The ratio of intensities at amxima and minima is `25 : 16`. What will be ratio of the widths of two slits in YDSE ?

Explanation

To solve the problem, we need to find the ratio of the widths of two slits in Young's Double Slit Experiment (YDSE) based on the given ratio of intensities at maxima and minima.

### Step-by-Step Solution:

1. **Understand the Relationship Between Intensity and Amplitude**:
The intensity (I) of light is proportional to the square of the amplitude (A) of the wave. Therefore, if we have two slits with amplitudes A1 and A2, the intensity ratio can be expressed as:
\[ \frac{I_1}{I_2} = \left(\frac{A_1}{A_2}\right)^2 \]

2. **Given Ratio of Intensities**:
We are given that the ratio of intensities at maxima and minima is:
\[ \frac{I_1}{I_2} = \frac{25}{16} \]

3. **Relate Amplitudes to Intensities**:
From the intensity ratio, we can express the amplitude ratio:
\[ \frac{A_1}{A_2} = \sqrt{\frac{I_1}{I_2}} = \sqrt{\frac{25}{16}} = \frac{5}{4} \]

4. **Express Amplitudes in Terms of Slit Widths**:
In YDSE, the amplitude of light from each slit is proportional to the width of the slit. If we let W1 and W2 be the widths of the two slits, we can write:
\[ \frac{A_1}{A_2} = \frac{W_1}{W_2} \]

5. **Set Up the Equation**:
From the amplitude ratio we found:
\[ \frac{W_1}{W_2} = \frac{5}{4} \]

6. **Find the Ratio of Widths**:
Rearranging gives us the ratio of the widths of the two slits:
\[ W_1 : W_2 = 5 : 4 \]

### Final Answer:
The ratio of the widths of the two slits is \( 5 : 4 \).