Energy Through a Rotating Polarizer – Detailed Solution

Energy Transmitted Through a Rotating Polarizer

A beam of plane polarised light of large cross-sectional area and uniform intensity \(\displaystyle I_0 = \frac{20}{7} \, \text{W/m}^2\) falls normally on a polarizer with cross‑sectional area \(\displaystyle A = 7 \times 10^{-4}\,\text{m}^2\). The polarizer rotates about its axis with an angular speed \(\displaystyle \omega = 15.7\,\text{rad/s}\). Calculate the energy of light passing through the polarizer per revolution.

• (1) \(3.0 \times 10^{-5}\, \text{J}\)
• (2) \(3.0 \times 10^{-4}\, \text{J}\)
• (3) \(4.0 \times 10^{-4}\, \text{J}\)
• (4) \(5.0 \times 10^{-4}\, \text{J}\)

Select the Correct Option

(1) \(3.0 \times 10^{-5}\, \text{J}\)

(2) \(3.0 \times 10^{-4}\, \text{J}\)

(3) \(4.0 \times 10^{-4}\, \text{J}\)

(4) \(5.0 \times 10^{-4}\, \text{J}\)

Step‑by‑Step Explanation

Step 1: Compute the Incident Power
The beam has an intensity \(I_0 = \frac{20}{7}\) W/m² and the polarizer has an area \(\displaystyle A = 7 \times 10^{-4}\,\text{m}^2\). So the power incident on the polarizer is: \[ P_{\text{incident}} = I_0 \times A = \frac{20}{7} \times 7 \times 10^{-4} = 20 \times 10^{-4} = 2.0 \times 10^{-3}\, \text{W}. \]

Step 2: Account for Polarization
When plane polarised light passes through a polarizer, the transmitted intensity is given by Malus’s law: \[ I_{\text{transmitted}} = I_0 \cos^2 \theta. \] As the polarizer rotates with angular speed \(\omega\), the angle \(\theta\) between the incident polarization direction and the polarizer’s axis changes uniformly over time. Over one full revolution, the average value of \(\cos^2 \theta\) is \(\frac{1}{2}\).

Therefore, the average transmitted power is: \[ P_{\text{transmitted}} = \frac{1}{2} P_{\text{incident}} = \frac{1}{2} \times 2.0 \times 10^{-3} = 1.0 \times 10^{-3}\,\text{W}. \]

Step 3: Determine the Time for One Revolution
The angular speed is \(\omega = 15.7\,\text{rad/s}\). The time for one revolution (angle \(2\pi\) radians) is: \[ T = \frac{2\pi}{\omega} \approx \frac{6.2832}{15.7} \approx 0.4\,\text{s}. \]

Step 4: Calculate the Energy Transmitted per Revolution
The energy \(E\) is the power transmitted multiplied by the time for one revolution: \[ E = P_{\text{transmitted}} \times T = 1.0 \times 10^{-3}\,\text{W} \times 0.4\,\text{s} = 4.0 \times 10^{-4}\,\text{J}. \]

Final Answer: The energy of light passing through the polarizer per revolution is approximately \(\displaystyle 4.0 \times 10^{-4}\,\text{J}\), which corresponds to option (3).

WhatsApp Group Join Now

Telegram Group Join Now