A source of light is placed in front of a screen. Intensity of light on the screen is I. Two polaroi

Q: Why does the first polaroid reduce the intensity to exactly half?

Unpolarized light has electric field vectors oscillating in all possible directions perpendicular to the direction of propagation. An ideal polaroid only allows the component of these vectors parallel to its pass axis to go through. When you average $\cos^2\theta$ over all possible angles from $0$ to $2\pi$, the result is exactly $1/2$.

Q: What if the initial light was already polarized?

If the initial light was already polarized, the first polaroid would not necessarily halve the intensity. The transmitted intensity would depend on the angle between the light's polarization direction and the polaroid's pass axis, governed by Malus's Law right from the start.

Q: Why did we assume the initial angle between $P_1$ and $P_2$ was $0^\circ$?

The problem states that after passing through $P_1$, the intensity is $I/2$. It also states that with both $P_1$ and $P_2$ in place (before any rotation), the intensity on the screen is still $I/2$. The only way the second polaroid doesn't reduce the intensity further is if its pass axis is perfectly aligned with the first one.

Q: Does the direction of rotation (clockwise vs counter-clockwise) matter?

No, it doesn't! Malus's Law depends on $\cos^2\phi$. Since $\cos(-\phi) = \cos(\phi)$, rotating the polaroid by $30^\circ$ clockwise or counter-clockwise will result in the exact same transmitted intensity.

Q: What happens to the light that doesn't pass through the polaroid?

In an ideal polaroid, the light component that is perpendicular to the pass axis is completely absorbed by the material (often long-chain molecules). This absorbed energy is typically converted into a tiny amount of heat.

Question

A source of light is placed in front of a screen. Intensity of light on the screen is $I$. Two polaroids $P_1$ and $P_2$ are so placed in between the source of light and screen that the intensity of light on screen is $I/2$. $P_2$ should be rotated by an angle of ........... (degrees), so that the intensity of light on the screen becomes $3I/8$.

Sigma Percentile · Accepted Answer

### The Magic of Polarization

Imagine a beam of light as a chaotic dance of electric fields vibrating in every possible direction perpendicular to its path. This is unpolarized light. When this light encounters a polaroid, something magical happens. The polaroid acts like a microscopic picket fence, allowing only the vibrations parallel to its 'pass axis' to slip through.

Because the original vibrations were completely random, exactly half of the light's energy makes it through this fence. Therefore, if the initial intensity is $I$, the intensity after the first polaroid, $P_1$, is perfectly halved to $I/2$.

### Analyzing the Setup

The problem presents a fascinating scenario. We place a second polaroid, $P_2$, behind the first one. Surprisingly, the intensity on the screen remains $I/2$. What does this tell us?

It means that the second polaroid didn't block any additional light! For this to happen, the 'picket fence' of $P_2$ must be perfectly aligned with the 'picket fence' of $P_1$. In physics terms, their pass axes are parallel, and the angle between them is $0^\circ$.

### The Master Equation: Malus's Law

Now, the real challenge begins. We are asked to rotate $P_2$ by an angle $\phi$ so that the final intensity drops to $3I/8$. To solve this, we invoke **Malus's Law**.

Malus's Law states that when completely polarized light of intensity $I_0$ passes through an analyzer (our second polaroid), the transmitted intensity $I$ is given by:

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

Here, $I_0$ is the intensity incident on $P_2$, which we know is $I/2$. The final desired intensity is $3I/8$. Let's set up our equation:

$$\frac{3I}{8} = \frac{I}{2} \cos^2\phi$$

### Final Calculation

This is where we execute the math. First, we can elegantly cancel the initial intensity $I$ from both sides of the equation.

$$\frac{3}{8} = \frac{1}{2} \cos^2\phi$$

Multiplying both sides by $2$, we isolate the trigonometric term:

$$\cos^2\phi = \frac{3}{4}$$

Taking the square root of both sides gives us:

$$\cos\phi = \frac{\sqrt{3}}{2}$$

From our fundamental knowledge of trigonometry, we recognize this value instantly. The angle whose cosine is $\frac{\sqrt{3}}{2}$ is $30^\circ$.

Therefore, to achieve the desired intensity of $3I/8$, the second polaroid must be rotated by exactly **$30^\circ$**.

Enter Numerical Value:

Visualized Solution

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

$I n t e n s i t y a f t er P_{1} an d P_{2}$

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

$A ppl y in g t h e C o n d i t i o n$

$S o l v in g f or ϕ$

$F ina l A n g l e$

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

The Sigma Insight: Polarization

The Magic of Polarization

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

Enter Numerical Value:

Visualized Solution

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