The axes of the polariser and analyser are inclined to each other at `45^(@)` . If the amplitude of the unpolarised light incident on the polariser is A , the amplitude of the light transmited through the analyser is

Options

A. `A/2`
B. `A/(sqrt(2))`
C. `(sqrt(3))/2` A
D. `(3A)/4`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We are given that the axes of the polariser and analyser are inclined at an angle of \(45^\circ\). The amplitude of the unpolarised light incident on the polariser is \(A\). We need to find the amplitude of the light transmitted through the analyser.

### Step 2: Use Malus's Law
Malus's Law states that when unpolarised light passes through a polariser, the amplitude of the transmitted light is given by:
\[ A' = A \cos(\theta) \]
where:
- \(A'\) is the amplitude of the transmitted light,
- \(A\) is the amplitude of the incident light,
- \(\theta\) is the angle between the light's initial polarisation direction and the axis of the polariser.

### Step 3: Calculate Amplitude After Polariser
In our case, the light is unpolarised initially, so when it passes through the polariser, the amplitude of the light transmitted through the polariser is:
\[ A_1 = A \cos(0^\circ) = A \]
This is because unpolarised light can be considered as having equal components in all directions, and the polariser will allow through the component aligned with its axis.

### Step 4: Calculate Amplitude After Analyser
Now, the light that comes out of the polariser is polarised along the axis of the polariser. When this light passes through the analyser, which is at an angle of \(45^\circ\) to the polariser, we again apply Malus's Law:
\[ A_2 = A_1 \cos(45^\circ) \]
Substituting \(A_1\) from the previous step:
\[ A_2 = A \cos(45^\circ) \]

### Step 5: Substitute the Value of \(\cos(45^\circ)\)
We know that:
\[ \cos(45^\circ) = \frac{1}{\sqrt{2}} \]
So, substituting this value, we get:
\[ A_2 = A \cdot \frac{1}{\sqrt{2}} = \frac{A}{\sqrt{2}} \]

### Final Answer
The amplitude of the light transmitted through the analyser is:
\[ \frac{A}{\sqrt{2}} \]

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