Upolarised light of intensity 32 W `m^(-2)` passes through three polarisers such that transmission axis of first is crossed with third. If intensity of emerging light is 2 W `m^(-2)`, what is the angle of transmission axis between the first two polarisers?

Options

A. `30^(@)`
B. `45^(@)`
C. `22.5^(@)`
D. `60^(@)`

Explanation

To solve the problem, 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 intensity of the transmitted light,
- \( I_0 \) is the intensity of the incident polarized light,
- \( \theta \) is the angle between the light's polarization direction and the axis of the polarizer.

### Step 1: Determine the intensity after the first polarizer
The initial intensity of the unpolarized light is given as \( I_0 = 32 \, \text{W/m}^2 \). When unpolarized light passes through the first polarizer, the intensity is halved:

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

### Step 2: Apply Malus's Law for the second polarizer
Let \( \theta \) be the angle between the first and second polarizer. The intensity after the second polarizer can be calculated using Malus's Law:

\[ I_2 = I_1 \cos^2(\theta) = 16 \cos^2(\theta) \]

### Step 3: Apply Malus's Law for the third polarizer
The third polarizer is crossed with the first, meaning the angle between the first and third polarizer is \( 90^\circ \). Therefore, the intensity after the third polarizer is:

\[ I_3 = I_2 \cos^2(90^\circ) = I_2 \cdot 0 = 0 \]

However, we know that the final intensity is given as \( I_3 = 2 \, \text{W/m}^2 \). This implies that the angle between the first and second polarizer is not \( 90^\circ \) but needs to be calculated.

### Step 4: Set up the equation using the known final intensity
Since we know that \( I_3 = 2 \, \text{W/m}^2 \), we can set up the equation:

\[ 2 = 16 \cos^2(\theta) \]

### Step 5: Solve for \( \cos^2(\theta) \)
Rearranging the equation gives:

\[ \cos^2(\theta) = \frac{2}{16} = \frac{1}{8} \]

### Step 6: Find \( \theta \)
Taking the square root gives:

\[ \cos(\theta) = \frac{1}{\sqrt{8}} = \frac{1}{2\sqrt{2}} \]

Now, we can find \( \theta \):

\[ \theta = \cos^{-1}\left(\frac{1}{2\sqrt{2}}\right) \]

To find the angle in degrees, we can calculate:

\[ \theta \approx 63.4^\circ \]

### Final Answer
The angle of transmission axis between the first two polarisers is approximately \( 63.4^\circ \).

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