If `theta` is the angle between the transmission axes of the polarizer and the analyser and `l_0` be the intensity of the polarized light incident on the analyser, then Intensity of transmitted light through analyser would be

Options

A. `(l_0 costheta)^2`
B. `sqrt2 l_0cos^2theta`
C. `l_0/sqrt2cos^2theta`
D. `l_0cos^2theta`

Answer

To solve the problem, we need to determine the intensity of light transmitted through an analyzer when polarized light is incident on it. The intensity of the polarized light incident on the analyzer is given as \( I_0 \), and the angle between the transmission axes of the polarizer and the analyzer is \( \theta \).

### Step-by-Step Solution:

1. **Understanding Polarization**: When unpolarized light passes through a polarizer, it becomes polarized. The intensity of the polarized light after passing through the polarizer is given by Malus's Law.

2. **Incident Intensity**: Let \( I_0 \) be the intensity of the polarized light incident on the analyzer.

3. **Analyzing the Angle**: The angle \( \theta \) is the angle between the transmission axis of the polarizer and the transmission axis of the analyzer.

4. **Components of Electric Field**: The electric field of the incident light can be represented as \( E_0 \). The component of the electric field that is parallel to the transmission axis of the analyzer is \( E_0 \cos \theta \). The component perpendicular to the transmission axis is \( E_0 \sin \theta \).

5. **Intensity and Amplitude Relationship**: The intensity of light is proportional to the square of the amplitude of the electric field. Therefore, the intensity transmitted through the analyzer can be expressed as:
\[ I = k (E_0 \cos \theta)^2 \]
where \( k \) is a proportionality constant.

6. **Intensity Relation**: Since the intensity of the incident light \( I_0 \) is proportional to \( E_0^2 \), we can write:
\[ I_0 = k E_0^2 \]
Thus, substituting \( E_0 \) in terms of \( I_0 \):
\[ I = k (E_0 \cos \theta)^2 = k E_0^2 \cos^2 \theta = I_0 \cos^2 \theta \]

7. **Final Expression**: Therefore, the intensity of the transmitted light through the analyzer is given by:
\[ I = I_0 \cos^2 \theta \]

### Conclusion:
The intensity of the transmitted light through the analyzer is \( I = I_0 \cos^2 \theta \).