A beam of natural light falls on a system of 6 polaroids, which are arranged in succession such that each polaroid is turned through `30^@` with respect to the preceding one. The percentage of incident intensity that passes through the system will be

Options

A. (a) 100%
B. (b) 50%
C. (c) 30%
D. (d) 12%

Explanation

To solve the problem of determining the percentage of incident intensity that passes through a system of 6 polaroids arranged at 30 degrees to each other, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Malus's Law**:
Malus's Law states that when unpolarized light passes through a polarizer, the intensity of the transmitted light is given by:
\[ I = I_0 \cdot \frac{1}{2} \cos^2(\theta) \]
where \(I_0\) is the intensity of the incident light, \(I\) is the intensity after passing through the polarizer, and \(\theta\) is the angle between the light's polarization direction and the axis of the polarizer.

2. **Setting Up the Problem**:
We have 6 polaroids, each rotated by \(30^\circ\) with respect to the previous one. Therefore, the angle \(\theta\) for each pair of successive polaroids is \(30^\circ\).

3. **Applying Malus's Law Successively**:
Since there are 6 polaroids, we will apply Malus's Law 5 times (from the first to the sixth polaroid). The intensity after passing through all 6 polaroids can be expressed as:
\[ I = I_0 \cdot \left(\frac{1}{2} \cos^2(30^\circ)\right)^5 \]

4. **Calculating \(\cos(30^\circ)\)**:
The cosine of \(30^\circ\) is:
\[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \]
Therefore:
\[ \cos^2(30^\circ) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \]

5. **Substituting Values**:
Now substituting \(\cos^2(30^\circ)\) into the intensity equation:
\[ I = I_0 \cdot \left(\frac{1}{2} \cdot \frac{3}{4}\right)^5 \]

6. **Simplifying the Expression**:
Simplifying the expression:
\[ I = I_0 \cdot \left(\frac{3}{8}\right)^5 \]

7. **Calculating \(\left(\frac{3}{8}\right)^5\)**:
Calculate \(\left(\frac{3}{8}\right)^5\):
\[ \left(\frac{3}{8}\right)^5 = \frac{3^5}{8^5} = \frac{243}{32768} \]

8. **Finding the Ratio \(\frac{I}{I_0}\)**:
Therefore, the ratio of transmitted intensity to incident intensity is:
\[ \frac{I}{I_0} = \frac{1}{2} \cdot \left(\frac{3}{4}\right)^5 = \frac{1}{2} \cdot \frac{243}{1024} = \frac{243}{2048} \]

9. **Calculating the Percentage**:
To find the percentage of the incident intensity that passes through the system:
\[ \frac{I}{I_0} \times 100 = \frac{243}{2048} \times 100 \approx 11.89\% \]
Rounding this gives approximately \(12\%\).

### Final Answer:
The percentage of incident intensity that passes through the system is **12%**.