Two polrizers are oriented with transmission planes making an angle of `60^(@)`. The percentage of incident unpolarized light that passes through the system is

Options

A. 0.5
B. 1
C. 0.125
D. 0.375

Answer

To solve the problem of finding the percentage of incident unpolarized light that passes through two polarizers oriented at an angle of \(60^\circ\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Intensity of Light:**
Let the intensity of the incident unpolarized light be \(I_0\).

2. **First Polarizer:**
When unpolarized light passes through the first polarizer, the intensity of the transmitted light \(I_1\) is given by:
\[ I_1 = \frac{I_0}{2} \]
This is because unpolarized light passing through a polarizer has its intensity halved.

3. **Second Polarizer:**
The second polarizer is oriented at an angle of \(60^\circ\) with respect to the first polarizer. According to Malus's Law, the intensity of light transmitted through the second polarizer \(I_2\) is given by:
\[ I_2 = I_1 \cos^2(\theta) \]
where \(\theta\) is the angle between the transmission axes of the two polarizers. Here, \(\theta = 60^\circ\).

4. **Substituting Values:**
Substitute \(I_1\) into the equation for \(I_2\):
\[ I_2 = \left(\frac{I_0}{2}\right) \cos^2(60^\circ) \]
We know that \(\cos(60^\circ) = \frac{1}{2}\), so:
\[ I_2 = \left(\frac{I_0}{2}\right) \left(\frac{1}{2}\right)^2 \]
Simplifying this gives:
\[ I_2 = \frac{I_0}{2} \cdot \frac{1}{4} = \frac{I_0}{8} \]

5. **Calculating the Percentage:**
To find the percentage of the incident light that passes through both polarizers, we calculate:
\[ \text{Percentage} = \left(\frac{I_2}{I_0}\right) \times 100\% \]
Substituting \(I_2 = \frac{I_0}{8}\):
\[ \text{Percentage} = \left(\frac{\frac{I_0}{8}}{I_0}\right) \times 100\% = \frac{1}{8} \times 100\% = 12.5\% \]

### Final Answer:
The percentage of incident unpolarized light that passes through the system is **12.5%**.

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