Two monochromatic (wavelength = a/5) and coherent sources of electromagnetic waves are placed on the x-axis at the points (2a,0) and (-a,0). A detector moves in a circle of radius R(gtgt2a) whose centre is at the origin. The number of maxima detected during one circular revolution by the detector are

Options

A. 60
B. 15
C. 64
D. None of these

Explanation

To solve the problem, we need to determine the number of maxima detected by a detector moving in a circular path around two coherent sources of electromagnetic waves. The sources are located at (2a, 0) and (-a, 0) on the x-axis, and the wavelength of the waves is given as λ = a/5.

### Step-by-Step Solution:

1. **Identify the positions of the sources**:
- The first source (S1) is at (2a, 0).
- The second source (S2) is at (-a, 0).

2. **Calculate the distance between the two sources**:
\[ \Delta x = 2a - (-a) = 2a + a = 3a \]

3. **Determine the wavelength**:
- Given that the wavelength \( \lambda = \frac{a}{5} \).

4. **Calculate the number of wavelengths in the distance between the sources**:
\[ \text{Number of wavelengths} = \frac{\Delta x}{\lambda} = \frac{3a}{\frac{a}{5}} = 3a \cdot \frac{5}{a} = 15 \]
This means that the distance between the two sources corresponds to 15 wavelengths.

5. **Understanding the maxima formation**:
- Maxima occur when the path difference between the two waves from the sources to the detector is an integer multiple of the wavelength.
- The path difference can range from 0 to \( 15\lambda \).

6. **Counting the maxima**:
- The path difference can take values: \( 0, \lambda, 2\lambda, \ldots, 15\lambda \).
- This gives us a total of \( 15 + 1 = 16 \) possible values (including 0).

7. **Considering the circular motion of the detector**:
- As the detector moves in a circle, it will encounter maxima at various angles corresponding to the path differences.
- The maxima will occur at the points where the path difference is \( n\lambda \) for \( n = 0, 1, 2, \ldots, 15 \).

8. **Final count of maxima**:
- Since there are 15 intervals between the maxima (from 0 to 15) and each interval can be counted at two points (one on either side of the midpoint), we multiply the number of intervals by 4 (for each quadrant).
- Therefore, the total number of maxima is:
\[ 15 \text{ intervals} \times 4 = 60 \]

### Conclusion:
The total number of maxima detected during one circular revolution by the detector is **60**.