The angle between polariser and analyser is `30^(@)`. The ratio of intensity of incident light and transmitted by the analyser is

Options

A. `3:4`
B. `4:3`
C. `sqrt(3):2`
D. `2: sqrt(3)`

Answer

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

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

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

Given:
- The angle \( \theta = 30^\circ \).

### Step-by-step Solution:

1. **Write down Malus's Law**:
According to Malus's Law, the transmitted intensity \( I \) can be expressed as:
\[ I = I_0 \cos^2(\theta) \]

2. **Substitute the angle**:
Substitute \( \theta = 30^\circ \) into the equation:
\[ I = I_0 \cos^2(30^\circ) \]

3. **Calculate \( \cos(30^\circ) \)**:
We know that:
\[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \]

4. **Square the cosine value**:
Now, square the cosine value:
\[ \cos^2(30^\circ) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \]

5. **Substitute back into the intensity equation**:
Now substitute this back into the equation for \( I \):
\[ I = I_0 \cdot \frac{3}{4} \]

6. **Find the ratio of intensities**:
The ratio of the intensity of incident light \( I_0 \) to the transmitted light \( I \) is:
\[ \frac{I_0}{I} = \frac{I_0}{I_0 \cdot \frac{3}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3} \]

### Final Answer:
The ratio of the intensity of incident light to the transmitted light by the analyzer is:
\[ \frac{4}{3} \]