If the width ratio of the two slits in Young's double slit experiment is 4:1, then the ratio of intensity at the maxima and minima in the interference patternn will be

Options

A. `1:4`
B. `3:1`
C. `9:1`
D. `16:1`

Explanation

To solve the problem of finding the ratio of intensity at the maxima and minima in Young's double slit experiment when the width ratio of the two slits is 4:1, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Width Ratio**:
- Let the widths of the two slits be \( W_1 \) and \( W_2 \).
- Given that \( W_1 : W_2 = 4 : 1 \), we can express this as:
\[ W_1 = 4k \quad \text{and} \quad W_2 = k \]
- Here, \( k \) is a constant.

2. **Relating Intensity to Width**:
- The intensity \( I \) at a point in the interference pattern is directly proportional to the square of the amplitude \( A \) of the wave, and the amplitude is proportional to the width of the slit.
- Therefore, we can write:
\[ \frac{I_1}{I_2} = \frac{W_1^2}{W_2^2} \]

3. **Calculating the Intensity Ratio**:
- Substituting the values of \( W_1 \) and \( W_2 \):
\[ \frac{I_1}{I_2} = \frac{(4k)^2}{(k)^2} = \frac{16k^2}{k^2} = 16 \]
- This means that the intensity ratio of the two slits is \( I_1 : I_2 = 16 : 1 \).

4. **Finding Amplitudes**:
- Let \( A_1 \) and \( A_2 \) be the amplitudes corresponding to \( I_1 \) and \( I_2 \).
- Since intensity is proportional to the square of the amplitude:
\[ \frac{A_1^2}{A_2^2} = 16 \implies \frac{A_1}{A_2} = 4 \]
- Thus, we can express the amplitudes as:
\[ A_1 = 4A \quad \text{and} \quad A_2 = A \]

5. **Calculating Maximum and Minimum Intensities**:
- The maximum intensity \( I_{max} \) is given by:
\[ I_{max} = (A_1 + A_2)^2 = (4A + A)^2 = (5A)^2 = 25A^2 \]
- The minimum intensity \( I_{min} \) is given by:
\[ I_{min} = (A_1 - A_2)^2 = (4A - A)^2 = (3A)^2 = 9A^2 \]

6. **Finding the Ratio of Maximum to Minimum Intensity**:
- Now we can find the ratio:
\[ \frac{I_{max}}{I_{min}} = \frac{25A^2}{9A^2} = \frac{25}{9} \]

### Final Answer:
The ratio of intensity at the maxima and minima in the interference pattern is:
\[ \frac{I_{max}}{I_{min}} = \frac{25}{9} \]