The axes of the polariser and analyser are inclined to each other at `60^(@)`. If the amplitude of the polarised light emergent through analyser is A. The amplitude of unpolarised light incident on polariser is

Options

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

Answer

To solve the problem step by step, we will use Malus's Law and the relationship between intensity and amplitude.

### Step 1: Understand the relationship between intensity and amplitude
The intensity \( I \) of light is related to the amplitude \( A \) by the formula:
\[ I \propto A^2 \]
This means that if we know the intensity, we can find the amplitude by taking the square root.

### Step 2: Apply Malus's Law
According to Malus's Law, the intensity of light emerging from a polariser when unpolarised light is incident on it is given by:
\[ I = I_0 \cos^2(\theta) \]
where \( I_0 \) is the intensity of the incident light, and \( \theta \) is the angle between the axes of the polariser and analyser.

### Step 3: Plug in the values
In this case, the angle \( \theta \) is \( 60^\circ \). Therefore, we have:
\[ I = I_0 \cos^2(60^\circ) \]
Knowing that \( \cos(60^\circ) = \frac{1}{2} \), we can substitute this value:
\[ I = I_0 \left(\frac{1}{2}\right)^2 = I_0 \cdot \frac{1}{4} \]

### Step 4: Relate intensity to amplitude
From the relationship between intensity and amplitude, we can express the intensity in terms of amplitude:
\[ I = A^2 \]
Let \( A_0 \) be the amplitude of the incident unpolarised light. Thus, we have:
\[ I_0 = A_0^2 \]
And for the emergent light:
\[ I = A^2 \]

### Step 5: Substitute the intensity expressions
From the previous steps, we can write:
\[ A^2 = \frac{A_0^2}{4} \]

### Step 6: Take the square root
Taking the square root of both sides gives:
\[ A = \frac{A_0}{2} \]

### Step 7: Solve for the incident amplitude
Now, rearranging the equation to find \( A_0 \):
\[ A_0 = 2A \]

### Final Answer
Thus, the amplitude of the unpolarised light incident on the polariser is:
\[ \boxed{2A} \]