Unpolarised light of intensity `32 Wm^(-2)` passes through three polarisers such that the transmission axis of the last polariser is crossed with first. If the intensity of the emerging light is `3Wm^(-2)`, the angle between the axes of the first two polarisers is

Options

A. `45^(@)`
B. `60^(@)`
C. `30^(@)`
D. Zero

Answer

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

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

where \( I_0 \) is the intensity of the incident light and \( \theta \) is the angle between the light's polarization direction and the polarizer's axis.

### Step 1: Understand the Setup
We have three polarizers:
- The first polarizer allows half of the unpolarized light to pass through, which is \( \frac{I_0}{2} \).
- The second polarizer is at an angle \( \theta \) with respect to the first.
- The third polarizer is at an angle of \( 90^\circ \) to the first polarizer, meaning it is at an angle of \( 90^\circ - \theta \) to the second polarizer.

### Step 2: Calculate the Intensity After the First Polarizer
The initial intensity of the unpolarized light is given as \( I_0 = 32 \, \text{W/m}^2 \). After passing through the first polarizer, the intensity becomes:
\[ I_1 = \frac{I_0}{2} = \frac{32}{2} = 16 \, \text{W/m}^2 \]

### Step 3: Calculate the Intensity After the Second Polarizer
Using Malus's Law for the second polarizer:
\[ I_2 = I_1 \cos^2(\theta) = 16 \cos^2(\theta) \]

### Step 4: Calculate the Intensity After the Third Polarizer
For the third polarizer, we apply Malus's Law again:
\[ I_3 = I_2 \cos^2(90^\circ - \theta) = I_2 \sin^2(\theta) \]
Substituting \( I_2 \):
\[ I_3 = 16 \cos^2(\theta) \sin^2(\theta) \]

### Step 5: Set Up the Equation Using the Final Intensity
We know that the final intensity \( I_3 \) is given as \( 3 \, \text{W/m}^2 \):
\[ 3 = 16 \cos^2(\theta) \sin^2(\theta) \]

### Step 6: Simplify the Equation
Rearranging gives:
\[ \cos^2(\theta) \sin^2(\theta) = \frac{3}{16} \]
Using the identity \( \sin^2(2\theta) = 4 \sin^2(\theta) \cos^2(\theta) \), we can write:
\[ \frac{1}{4} \sin^2(2\theta) = \frac{3}{16} \]
Multiplying both sides by 4:
\[ \sin^2(2\theta) = \frac{3}{4} \]

### Step 7: Solve for \( 2\theta \)
Taking the square root:
\[ \sin(2\theta) = \frac{\sqrt{3}}{2} \]
This implies:
\[ 2\theta = \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = 60^\circ \]

### Step 8: Find \( \theta \)
Dividing by 2 gives:
\[ \theta = 30^\circ \]

### Final Answer
The angle between the axes of the first two polarizers is \( 30^\circ \).