When a plane polarised light is passed through an analyser and analyser is rotated through `90^(@)`, the intensity of the emerging light

Options

A. `I_(0) `
B. `2I_(0)`
C. `I_(0)//4`
D. Zero

Answer

**Step-by-Step Solution:**

1. **Understanding the Problem:**
- We are dealing with plane polarized light passing through an analyzer. The intensity of light changes as the analyzer is rotated.

2. **Initial Intensity:**
- Let the initial intensity of the polarized light be \( I_0 \).

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

4. **Analyzing the Rotation:**
- Initially, let’s assume the analyzer is aligned with the plane of polarization of the incoming light (i.e., \( \theta = 0^\circ \)). Thus, the intensity after passing through the analyzer is:
\[ I = I_0 \cos^2(0^\circ) = I_0 \]

5. **After Rotating the Analyzer:**
- When the analyzer is rotated through \( 90^\circ \), the angle \( \theta \) becomes \( 90^\circ \):
\[ I = I_0 \cos^2(90^\circ) = I_0 \cdot 0 = 0 \]
- Therefore, the intensity of the emerging light after the analyzer is rotated through \( 90^\circ \) is \( 0 \).

6. **Conclusion:**
- The intensity of the light emerging from the analyzer after it has been rotated through \( 90^\circ \) is \( 0 \).

**Final Answer:**
The intensity of the emerging light after the analyzer is rotated through \( 90^\circ \) is \( 0 \).

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