A light has amplitude A and angle between analyser and polariser is `60^@`. Light is reflected by analyser has amplitude

Options

A. (a) `Asqrt2`
B. (b) `A//sqrt2`
C. (c) `sqrt3A//2`
D. (d) `A//2`

Explanation

To solve the problem of finding the amplitude of light reflected by the analyzer when the angle between the analyzer and the polarizer is \(60^\circ\), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have a light wave with an initial amplitude \(A\).
- The light passes through a polarizer and then through an analyzer, with an angle of \(60^\circ\) between the two.

2. **Apply Malus's Law**:
- According to Malus's Law, when polarized light passes through a polarizer or analyzer, the amplitude of the transmitted light is given by:
\[ A' = A \cos(\theta) \]
- Here, \(A'\) is the amplitude of the light after passing through the analyzer, \(A\) is the initial amplitude, and \(\theta\) is the angle between the light's polarization direction and the axis of the analyzer.

3. **Substitute the Values**:
- In this case, the angle \(\theta\) is \(60^\circ\).
- Therefore, we can write:
\[ A' = A \cos(60^\circ) \]

4. **Calculate \(\cos(60^\circ)\)**:
- We know that:
\[ \cos(60^\circ) = \frac{1}{2} \]
- Substituting this value into the equation gives:
\[ A' = A \cdot \frac{1}{2} \]

5. **Final Result**:
- Thus, the amplitude of the light reflected by the analyzer is:
\[ A' = \frac{A}{2} \]

### Conclusion:
The amplitude of the light reflected by the analyzer is \(\frac{A}{2}\).