Two slits in Young's experiment have widths in the ratio `1:25`. The ratio of intensity at the maxima and minima in the interference pattern `I_(max)/I_(min)` is

Options

A. (a) `4/9`
B. (b) `9/4`
C. (c) `(121)/(49)`
D. (d) `(49)/(121)`

Explanation

To solve the problem of finding the ratio of intensity at the maxima and minima in Young's double-slit experiment given the widths of the slits in the ratio of 1:25, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Widths of the Slits**:
Let the widths of the two slits be \( W_1 \) and \( W_2 \) such that \( W_1 : W_2 = 1 : 25 \). We can express this as:
\[ W_1 = x \quad \text{and} \quad W_2 = 25x \]

2. **Relate Intensity to Width**:
The intensity of light from each slit is proportional to the square of the amplitude of the light wave, which in turn is proportional to the width of the slit. Therefore, we have:
\[ \frac{I_1}{I_2} = \frac{W_1^2}{W_2^2} \]
Substituting the values of \( W_1 \) and \( W_2 \):
\[ \frac{I_1}{I_2} = \frac{x^2}{(25x)^2} = \frac{x^2}{625x^2} = \frac{1}{625} \]

3. **Express Intensities**:
Let \( I_1 = I \) and \( I_2 = 625I \). Thus, we have:
\[ I_1 = I \quad \text{and} \quad I_2 = 625I \]

4. **Calculate Intensity at Maxima**:
The intensity at maxima \( I_{max} \) is given by:
\[ I_{max} = I_1 + I_2 = I + 625I = 626I \]

5. **Calculate Intensity at Minima**:
The intensity at minima \( I_{min} \) is given by:
\[ I_{min} = |I_1 - I_2| = |I - 625I| = | -624I | = 624I \]

6. **Find the Ratio of Intensities**:
Now, we can find the ratio of the intensities at maxima and minima:
\[ \frac{I_{max}}{I_{min}} = \frac{626I}{624I} = \frac{626}{624} = \frac{313}{312} \]

### Final Answer:
The ratio of intensity at the maxima to the intensity at the minima is:
\[ \frac{I_{max}}{I_{min}} = \frac{313}{312} \]