Unpolarized light of intensity `32Wm^(-2)` passes through three polarizers such that transmission axes of the first and second polarizer make an angle `30^@` with each other and the transmission axis of the last polarizer is crossed with that of the first . The intensity of the final emerging light will be

Options

A. `32 Wm^(-2)`
B. `3 Wm^(-2)`
C. `8 Wm^(-2)`
D. `4 Wm^(-2)`

Answer

To solve the problem step by step, we will use Malus's Law, which states that when polarized light passes through a polarizer, the intensity of the transmitted light is given by:

\[ I = I_0 \cos^2(\theta) \]

where:
- \( I \) is the transmitted intensity,
- \( I_0 \) is the initial intensity,
- \( \theta \) is the angle between the light's polarization direction and the polarizer's axis.

### Step 1: Initial Intensity
The initial intensity of the unpolarized light is given as:

\[ I_0 = 32 \, \text{W/m}^2 \]

### Step 2: First Polarizer
When unpolarized light passes through the first polarizer, the intensity is reduced to half:

\[ I_1 = \frac{I_0}{2} = \frac{32}{2} = 16 \, \text{W/m}^2 \]

### Step 3: Second Polarizer
The angle between the first and second polarizer is \( 30^\circ \). We apply Malus's Law to find the intensity after the second polarizer:

\[ I_2 = I_1 \cos^2(30^\circ) \]

Calculating \( \cos(30^\circ) \):

\[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \]

Now substituting this into the equation:

\[ I_2 = 16 \cdot \left(\frac{\sqrt{3}}{2}\right)^2 \]
\[ I_2 = 16 \cdot \frac{3}{4} \]
\[ I_2 = 12 \, \text{W/m}^2 \]

### Step 4: Third Polarizer
The third polarizer is crossed with the first polarizer, meaning the angle between the second and third polarizer is \( 90^\circ - 30^\circ = 60^\circ \). We again apply Malus's Law:

\[ I_3 = I_2 \cos^2(60^\circ) \]

Calculating \( \cos(60^\circ) \):

\[ \cos(60^\circ) = \frac{1}{2} \]

Now substituting this into the equation:

\[ I_3 = 12 \cdot \left(\frac{1}{2}\right)^2 \]
\[ I_3 = 12 \cdot \frac{1}{4} \]
\[ I_3 = 3 \, \text{W/m}^2 \]

### Final Result
The intensity of the final emerging light is:

\[ I_3 = 3 \, \text{W/m}^2 \]