Four perfect polarising plates are stacked so that the axis of each is turned `30^@` clockwise to the preceding plate, the last plate therefore being crossed with the first. A beam of unpolarised light of intensity 1 passes through the stack perpendicularly. The transmitted beam has intensity

Options

A. `27/128`I
B. `81/256` I
C. `1/8`I
D. `27/64`I

Explanation

To solve the problem of determining the transmitted intensity of a beam of unpolarised light passing through four polarising plates, we will follow these steps:

### Step 1: Initial Setup
We start with unpolarised light of intensity \( I_0 = 1 \). When unpolarised light passes through the first polarising plate, it becomes polarised.

### Step 2: First Polarising Plate
According to Malus's Law, the intensity of polarised light after passing through a polariser is given by:
\[ I_1 = I_0 \cdot \frac{1}{2} \]
This is because the average value of \( \cos^2 \theta \) for unpolarised light is \( \frac{1}{2} \). Therefore,
\[ I_1 = 1 \cdot \frac{1}{2} = \frac{1}{2} \]

### Step 3: Second Polarising Plate
The second plate is oriented at \( 30^\circ \) to the first. Using Malus's Law again:
\[ I_2 = I_1 \cdot \cos^2(30^\circ) \]
Substituting \( I_1 \):
\[ I_2 = \frac{1}{2} \cdot \cos^2(30^\circ) \]
Since \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \):
\[ I_2 = \frac{1}{2} \cdot \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{2} \cdot \frac{3}{4} = \frac{3}{8} \]

### Step 4: Third Polarising Plate
The third plate is also oriented at \( 30^\circ \) to the second plate:
\[ I_3 = I_2 \cdot \cos^2(30^\circ) \]
Substituting \( I_2 \):
\[ I_3 = \frac{3}{8} \cdot \cos^2(30^\circ) = \frac{3}{8} \cdot \frac{3}{4} = \frac{9}{32} \]

### Step 5: Fourth Polarising Plate
The fourth plate is crossed with the first plate, meaning it is oriented at \( 90^\circ \) to the first plate. The angle between the third and fourth plates is \( 60^\circ \):
\[ I_4 = I_3 \cdot \cos^2(60^\circ) \]
Substituting \( I_3 \):
\[ I_4 = \frac{9}{32} \cdot \cos^2(60^\circ) = \frac{9}{32} \cdot \left(\frac{1}{2}\right)^2 = \frac{9}{32} \cdot \frac{1}{4} = \frac{9}{128} \]

### Final Result
Thus, the transmitted intensity \( I_4 \) is:
\[ I_4 = \frac{9}{128} \cdot I_0 \]
Since \( I_0 = 1 \):
\[ I_4 = \frac{9}{128} \]