If `theta` is the angle between the transmission axes of the polarizer and the analyser and `I_0` be the intensity of the polarized light incident on the analyser, then Intensity of transmitted light through analyser would be

Options

A. `(I_0 costheta)^2`
B. `sqrt2 I_0cos^2theta`
C. `I_0/sqrt2cos^2theta`
D. `I_0cos^2theta`

Answer

To find the intensity of the transmitted light through the analyzer when polarized light passes through it at an angle θ, we can use Malus's Law. Here's the step-by-step solution:

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have polarized light with an initial intensity \( I_0 \).
- The light passes through a polarizer (which is not mentioned in the question but is implied) and then through an analyzer at an angle \( \theta \) relative to the transmission axis of the polarizer.

2. **Apply Malus's Law**:
- Malus's Law states that when polarized light passes through a polarizer (or analyzer), the intensity of the transmitted light \( I \) is given by:
\[ I = I_0 \cos^2(\theta) \]
- Here, \( I_0 \) is the intensity of the incident polarized light, and \( \theta \) is the angle between the light's polarization direction and the transmission axis of the analyzer.

3. **Write the Final Expression**:
- Therefore, the intensity of the transmitted light through the analyzer is:
\[ I = I_0 \cos^2(\theta) \]

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

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