When an unpolarised light of intensity I_0 is incident on a polarising sheet, the intensity of the l

Q: Why is the transmitted intensity exactly half?

Unpolarised light is a random mixture of all polarizations. When you average the transmission factor $\cos^2\theta$ over all possible angles from $0$ to $2\pi$, the mathematical average is exactly $1/2$.

Q: Does Malus's Law apply here?

Malus's law ($I = I_0 \cos^2\theta$) strictly applies when the incident light is already polarised. For unpolarised light, we integrate Malus's law over all angles, which gives us the constant factor of $1/2$.

Q: What happens to the untransmitted light?

The light that is blocked by the polariser doesn't just disappear; its energy is absorbed by the polarising material, which is typically converted into a tiny amount of heat.

Q: Would the answer change if the polariser was rotated?

No. Because unpolarised light has no preferred direction, rotating the polariser will not change the fact that exactly half of the light is transmitted and half is absorbed.

Q: What if the incident light was already polarised?

If the incident light was polarised, we would use Malus's Law directly. The transmitted intensity would be $I_0 \cos^2\theta$, and the untransmitted (absorbed) part would be $I_0 - I_0 \cos^2\theta = I_0 \sin^2\theta$.

Question

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

Options:

(A) $\frac{1}{2} I_0$

(B) $\frac{1}{4} I_0$

(C) zero

(D) $I_0$

Sigma Percentile · Accepted Answer

### The Setup: Unpolarised Light Meets a Polariser

Imagine an unpolarised light beam with an initial intensity of $I_0$ traveling towards a polarising sheet. In unpolarised light, the electric field vibrates in all possible directions perpendicular to its path of propagation. It is a chaotic mix of waves.

When this light hits the polariser, the polariser acts like a strict bouncer at a club. It has a specific "pass axis." The polariser only allows the component of the electric field that is perfectly parallel to this pass axis to go through. Any component perpendicular to it is completely blocked.

### The Filter: What Gets Through?

Because unpolarised light is a random mixture of all possible polarization angles, we can think of it as having equal components in all directions. Mathematically, if we average the transmission over all possible angles using Malus's Law ($I = I_0 \cos^2	heta$), the average value of $\cos^2	heta$ is exactly $\frac{1}{2}$.

Therefore, the intensity of the light that successfully makes it through the polariser is exactly half of the original intensity.

$$I_t = \frac{I_0}{2}$$

### The Catch: What Gets Left Behind?

Now, here is where many students make a silly mistake. They see $\frac{I_0}{2}$, find it in the options, and immediately tick it. But we must read the question carefully! The question asks for the intensity of the light which does **not** get transmitted.

This is a simple application of the law of conservation of energy. The total energy (or intensity) of the incident light must equal the sum of the transmitted light and the absorbed (untransmitted) light.

$$I_{	ext{incident}} = I_{	ext{transmitted}} + I_{	ext{untransmitted}}$$

We know the incident intensity is $I_0$ and the transmitted intensity is $\frac{I_0}{2}$. Let's substitute these values to find the untransmitted intensity $I_u$:

$$I_u = I_0 - I_t$$

$$I_u = I_0 - \frac{I_0}{2}$$

$$I_u = \frac{I_0}{2}$$

So, exactly half of the light is transmitted, and the other half is absorbed by the polarising sheet. The intensity of the untransmitted light is **$\frac{1}{2} I_0$**.

Animated Solution for Physics - Optics: 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

Unpolarised Light and Polariser

Transmitted Intensity

Calculating Transmitted Intensity

Untransmitted Intensity

Final Calculation

The Way Forward

The Sigma Insight: Polarization

The Setup

The Filter

The Catch

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