If an interference experiment is performed using two wavelengths close to each other, two interference patterns corresponding to the two wavelengths are obtained on the screen. The frings system remains distinct up to a point on the screen where the `n^("th")` order maximum of one wavelength falls on the `n^("th")` order minimum of the other wavelength. Suppose in Young's double sit experiment, sodium light composed of two wavelengths `lambda_(1)and lambda_(2)` close to each other (with `lambda_(2)` greater than `lambda_(1)`) is used. The order n, up to which the fringes can be seen on the screen, is given by

Options

A. `n=(lambda_(2))/(lambda_(2)-lambda_(1))`
B. `n=(lambda_(1))/(lambda_(2)-lambda_(1))`
C. `n=(lambda_(2))/(2(lambda_(2)-lambda_(1)))`
D. `n=(lambda_(1))/(2(lambda_(2)-lambda_(1)))`

Answer

To solve the problem, we need to analyze the situation where two wavelengths, \( \lambda_1 \) and \( \lambda_2 \) (with \( \lambda_2 > \lambda_1 \)), create interference patterns in a Young's double slit experiment. The goal is to find the maximum order \( n \) up to which the fringes can be observed before they overlap.

### Step-by-Step Solution:

1. **Understanding the Conditions for Maxima and Minima**:
- For a maximum (bright fringe) of wavelength \( \lambda \), the condition is given by:
\[ d \sin \theta = n \lambda \]
- For a minimum (dark fringe) of wavelength \( \lambda \), the condition is:
\[ d \sin \theta = \left( n - \frac{1}{2} \right) \lambda \]

2. **Setting Up the Problem**:
- We have two wavelengths \( \lambda_1 \) and \( \lambda_2 \).
- The \( n^{th} \) order maximum of \( \lambda_1 \) coincides with the \( n^{th} \) order minimum of \( \lambda_2 \).

3. **Equating the Conditions**:
- For \( \lambda_1 \) at maximum:
\[ d \sin \theta = n \lambda_1 \]
- For \( \lambda_2 \) at minimum:
\[ d \sin \theta = \left( n - \frac{1}{2} \right) \lambda_2 \]

4. **Setting the Two Conditions Equal**:
- Since both expressions equal \( d \sin \theta \), we can set them equal to each other:
\[ n \lambda_1 = \left( n - \frac{1}{2} \right) \lambda_2 \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ n \lambda_1 = n \lambda_2 - \frac{1}{2} \lambda_2 \]
- This can be rewritten as:
\[ n \lambda_2 - n \lambda_1 = \frac{1}{2} \lambda_2 \]
- Factoring out \( n \):
\[ n (\lambda_2 - \lambda_1) = \frac{1}{2} \lambda_2 \]

6. **Solving for \( n \)**:
- We can now solve for \( n \):
\[ n = \frac{\frac{1}{2} \lambda_2}{\lambda_2 - \lambda_1} \]
- This simplifies to:
\[ n = \frac{\lambda_2}{2(\lambda_2 - \lambda_1)} \]

### Final Answer:
The order \( n \) up to which the fringes can be seen on the screen is given by:
\[ n = \frac{\lambda_2}{2(\lambda_2 - \lambda_1)} \]