If the amplitude ratio of two sources producing interference is 3 : 5, the ratio of intensities at maxima and minima is

Options

A. `25 : 16`
B. `5:3`
C. `16:1`
D. `25:9`

Answer

To solve the problem of finding the ratio of intensities at maxima and minima given the amplitude ratio of two sources producing interference is 3:5, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Amplitudes**:
Let the amplitudes of the two sources be \( A_1 \) and \( A_2 \).
Given the ratio of amplitudes:
\[ \frac{A_1}{A_2} = \frac{3}{5} \]
We can express \( A_1 \) in terms of \( A_2 \):
\[ A_1 = \frac{3}{5} A_2 \]

2. **Calculate the Amplitude at Maxima**:
The amplitude at the point of maximum intensity (constructive interference) is given by:
\[ A_{max} = A_1 + A_2 \]
Substituting \( A_1 \):
\[ A_{max} = \frac{3}{5} A_2 + A_2 = \frac{3}{5} A_2 + \frac{5}{5} A_2 = \frac{8}{5} A_2 \]

3. **Calculate the Amplitude at Minima**:
The amplitude at the point of minimum intensity (destructive interference) is given by:
\[ A_{min} = |A_1 - A_2| \]
Substituting \( A_1 \):
\[ A_{min} = \left| \frac{3}{5} A_2 - A_2 \right| = \left| \frac{3}{5} A_2 - \frac{5}{5} A_2 \right| = \left| -\frac{2}{5} A_2 \right| = \frac{2}{5} A_2 \]

4. **Determine the Intensities**:
The intensity \( I \) is proportional to the square of the amplitude:
\[ I \propto A^2 \]
Thus, the maximum intensity \( I_{max} \) and minimum intensity \( I_{min} \) can be expressed as:
\[ I_{max} \propto (A_{max})^2 = \left(\frac{8}{5} A_2\right)^2 = \frac{64}{25} A_2^2 \]
\[ I_{min} \propto (A_{min})^2 = \left(\frac{2}{5} A_2\right)^2 = \frac{4}{25} A_2^2 \]

5. **Calculate the Ratio of Intensities**:
Now, we can find the ratio of maximum intensity to minimum intensity:
\[ \frac{I_{max}}{I_{min}} = \frac{\frac{64}{25} A_2^2}{\frac{4}{25} A_2^2} = \frac{64}{4} = 16 \]

6. **Final Result**:
Therefore, the ratio of intensities at maxima and minima is:
\[ I_{max} : I_{min} = 16 : 1 \]