Plane polarised light is incident on an analyser . The intensity then becomes three fourth . The angle of the axis of the analyser with the beam is

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `0^(@)`

Answer

To solve the problem, we need to find the angle of the axis of the analyser with respect to the beam of plane polarized light. We will use Malus's Law, which states that the intensity of polarized light passing through an analyser 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 analyser.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- The transmitted intensity \( I \) is three-fourths of the incident intensity \( I_0 \).
- This can be expressed as:
\[ I = \frac{3}{4} I_0 \]

2. **Apply Malus's Law:**
- According to Malus's Law:
\[ I = I_0 \cos^2 \theta \]
- Substitute the expression for \( I \):
\[ \frac{3}{4} I_0 = I_0 \cos^2 \theta \]

3. **Simplify the Equation:**
- Since \( I_0 \) is common on both sides, we can divide both sides by \( I_0 \) (assuming \( I_0 \neq 0 \)):
\[ \frac{3}{4} = \cos^2 \theta \]

4. **Solve for \( \cos \theta \):**
- Take the square root of both sides:
\[ \cos \theta = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

5. **Find the Angle \( \theta \):**
- The angle whose cosine is \( \frac{\sqrt{3}}{2} \) is:
\[ \theta = 30^\circ \]

### Final Answer:
The angle of the axis of the analyser with the beam is \( 30^\circ \).

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