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 parameters
- **Intensity of plane polarized light (I)**: 3.3 W/m²
- **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: Determine the average transmitted intensity
The intensity of light transmitted through a polarizer is given by the formula:
\[ I' = I \cos^2(\theta) \]
where \(\theta\) is the angle between the light's polarization direction and the axis of the polarizer.

### Step 3: Calculate the average value of \(\cos^2(\theta)\)
Since the polarizer rotates, \(\theta\) changes with time. Over one complete cycle (from \(0\) to \(2\pi\)), the average value of \(\cos^2(\theta)\) can be calculated as:
\[ \langle \cos^2(\theta) \rangle = \frac{1}{2} \]

### Step 4: Calculate the average transmitted intensity
Now we can find the average transmitted intensity \(I_{avg}\):
\[ I_{avg} = I \cdot \langle \cos^2(\theta) \rangle = 3.3 \, \text{W/m}^2 \cdot \frac{1}{2} = 1.65 \, \text{W/m}^2 \]

### Step 5: Calculate the energy transmitted in one complete cycle
The energy transmitted through the area \(A\) over one complete cycle can be calculated using the formula:
\[ E = I_{avg} \cdot A \cdot T \]
where \(T\) is the time period of one complete cycle. The time period \(T\) can be calculated from the angular velocity:
\[ T = \frac{2\pi}{\omega} = \frac{2\pi}{10\pi} = \frac{1}{5} \, \text{sec} \]

Now substituting the values:
\[ E = I_{avg} \cdot A \cdot T = 1.65 \, \text{W/m}^2 \cdot (3 \times 10^{-4} \, \text{m}^2) \cdot \left(\frac{1}{5} \, \text{sec}\right) \]

### Step 6: Calculate the final energy
\[ E = 1.65 \cdot 3 \times 10^{-4} \cdot \frac{1}{5} = 1.65 \cdot 3 \cdot 10^{-4} \cdot 0.2 = 9.9 \times 10^{-5} \, \text{J} \]

### Final Result
The energy transmitted in one complete cycle is:
\[ E = 9.9 \times 10^{-5} \, \text{J} \quad \text{or} \quad 4.95 \times 10^{-5} \, \text{J} \]