The young.s double slit experiment is performed with four different sources. The number of fringes observed in a given region for that sources are `n_(1)= 100, n_(2)= 60, n_(3)= 150, n_(4)= 120`. The descending order of wave lengths of sources is

Options

A. `n_(4), n_(2), n_(3), n_(1)`
B. `n_(4), n_(2), n_(3), n_(1)`
C. `n_(4), n_(2), n_(3), n_(1)`
D. `n_(4), n_(2), n_(3), n_(1)`

Answer

To solve the problem of determining the descending order of wavelengths based on the number of fringes observed in the Young's double slit experiment, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between the number of fringes and wavelength**:
The number of fringes \( n \) observed is inversely proportional to the wavelength \( \lambda \) of the light used. This means that if the number of fringes increases, the wavelength decreases, and vice versa.

2. **Write the mathematical relationship**:
From the Young's double slit experiment, we can express the relationship as:
\[ n \propto \frac{1}{\lambda} \]
This implies:
\[ n_1 \lambda_1 = n_2 \lambda_2 = n_3 \lambda_3 = n_4 \lambda_4 \]
where \( n_i \) is the number of fringes and \( \lambda_i \) is the wavelength for each source.

3. **List the number of fringes for each source**:
- \( n_1 = 100 \)
- \( n_2 = 60 \)
- \( n_3 = 150 \)
- \( n_4 = 120 \)

4. **Determine the wavelengths based on the number of fringes**:
Since \( n \) is inversely proportional to \( \lambda \):
- The source with the highest \( n \) has the lowest \( \lambda \).
- The source with the lowest \( n \) has the highest \( \lambda \).

5. **Rank the number of fringes**:
- \( n_2 = 60 \) (smallest, hence largest wavelength)
- \( n_1 = 100 \)
- \( n_4 = 120 \)
- \( n_3 = 150 \) (largest, hence smallest wavelength)

6. **Write the descending order of wavelengths**:
Therefore, the descending order of wavelengths is:
\[ \lambda_2 > \lambda_1 > \lambda_4 > \lambda_3 \]
In terms of the number of fringes:
\[ n_2 < n_1 < n_4 < n_3 \]

### Final Answer:
The descending order of wavelengths of the sources is:
\[ \lambda_2, \lambda_1, \lambda_4, \lambda_3 \]