Intensity of plane polarized light is 3.3 W/m. Area of plane `3 x 10^(-4) m^2` and polarizer rotates with `10pi` rad/sec. Energy transmitted in 1 complete cycle

Options

A. 4.95 x 10^(-4)
B. 3.95 x 10^(-4)
C. 2.95 x 10^(-4)
D. 6.95 x 10^(-4)

Answer

To solve the problem, we need to calculate the energy transmitted through a rotating polarizer in one complete cycle. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Given Values
We are given:
- Intensity of plane polarized light, \( I_0 = 3.3 \, \text{W/m}^2 \)
- Area of the plane, \( A = 3 \times 10^{-4} \, \text{m}^2 \)
- Angular velocity of the polarizer, \( \omega = 10\pi \, \text{rad/sec} \)

### Step 2: Calculate the Average Energy Transmitted
The energy transmitted through the polarizer can be calculated using the formula:
\[ E = I_0 \cdot A \cdot \langle \cos^2 \theta \rangle \]
where \( \langle \cos^2 \theta \rangle \) is the average value of \( \cos^2 \theta \) over one complete cycle.

### Step 3: Find the Average of \( \cos^2 \theta \)
The average value of \( \cos^2 \theta \) over one complete cycle (from \( 0 \) to \( 2\pi \)) is:
\[ \langle \cos^2 \theta \rangle = \frac{1}{2} \]

### Step 4: Substitute the Values into the Formula
Now, substituting the known values into the energy formula:
\[ E = I_0 \cdot A \cdot \langle \cos^2 \theta \rangle = 3.3 \, \text{W/m}^2 \cdot (3 \times 10^{-4} \, \text{m}^2) \cdot \frac{1}{2} \]

### Step 5: Calculate the Energy
Calculating the above expression:
\[ E = 3.3 \cdot 3 \times 10^{-4} \cdot \frac{1}{2} = \frac{9.9 \times 10^{-4}}{2} = 4.95 \times 10^{-4} \, \text{J} \]

### Conclusion
The energy transmitted in one complete cycle is:
\[ E = 4.95 \times 10^{-4} \, \text{J} \]

### Final Answer
Thus, the correct option is:
**Option 1: \( 4.95 \times 10^{-4} \, \text{J} \)**

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