A system of three polarisers P1, P2, P3 is set up such that the pass axis of P3 is crossed with resp

Q: Why does the first polariser reduce the intensity by exactly half?

Unpolarised light consists of electric field vectors vibrating randomly in all directions. When resolved into components parallel and perpendicular to the pass axis, the average of $\cos^2\theta$ over all random angles is exactly $1/2$. Thus, half the light is transmitted.

Q: Why do we use the angle between $P_1$ and $P_2$ instead of the angle with the horizontal?

Malus's Law depends on the angle between the polarisation direction of the *incident* light and the pass axis of the *current* polariser. Since the light hitting $P_2$ was polarised by $P_1$, its polarisation direction is along $P_1$.

Q: What would happen if we removed the middle polariser $P_2$?

If $P_2$ is removed, the light from $P_1$ would directly hit $P_3$. Since $P_1$ and $P_3$ are crossed (angle is $90^\circ$), the transmitted intensity would be $I_1 \cos^2(90^\circ) = 0$. The middle polariser actually *enables* light to pass through!

Q: Is it possible to get an intensity greater than $I_0/2$ after the first polariser?

No. Ideal polarisers only absorb or transmit light; they do not amplify it. The maximum intensity you can ever have after the first polariser is $I_0/2$, and it will only decrease or stay the same as it passes through subsequent polarisers.

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

The polariser transmits the *amplitude* component of the electric field parallel to its axis, which is $E = E_0 \cos\theta$. Since intensity is proportional to the square of the amplitude ($I \propto E^2$), the transmitted intensity becomes proportional to $\cos^2\theta$.

Question

A system of three polarisers $P_1, P_2, P_3$ is set up such that the pass axis of $P_3$ is crossed with respect to that of $P_1$. The pass axis of $P_2$ is inclined at $60^\circ$ to the pass axis of $P_3$. When a beam of unpolarised light of intensity $I_0$ is incident on $P_1$, the intensity of light transmitted by the three polarisers is $I$. The ratio $(I_0 / I)$ equals (nearly)

Options:

(A) 5.33

(B) 16.00

(C) 10.67

(D) 1.80

Sigma Percentile · Accepted Answer

Imagine a beam of chaotic, unpolarised light—its electric field vectors vibrating in every possible direction perpendicular to its path. This is the starting point of our journey through a system of three polarisers, a classic setup that beautifully demonstrates the principles of wave optics and Malus's Law.

### Analyzing the Setup
Before we dive into the math, we must understand the geometry of our polarisers. We are given three polarisers: $P_1$, $P_2$, and $P_3$. The problem states that $P_1$ and $P_3$ are **crossed**. In the language of optics, this means their pass axes are perpendicular to each other, creating an angle of $90^\circ$.

Next, we introduce the middleman, $P_2$. Its pass axis is inclined at $60^\circ$ to the pass axis of $P_3$. To apply Malus's Law correctly, we need the angle between consecutive polarisers. Since $P_1$ is at $90^\circ$ and $P_2$ is at $60^\circ$ relative to $P_3$, the angle between $P_1$ and $P_2$ is simply the difference: $90^\circ - 60^\circ = 30^\circ$. This geometric deduction is the most critical step; get this wrong, and the entire calculation collapses.

### The First Encounter
Our unpolarised light, with an initial intensity of $I_0$, strikes the first polariser, $P_1$. There is a fundamental rule in optics: when unpolarised light passes through an ideal polariser, exactly half of its intensity is absorbed, and the transmitted light becomes plane-polarised parallel to the pass axis.

Therefore, the intensity of the light emerging from $P_1$ is:
$$I_1 = \frac{I_0}{2}$$

### The Master Equation: Malus's Law
Now, this plane-polarised light encounters $P_2$. Whenever polarised light passes through a subsequent polariser, its transmitted intensity is governed by **Malus's Law**, which states $I = I_{	ext{initial}} \cos^2	heta$, where $	heta$ is the angle between the polarisation direction of the incident light and the pass axis of the polariser.

For $P_2$, the incident light is polarised along $P_1$, and the angle between $P_1$ and $P_2$ is $30^\circ$. Applying Malus's Law:
$$I_2 = I_1 \cos^2(30^\circ)$$

Substituting $I_1 = \frac{I_0}{2}$ and $\cos(30^\circ) = \frac{\sqrt{3}}{2}$:
$$I_2 = \left(\frac{I_0}{2}ight) \left(\frac{\sqrt{3}}{2}ight)^2$$

$$I_2 = \left(\frac{I_0}{2}ight) \left(\frac{3}{4}ight) = \frac{3I_0}{8}$$

### The Final Gate
The light, now with intensity $I_2$ and polarised along $P_2$, reaches the final polariser, $P_3$. We apply Malus's Law one last time. The angle between $P_2$ and $P_3$ is given as $60^\circ$.

$$I_3 = I_2 \cos^2(60^\circ)$$

Substituting our value for $I_2$ and $\cos(60^\circ) = \frac{1}{2}$:
$$I_3 = \left(\frac{3I_0}{8}ight) \left(\frac{1}{2}ight)^2$$

$$I_3 = \left(\frac{3I_0}{8}ight) \left(\frac{1}{4}ight) = \frac{3I_0}{32}$$

This $I_3$ is our final transmitted intensity, $I$.

### Final Calculation
The problem asks for the ratio of the initial intensity $I_0$ to the final intensity $I$.

$$\frac{I_0}{I} = \frac{I_0}{\frac{3I_0}{32}} = \frac{32}{3}$$

Calculating the decimal value:
$$\frac{32}{3} \approx 10.67$$

This elegant result shows how a seemingly opaque system (crossed polarisers) can transmit light simply by introducing a third polariser at an intermediate angle. It's a beautiful demonstration of vector resolution in physics!

Select Answer:

Visualized Solution

$Geometry of Pass Axes$

$Intensity after P_{1}$

$Malus’s Law for P_{2}$

$Substituting Values for P_{2}$

$Calculating I_{2}$

$Malus’s Law for P_{3}$

$Substituting Values for P_{3}$

$Calculating I_{3}$

$Final Ratio$

$The Way Forward$

The Sigma Insight: Polarization

Analyzing the Setup Before we dive into the math, we must understand the geometry of our polarisers

The First Encounter Our unpolarised light, with an initial intensity of $I_{0}$ , strikes the first polariser, $P_{1}$

The Master Equation

The Final Gate The light, now with intensity $I_{2}$ and polarised along $P_{2}$ , reaches the final polariser, $P_{3}$

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

Select Answer:

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Intensity after P1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Malus’s Law for P2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values for P2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating I2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Malus’s Law for P3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values for P3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating I3​

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

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