Light waves producing interference have their amplitudes in the ratio `3:2`. The intensity ratio of maximum and minimum interference fringes is

Options

A. `36:1`
B. `9:4`
C. `25:1`
D. `6:4`

Explanation

To solve the problem of finding the intensity ratio of maximum and minimum interference fringes given the amplitude ratio of light waves, we can follow these steps:

### Step-by-Step Solution

1. **Understand the Relationship Between Amplitude and Intensity**:
The intensity (I) of a wave is proportional to the square of its amplitude (A). Therefore, if the amplitudes of two waves are in the ratio \( A_1 : A_2 = 3 : 2 \), the intensities will be in the ratio:
\[ I_1 : I_2 = A_1^2 : A_2^2 \]

2. **Calculate the Intensities**:
Let \( A_1 = 3k \) and \( A_2 = 2k \) for some constant \( k \). Then:
\[ I_1 = (3k)^2 = 9k^2 \]
\[ I_2 = (2k)^2 = 4k^2 \]

3. **Determine Maximum and Minimum Intensity**:
The maximum intensity \( I_{max} \) and minimum intensity \( I_{min} \) in interference are given by:
\[ I_{max} = I_1 + I_2 + 2\sqrt{I_1 I_2} \]
\[ I_{min} = I_1 + I_2 - 2\sqrt{I_1 I_2} \]

4. **Calculate \( I_{max} \)**:
First, find \( I_1 + I_2 \):
\[ I_1 + I_2 = 9k^2 + 4k^2 = 13k^2 \]
Now calculate \( \sqrt{I_1 I_2} \):
\[ \sqrt{I_1 I_2} = \sqrt{9k^2 \cdot 4k^2} = \sqrt{36k^4} = 6k^2 \]
Thus,
\[ I_{max} = 13k^2 + 2 \cdot 6k^2 = 13k^2 + 12k^2 = 25k^2 \]

5. **Calculate \( I_{min} \)**:
Now calculate \( I_{min} \):
\[ I_{min} = 13k^2 - 2 \cdot 6k^2 = 13k^2 - 12k^2 = k^2 \]

6. **Find the Intensity Ratio**:
The ratio of maximum to minimum intensity is:
\[ \frac{I_{max}}{I_{min}} = \frac{25k^2}{k^2} = 25 \]

### Final Answer
The intensity ratio of maximum to minimum interference fringes is \( 25 : 1 \).