In Young's double slit experiment, if the slit widths are in the ratio `1:9`, then the ratio of the intensity at minima to that at maxima will be

Options

A. (a)1
B. (b)`1/9`
C. (c)`1/4`
D. (d)`1/3`

Answer

To solve the problem of finding the ratio of the intensity at minima to that at maxima in Young's double slit experiment with slit widths in the ratio of 1:9, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Slit Widths**:
Let the widths of the two slits be \( w_1 \) and \( w_2 \) such that \( w_1:w_2 = 1:9 \). We can assign \( w_1 = 1 \) and \( w_2 = 9 \).

2. **Relating Width to Intensity**:
In Young's double slit experiment, the intensity at each slit is proportional to the square of the amplitude of the wave emerging from that slit. The amplitude is proportional to the square root of the slit width. Therefore, we have:
\[ I_1 \propto w_1 \quad \text{and} \quad I_2 \propto w_2 \]
Thus, we can express the intensities as:
\[ I_1 = k \cdot w_1 \quad \text{and} \quad I_2 = k \cdot w_2 \]
where \( k \) is a proportionality constant.

3. **Calculating Intensities**:
Substituting the values of \( w_1 \) and \( w_2 \):
\[ I_1 = k \cdot 1 = k \quad \text{and} \quad I_2 = k \cdot 9 = 9k \]

4. **Finding Amplitudes**:
The amplitudes \( A_1 \) and \( A_2 \) corresponding to the intensities can be found as:
\[ A_1 = \sqrt{I_1} = \sqrt{k} \quad \text{and} \quad A_2 = \sqrt{I_2} = \sqrt{9k} = 3\sqrt{k} \]

5. **Calculating Maximum Intensity**:
The maximum intensity \( I_{\text{max}} \) occurs when the waves interfere constructively:
\[ I_{\text{max}} = (A_1 + A_2)^2 = (\sqrt{k} + 3\sqrt{k})^2 = (4\sqrt{k})^2 = 16k \]

6. **Calculating Minimum Intensity**:
The minimum intensity \( I_{\text{min}} \) occurs when the waves interfere destructively:
\[ I_{\text{min}} = (A_1 - A_2)^2 = (\sqrt{k} - 3\sqrt{k})^2 = (-2\sqrt{k})^2 = 4k \]

7. **Finding the Ratio**:
Now, we can find the ratio of the intensity at minima to that at maxima:
\[ \frac{I_{\text{min}}}{I_{\text{max}}} = \frac{4k}{16k} = \frac{4}{16} = \frac{1}{4} \]

### Final Answer:
The ratio of the intensity at minima to that at maxima is \( \frac{1}{4} \).