The ratio of intensity at maxima and minima in the interference pattern is 25:9. What will be the widths of the two slits in Young's interference experiment ?

Options

A. `18:3`
B. `4:1`
C. `8:1`
D. `16:1`

Explanation

To solve the problem, we need to find the widths of the two slits in Young's double slit experiment given the ratio of intensity at maxima and minima is 25:9.

### Step-by-Step Solution:

1. **Understanding the Ratios**:
We are given that the ratio of intensity at maxima (I_max) to intensity at minima (I_min) is 25:9. This can be expressed mathematically as:
\[ \frac{I_{max}}{I_{min}} = \frac{25}{9} \]

2. **Relating Intensity to Amplitude**:
The intensity of light is proportional to the square of the amplitude. If we denote the amplitudes corresponding to the two slits as \( A_1 \) and \( A_2 \), we can express the intensities as:
\[ I_{max} \propto (A_1 + A_2)^2 \]
\[ I_{min} \propto (A_1 - A_2)^2 \]

3. **Setting Up the Equation**:
Using the above relationships, we can write:
\[ \frac{(A_1 + A_2)^2}{(A_1 - A_2)^2} = \frac{25}{9} \]

4. **Taking Square Roots**:
Taking the square root of both sides gives:
\[ \frac{A_1 + A_2}{A_1 - A_2} = \frac{5}{3} \]

5. **Cross-Multiplying**:
Cross-multiplying gives us:
\[ 3(A_1 + A_2) = 5(A_1 - A_2) \]
Expanding this, we have:
\[ 3A_1 + 3A_2 = 5A_1 - 5A_2 \]

6. **Rearranging the Equation**:
Rearranging the terms leads to:
\[ 3A_2 + 5A_2 = 5A_1 - 3A_1 \]
Simplifying gives:
\[ 8A_2 = 2A_1 \]
Thus, we can express the relationship as:
\[ \frac{A_1}{A_2} = \frac{8}{2} = 4 \]

7. **Finding the Widths of the Slits**:
Since the amplitude is proportional to the square root of the width of the slits, we can write:
\[ \frac{A_1}{A_2} = \sqrt{\frac{\omega_1}{\omega_2}} = 4 \]
Squaring both sides gives:
\[ \frac{\omega_1}{\omega_2} = 16 \]

8. **Final Ratio of Widths**:
Therefore, the ratio of the widths of the two slits is:
\[ \omega_1 : \omega_2 = 16 : 1 \]

### Conclusion:
The widths of the two slits in Young's interference experiment are in the ratio of 16:1.