A polariser-analyser set is adjusted such that the intensity of light coming out of the analyser is

Q: Why did we assume the original intensity is the maximum transmitted intensity?

In problems involving a 'polariser-analyser set', the reference to 'original intensity' typically means the maximum intensity that the set can transmit (when axes are parallel), which serves as $I_{max}$ in Malus's Law.

Q: What if the light entering the polariser was unpolarised and the 10% referred to that?

If the 10% referred to the unpolarised intensity before the polariser, the equation would be $I_{out} = \frac{I_{unpol}}{2} \cos^2 \theta = 0.1 I_{unpol}$, leading to $\cos^2 \theta = 0.2$. However, solving this yields an angle that doesn't match the given options, confirming our interpretation is the intended one.

Q: How do I calculate $\cos^{-1}(1/\sqrt{10})$ without a calculator?

You can use approximations. $\sqrt{10} \approx 3.16$, so $\cos \theta \approx 1/3.16 \approx 0.316$. Knowing $\cos 72^\circ \approx 0.309$, you can estimate that the angle is very close to $71.6^\circ$.

Q: Why do we subtract the angle from $90^\circ$?

The question asks for the *further* rotation needed to reach zero intensity. Zero intensity happens when the axes are perpendicular, which means they are exactly $90^\circ$ apart.

Q: Can the analyser be rotated in the opposite direction to get zero intensity?

Yes, rotating it to $-90^\circ$ would also yield zero intensity, but the shortest rotation is towards the nearest perpendicular axis, which is $90^\circ - 71.6^\circ = 18.4^\circ$.

Question

A polariser-analyser set is adjusted such that the intensity of light coming out of the analyser is just 10% of the original intensity. Assuming that the polariser-analyser set does not absorb any light, the angle by which the analyser need to be rotated further to reduce the output intensity to be zero, is

Options:

(A) $71.6^\circ$

(B) $90^\circ$

(C) $45^\circ$

(D) $18.4^\circ$

Sigma Percentile · Accepted Answer

### The Setup: Polariser and Analyser

Imagine a beam of light passing through a two-stage optical system: a **polariser** followed by an **analyser**. The polariser's job is to take unpolarised light and restrict its electric field vibrations to a single plane. The analyser, which is essentially a second polariser, then acts as a gatekeeper. The amount of light it lets through depends entirely on the angle between its transmission axis and the transmission axis of the first polariser.

### Applying Malus's Law

To mathematically determine the transmitted intensity, we rely on **Malus's Law**. This fundamental principle of optics states that the output intensity $I_{out}$ is equal to the maximum incident intensity $I_{max}$ multiplied by the square of the cosine of the angle $	heta$ between the two transmission axes:

$$I_{out} = I_{max} \cos^2 	heta$$

The problem states that the intensity coming out of the analyser is exactly $10\%$ of the original maximum intensity. By substituting this condition into Malus's Law, we can set up our core equation:

$$0.1 I_{max} = I_{max} \cos^2 	heta$$

### Calculating the Initial Angle

We can simplify the equation by dividing both sides by $I_{max}$, which leaves us with a straightforward trigonometric relationship:

$$\cos^2 	heta = 0.1 = \frac{1}{10}$$

Taking the square root of both sides gives us the value of $\cos 	heta$:

$$\cos 	heta = \frac{1}{\sqrt{10}}$$

To find the actual angle $	heta$, we take the inverse cosine. While you might not know $\cos^{-1}(1/\sqrt{10})$ off the top of your head, recognizing that $\sqrt{10} \approx 3.16$ means $\cos 	heta \approx 0.316$. This corresponds to an angle of approximately:

$$	heta \approx 71.6^\circ$$

### The Final Rotation to Zero

Now, we must address the final part of the question: *How much further must the analyser be rotated to reduce the output intensity to zero?*

We know from Malus's Law that the intensity drops to zero when $\cos 	heta = 0$. This occurs when the transmission axes are perfectly perpendicular to each other, meaning the angle between them is $90^\circ$.

Since our analyser is currently sitting at $71.6^\circ$ relative to the polariser, the additional rotation $\phi$ required to reach the $90^\circ$ mark is simply the difference between the two angles:

$$\phi = 90^\circ - 71.6^\circ = 18.4^\circ$$

Thus, a further rotation of **$18.4^\circ$** will completely extinguish the transmitted light.

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Visualized Solution

$Setup: Polariser and Analyser$

$Malus’s Law$

$Applying the Condition$

$Solving for θ$

$Calculating the Angle$

$Condition for Zero Intensity$

$Further Rotation Required$

$Conceptual Check$

The Sigma Insight: Polarization

The Setup

Applying Malus's Law

Calculating the Initial Angle

The Final Rotation to Zero

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Visualized Solution

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

.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; } Applying the Condition

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

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

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

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

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