A beam of plane polarized light having flux `10^(-3)` Watt falls normally on polarizer of cross sectional area `3xx10^(-4) m^2`. Polarizer rotates with angular frequency of 31.4 rad/s. Energy of light passes through the polarizer per revolution will be

Options

A. `10^(-4) J`
B. `10^(-3) J`
C. `10^(-2) J`
D. `10^(-1) J`

Answer

To find the energy of light that passes through the polarizer per revolution, we can follow these steps:

### Step 1: Understand the relationship between power, time, and energy.
The energy \( E \) passing through the polarizer can be calculated using the formula:
\[ E = P \times t \]
where \( P \) is the power of the light and \( t \) is the time duration for one complete revolution of the polarizer.

### Step 2: Identify the given values.
From the problem statement, we have:
- Power \( P = 10^{-3} \) Watts
- Angular frequency \( \omega = 31.4 \) rad/s

### Step 3: Calculate the time period \( t \) of one revolution.
The time period \( T \) is given by the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{31.4} \]
Calculating this gives:
\[ T \approx \frac{6.28}{31.4} \approx 0.2 \text{ seconds} \]

### Step 4: Calculate the energy \( E \) passing through the polarizer.
Now that we have the time period, we can substitute \( P \) and \( T \) into the energy formula:
\[ E = P \times T = 10^{-3} \times 0.2 \]
Calculating this gives:
\[ E = 0.2 \times 10^{-3} = 2 \times 10^{-4} \text{ Joules} \]

### Step 5: Final answer.
Thus, the energy of light that passes through the polarizer per revolution is:
\[ E = 2 \times 10^{-4} \text{ Joules} \]