In a hall, a person receives direct sound waves from a source `120 m` away. He also receives waves from the same source which reach him after being reflected from the `25 m` high ceiling at a point halfway between them. The two waves interfere constructively for wavelengths (in metres)

Options

A. `20,20/3,20/3 , ` etc.
B. `10, 5, 3.33,` etc
C. `10,20 ,30 ,` etc
D. `15,25,35, ` etc

Answer

To solve the problem step by step, we need to find the wavelengths for which the two sound waves interfere constructively. Here’s how we can approach this problem:

### Step 1: Understand the Geometry of the Problem
We have a sound source and a person receiving sound waves. The direct sound wave travels a distance of 120 m, while the reflected sound wave travels from the source to the ceiling and then to the person. The ceiling height is 25 m.

### Step 2: Determine the Path Lengths
1. The direct path (AC) is 120 m.
2. The path of the reflected wave (AB + BC) can be calculated using the right triangle formed by the height of the ceiling and the horizontal distance to the halfway point (M).

### Step 3: Calculate the Distance AB
- The distance from the source to the ceiling (AB) can be calculated using the Pythagorean theorem. The horizontal distance from the source to the halfway point is half of 120 m, which is 60 m.
- Therefore, the distance AB can be calculated as:
\[ AB = \sqrt{(25^2) + (60^2)} = \sqrt{625 + 3600} = \sqrt{4225} = 65 \text{ m} \]

### Step 4: Calculate the Distance BC
- Since the path BC is the same as AB (the sound wave travels the same distance after reflecting), we have:
\[ BC = 65 \text{ m} \]

### Step 5: Calculate the Total Path Length for the Reflected Wave
- The total path length for the reflected wave (AB + BC) is:
\[ AB + BC = 65 + 65 = 130 \text{ m} \]

### Step 6: Calculate the Path Difference
- The path difference (Δ) between the two waves is given by:
\[ \Delta = (AB + BC) - AC = 130 - 120 = 10 \text{ m} \]

### Step 7: Determine the Condition for Constructive Interference
- For constructive interference, the path difference must be an integral multiple of the wavelength (λ):
\[ \Delta = n\lambda \quad (n = 0, 1, 2, 3, \ldots) \]
- Therefore, we can write:
\[ 10 = n\lambda \]

### Step 8: Calculate Possible Wavelengths
- Rearranging gives:
\[ \lambda = \frac{10}{n} \]
- For different values of n:
- For \( n = 1 \): \( \lambda = 10 \text{ m} \)
- For \( n = 2 \): \( \lambda = 5 \text{ m} \)
- For \( n = 3 \): \( \lambda = \frac{10}{3} \approx 3.33 \text{ m} \)

### Conclusion
The wavelengths for which the two waves interfere constructively are:
- \( \lambda = 10 \text{ m}, 5 \text{ m}, \text{ and } 3.33 \text{ m} \)