In YDSE, if a bichromatic light having wavelengths `lambda_(1)` and `lambda_(2)` is used, then maxima due to both lights will overlaps at a certain distance y from the central maxima. Take separation between slits as d and distance between screen and slits as D. Then the value of y will be

Options

A. `((lambda_(1) + lambda_(2))/(2D))d`
B. `(lambda_(1) - lambda_(2))/(D) xx 2d`
C. LCM of `(lambda_(1) D)/(d)` and `(lambda_(2) D)/(d)`
D. HCF of `(lambda_(1) D)/(d)` and `(lambda_(2) D)/(d)`

Explanation

To solve the problem, we need to find the distance \( y \) from the central maxima where the maxima due to two different wavelengths \( \lambda_1 \) and \( \lambda_2 \) overlap in Young's Double Slit Experiment (YDSE).

### Step-by-Step Solution:

1. **Understanding the formula for maxima in YDSE**:
In YDSE, the position of the maxima on the screen for a single wavelength is given by the formula:
\[ y_n = \frac{n \cdot D \cdot \lambda}{d} \]
where:
- \( y_n \) is the position of the nth maxima,
- \( D \) is the distance from the slits to the screen,
- \( d \) is the separation between the slits,
- \( \lambda \) is the wavelength of the light used,
- \( n \) is the order of the maxima (0, 1, 2, ...).

2. **Position of maxima for each wavelength**:
For the first wavelength \( \lambda_1 \):
\[ y_{n1} = \frac{n_1 \cdot D \cdot \lambda_1}{d} \]
For the second wavelength \( \lambda_2 \):
\[ y_{n2} = \frac{n_2 \cdot D \cdot \lambda_2}{d} \]

3. **Setting the positions equal for overlap**:
To find the distance \( y \) where the maxima overlap, we set \( y_{n1} \) equal to \( y_{n2} \):
\[ \frac{n_1 \cdot D \cdot \lambda_1}{d} = \frac{n_2 \cdot D \cdot \lambda_2}{d} \]

4. **Simplifying the equation**:
Since \( D \) and \( d \) are common in both terms, we can cancel them out:
\[ n_1 \cdot \lambda_1 = n_2 \cdot \lambda_2 \]

5. **Finding the relationship between orders of maxima**:
Rearranging gives:
\[ \frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} \]
This means that the orders of the maxima are in the ratio of the wavelengths.

6. **Finding the minimum distance of overlap**:
The positions \( y \) where the maxima overlap will be at the least common multiple (LCM) of the individual distances for each wavelength:
\[ y = \text{LCM}\left(\frac{D \cdot \lambda_1}{d}, \frac{D \cdot \lambda_2}{d}\right) \]

7. **Final expression for \( y \)**:
Thus, the distance \( y \) where the maxima overlap is given by:
\[ y = \frac{D}{d} \cdot \text{LCM}(\lambda_1, \lambda_2) \]