Unpolarised light of intensity 32 `W//m^(2)` passes through a polariser and analyser which are at an angle of `30^(@)` with respect to each other. The intensity of the light coming from analyser is

Options

A. `16sqrt(3)W//m^(2)`
B. `12W//m^(2)`
C. `16W//m^(2)`
D. `14W//m^(2)`

Explanation

To solve the problem of finding the intensity of light coming from the analyzer when unpolarized light passes through a polarizer and analyzer at an angle of 30 degrees, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Intensity of Unpolarized Light:**
The intensity of the unpolarized light is given as \( I_0 = 32 \, \text{W/m}^2 \).

2. **Calculate the Intensity After the Polarizer:**
When unpolarized light passes through a polarizer, the intensity of the transmitted light is reduced to half of the incident intensity. Therefore, the intensity after the polarizer \( I_1 \) is:
\[ I_1 = \frac{I_0}{2} = \frac{32 \, \text{W/m}^2}{2} = 16 \, \text{W/m}^2 \]

3. **Apply Malus's Law for the Analyzer:**
The intensity of light transmitted through the analyzer is given by Malus's Law, which states:
\[ I = I_1 \cos^2(\theta) \]
where \( \theta \) is the angle between the light's polarization direction (after the polarizer) and the axis of the analyzer. Here, \( \theta = 30^\circ \).

4. **Calculate the Cosine Value:**
Calculate \( \cos(30^\circ) \):
\[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \]

5. **Substitute into Malus's Law:**
Now substitute \( I_1 \) and \( \cos(30^\circ) \) into Malus's Law:
\[ I = I_1 \cos^2(30^\circ) = 16 \, \text{W/m}^2 \left( \frac{\sqrt{3}}{2} \right)^2 \]
\[ I = 16 \, \text{W/m}^2 \cdot \frac{3}{4} = 16 \cdot 0.75 = 12 \, \text{W/m}^2 \]

6. **Final Result:**
The intensity of the light coming from the analyzer is:
\[ I = 12 \, \text{W/m}^2 \]

### Summary:
The intensity of the light coming from the analyzer is \( 12 \, \text{W/m}^2 \).

---