Two slits in Youngs experiment have widths in the ratio 1: 25.h The ratio of intensity at the macxima and minima in the interderence pattern `I_(max)/(I_(min))is` is

Options

A. `9/4`
B. `121/49`
C. `49/121`
D. `4/9`

Answer

To find the ratio of intensity at the maxima and minima in Young's double-slit experiment when the widths of the two slits are in the ratio of 1:25, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have two slits with widths in the ratio \( W_1 : W_2 = 1 : 25 \). We need to find the ratio of maximum intensity \( I_{max} \) to minimum intensity \( I_{min} \) in the interference pattern.

2. **Relating Intensity to Width**:
The intensity of light from each slit is proportional to the square of the width of the slit. Therefore, we can write:
\[ \frac{I_1}{I_2} = \frac{W_1^2}{W_2^2} \]
Given \( W_1 : W_2 = 1 : 25 \), we can express this as:
\[ \frac{I_1}{I_2} = \frac{1^2}{25^2} = \frac{1}{625} \]

3. **Expressing Intensities**:
Let \( I_1 = 1 \) and \( I_2 = 625 \) (based on the ratio derived).

4. **Calculating Maximum and Minimum Intensities**:
The maximum intensity \( I_{max} \) and minimum intensity \( I_{min} \) can be calculated as follows:
\[ I_{max} = I_1 + I_2 = 1 + 625 = 626 \]
\[ I_{min} = I_1 - I_2 = 1 - 625 = -624 \]
However, since intensity cannot be negative, we take the absolute value:
\[ I_{min} = |I_1 - I_2| = |1 - 625| = 624 \]

5. **Finding the Ratio**:
Now, we find the ratio \( \frac{I_{max}}{I_{min}} \):
\[ \frac{I_{max}}{I_{min}} = \frac{626}{624} \]
Simplifying this gives:
\[ \frac{626}{624} = \frac{313}{312} \]

6. **Final Result**:
The ratio \( \frac{I_{max}}{I_{min}} \) simplifies to \( \frac{313}{312} \), which does not match any of the provided options. Thus, we need to re-evaluate the calculations or the interpretation of the problem.