The polariser and analyser are inclined to each other at `60^(@)`. If I/2 is the intensity of the polarised light emergent from analyser. Then the intensity of the unpolarised light incident on the polariser is

Options

A. 8I
B. 4I
C. 2I
D. I

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the scenario
We have a polarizer and an analyzer inclined at an angle of \(60^\circ\). The intensity of the polarized light emerging from the analyzer is given as \(I/2\).

### Step 2: Determine the intensity after passing through the polarizer
When unpolarized light with intensity \(I_0\) passes through a polarizer, the intensity of the light that emerges from the polarizer is given by:
\[ I_p = \frac{I_0}{2} \]
This is because a polarizer transmits half of the intensity of unpolarized light.

### Step 3: Apply Malus's Law
After passing through the polarizer, the polarized light then passes through the analyzer. According to Malus's Law, the intensity of light \(I_a\) after passing through the analyzer is given by:
\[ I_a = I_p \cdot \cos^2(\theta) \]
where \(\theta\) is the angle between the light's polarization direction and the analyzer's axis. Here, \(\theta = 60^\circ\).

### Step 4: Substitute the known values
From Step 2, we know:
\[ I_p = \frac{I_0}{2} \]
Substituting this into Malus's Law gives:
\[ I_a = \left(\frac{I_0}{2}\right) \cdot \cos^2(60^\circ) \]

### Step 5: Calculate \(\cos^2(60^\circ)\)
We know that:
\[ \cos(60^\circ) = \frac{1}{2} \]
Thus:
\[ \cos^2(60^\circ) = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

### Step 6: Substitute \(\cos^2(60^\circ)\) into the equation
Now substituting this value back into the equation for \(I_a\):
\[ I_a = \left(\frac{I_0}{2}\right) \cdot \frac{1}{4} = \frac{I_0}{8} \]

### Step 7: Set the equation equal to the given intensity
We know from the problem that:
\[ I_a = \frac{I}{2} \]
Setting this equal to our expression for \(I_a\):
\[ \frac{I_0}{8} = \frac{I}{2} \]

### Step 8: Solve for \(I_0\)
To find \(I_0\), we multiply both sides by 8:
\[ I_0 = 8 \cdot \frac{I}{2} = 4I \]

### Conclusion
Thus, the intensity of the unpolarized light incident on the polarizer is:
\[ I_0 = 4I \]