{unpol}$ hits a polarizer, the transmitted intensity is exactly half, $I_{unpol}/2$. However, in this specific question, we are already given the intensity *after* the light has been polarized.

Q: Why is the formula $\cos^2 \theta$ and not just $\cos \theta$?

The electric field amplitude of the light passes through the filter as $E = E_0 \cos \theta$. Since the physical intensity of light is proportional to the square of its electric field amplitude ($I \propto E^2$), we must square the cosine term, resulting in $\cos^2 \theta$.

Q: What happens if we rotate the analyser by $90^\circ$?

If $\theta = 90^\circ$, then $\cos(90^\circ) = 0$. Therefore, the transmitted intensity becomes zero. This is known as the 'crossed polarizer' configuration, which blocks all light.

Question

An unpolarised light beam is incident on the polariser of a polarisation experiment and the intensity of light beam emerging from the analyser is measured as $100$ lumens. Now, if the analyser is rotated around the horizontal axis (direction of light) by $30^{\circ}$ in clockwise direction, the intensity of emerging light will be ....... lumens.

Sigma Percentile · Accepted Answer

### Unveiling the Magic of Malus's Law

Have you ever wondered how polarized sunglasses manage to cut out the harsh glare from the road or water? The secret lies in the beautiful physics of polarization. In this problem, we are going to explore exactly how rotating a polarizing filter controls the amount of light that passes through it.

### Analyzing the Setup

Imagine you are looking straight down the barrel of a polarization experiment. The light has already passed through an initial polarizer and is now emerging from an analyser.

We are told that the initial intensity of the light emerging from this analyser is $100 	ext{ lumens}$. This is our baseline. Because the light has already passed through the analyser, we know it is perfectly polarized along the analyser's current pass axis. Let's call this initial intensity $I_0 = 100 	ext{ lm}$.

### The Master Equation

Now, the experiment gets interesting. We rotate the analyser clockwise by an angle of $	heta = 30^\circ$. What happens to the light intensity?

To answer this, we invoke **Malus's Law**. This fundamental law of optics states that when completely polarized light is incident on an analyser, the intensity $I$ of the light transmitted by the analyser is directly proportional to the square of the cosine of the angle between the transmission axes of the polarizer and the analyser.

Mathematically, it is expressed as:

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

### Final Calculation

Let's plug our specific numbers into Malus's Law. We know our initial intensity $I_0$ is $100$, and our angle $	heta$ is $30^\circ$.

$$I = 100 	imes \cos^2(30^\circ)$$

Do you remember the exact value of $\cos(30^\circ)$ from trigonometry? It is $\frac{\sqrt{3}}{2}$. We need to carefully square this entire fraction.

$$I = 100 	imes \left(\frac{\sqrt{3}}{2}ight)^2$$

Squaring the numerator ($\sqrt{3}$) gives us $3$, and squaring the denominator ($2$) gives us $4$.

$$I = 100 	imes \frac{3}{4}$$

Finally, we just do the basic arithmetic. $100$ divided by $4$ is $25$, and $25$ multiplied by $3$ gives us exactly $75$.

$$I = 75 	ext{ lumens}$$

And there we have it! By rotating the analyser just $30^\circ$, we have reduced the transmitted light intensity from $100 	ext{ lumens}$ down to $75 	ext{ lumens}$. This simple yet powerful principle is exactly how adjustable light filters and LCD screens operate in the real world.

Enter Numerical Value:

Visualized Solution

Initial Setup

Rotation and Malus's Law

Substituting Values

Evaluating Cosine

Squaring the Fraction

Final Intensity

Conceptual Reflection

The Sigma Insight: Polarization

Unveiling the Magic of Malus's Law

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

A beam of unpolarised light of intensity $I_{0}$ is passed through a polaroid $A$ and then through another polaroid $B$ which is oriented so that its principal plane makes an angle of $4 5^{\circ}$ relative to that of $A$ . The intensity of the emergent light is

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