The amplitude of polarised light transmitted through a polariser is A. The amplitude of unpolarised light incident on it is

Options

A. `A//2`
B. `A//sqrt(2)`
C. `2A`
D. `sqrt(2)A`

Explanation

To find the amplitude of the unpolarized light incident on a polarizer that transmits polarized light with an amplitude \( A \), we can follow these steps:

### Step-by-Step Solution

1. **Understanding the Relationship Between Intensity and Amplitude**:
- The intensity \( I \) of a wave is proportional to the square of its amplitude \( A \). This can be expressed mathematically as:
\[ I \propto A^2 \]

2. **Intensity of Unpolarized Light**:
- When unpolarized light passes through a polarizer, only half of its intensity is transmitted. If the intensity of the unpolarized light is \( I_0 \), the transmitted intensity \( I_t \) becomes:
\[ I_t = \frac{I_0}{2} \]

3. **Relating Intensities to Amplitudes**:
- Let \( A_0 \) be the amplitude of the unpolarized light incident on the polarizer, and \( A \) be the amplitude of the transmitted polarized light. According to the relationship between intensity and amplitude:
\[ I_0 \propto A_0^2 \quad \text{and} \quad I_t \propto A^2 \]

4. **Setting Up the Equation**:
- From the previous steps, we can write:
\[ I_t = \frac{I_0}{2} \implies A^2 = \frac{A_0^2}{2} \]

5. **Substituting the Intensities**:
- Since \( I_t = A^2 \) and \( I_0 = A_0^2 \), we can substitute:
\[ A^2 = \frac{A_0^2}{2} \]

6. **Solving for the Amplitude of the Incident Light**:
- Rearranging gives:
\[ A_0^2 = 2A^2 \]
- Taking the square root of both sides, we find:
\[ A_0 = \sqrt{2} A \]

### Final Answer
The amplitude of the unpolarized light incident on the polarizer is:
\[ A_0 = \sqrt{2} A \]

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