A source emitting light of wavelengths `lambda_1 and lambda_2` is used in Young's double-slit experiment. If fourth bright of `lambda_1` coincides with sixth bright of `lambda_2`, then

Options

A. `2 lambda_1 - 3lambda_2`
B. `3lambda_1 - 2lambda_2`
C. `4lambda_1 - 3lambda_2`
D. `3lambda_1 - 4lambda_2`

Explanation

To solve the problem, we need to analyze the conditions given in the Young's double-slit experiment (YDSE) where the fourth bright fringe of wavelength \( \lambda_1 \) coincides with the sixth bright fringe of wavelength \( \lambda_2 \).

### Step-by-Step Solution:

1. **Understanding the Bright Fringe Position**:
The position of the \( n \)-th bright fringe in YDSE is given by the formula:
\[ y_n = \frac{n \lambda D}{d} \]
where:
- \( y_n \) is the position of the \( n \)-th bright fringe,
- \( n \) is the order of the fringe,
- \( \lambda \) is the wavelength of the light used,
- \( D \) is the distance from the slits to the screen,
- \( d \) is the distance between the slits.

2. **Position of the Fourth Bright Fringe for \( \lambda_1 \)**:
For the fourth bright fringe of wavelength \( \lambda_1 \):
\[ y_{4, \lambda_1} = \frac{4 \lambda_1 D}{d} \]

3. **Position of the Sixth Bright Fringe for \( \lambda_2 \)**:
For the sixth bright fringe of wavelength \( \lambda_2 \):
\[ y_{6, \lambda_2} = \frac{6 \lambda_2 D}{d} \]

4. **Setting the Positions Equal**:
Since the fourth bright fringe of \( \lambda_1 \) coincides with the sixth bright fringe of \( \lambda_2 \), we can set their positions equal:
\[ \frac{4 \lambda_1 D}{d} = \frac{6 \lambda_2 D}{d} \]

5. **Cancelling Common Terms**:
We can cancel \( D \) and \( d \) from both sides (assuming they are not zero):
\[ 4 \lambda_1 = 6 \lambda_2 \]

6. **Rearranging the Equation**:
Rearranging gives:
\[ 2 \lambda_1 = 3 \lambda_2 \]
or
\[ 2 \lambda_1 - 3 \lambda_2 = 0 \]

### Conclusion:
Thus, the relationship between the wavelengths is:
\[ 2 \lambda_1 = 3 \lambda_2 \]