An analyser is inclined to a polariser at an angle of `30^(@)`. The intensity of light emerging from the analyser is `1"/"n`th of that is incident on the polariser. Then n is equal to

Options

A. `4`
B. `4"/"3`
C. `8"/"3`
D. `1"/"4`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have a polarizer and an analyzer inclined at an angle of \(30^\circ\). The intensity of light emerging from the analyzer is \( \frac{1}{n} \) times the intensity of the light incident on the polarizer. We need to find the value of \( n \).

### Step 2: Define the Variables
Let:
- \( I_0 \) = Intensity of the incident light on the polarizer.
- \( I_a \) = Intensity of light emerging from the analyzer.
- The angle \( \theta = 30^\circ \).

### Step 3: Apply Malus's Law
According to Malus's Law, the intensity of light after passing through a polarizer is given by:
\[ I' = I \cos^2(\theta) \]
For the light passing through the polarizer, we have:
\[ I' = I_0 \cos^2(30^\circ) \]

### Step 4: Calculate \( \cos(30^\circ) \)
We know that:
\[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \]
Thus,
\[ \cos^2(30^\circ) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \]

### Step 5: Substitute into Malus's Law
Now substituting this back into the equation for \( I' \):
\[ I' = I_0 \cdot \frac{3}{4} \]

### Step 6: Relate \( I_a \) and \( I' \)
The intensity of light emerging from the analyzer is given as:
\[ I_a = \frac{I_0}{n} \]
From the previous step, we also have:
\[ I_a = I' = I_0 \cdot \frac{3}{4} \]

### Step 7: Set the Two Expressions Equal
Now we can set the two expressions for \( I_a \) equal to each other:
\[ \frac{I_0}{n} = I_0 \cdot \frac{3}{4} \]

### Step 8: Solve for \( n \)
We can cancel \( I_0 \) from both sides (assuming \( I_0 \neq 0 \)):
\[ \frac{1}{n} = \frac{3}{4} \]
Taking the reciprocal gives:
\[ n = \frac{4}{3} \]

### Conclusion
Thus, the value of \( n \) is \( \frac{4}{3} \).

### Final Answer
The correct answer is \( n = \frac{4}{3} \).

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