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 determine the order \( n \) up to which the interference fringes can be observed on the screen when using two wavelengths \( \lambda_1 \) and \( \lambda_2 \) (where \( \lambda_2 > \lambda_1 \)). The key point is that the \( n^{th} \) order maximum of one wavelength coincides with the \( n^{th} \) order minimum of the other wavelength.

### Step-by-Step Solution:

1. **Understand the Conditions for Maxima and Minima**:
- The condition for maxima in a double-slit experiment is given by:
\[ d \sin \theta = n \lambda \]
- The condition for minima is given by:
\[ d \sin \theta = \left( n + \frac{1}{2} \right) \lambda \]

2. **Set Up the Equations**:
- For wavelength \( \lambda_1 \) (maxima):
\[ d \sin \theta = n \lambda_1 \]
- For wavelength \( \lambda_2 \) (minima):
\[ d \sin \theta = \left( n - \frac{1}{2} \right) \lambda_2 \]

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

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

5. **Solve for \( n \)**:
- Now, solving for \( n \):
\[ n = \frac{\frac{1}{2} \lambda_2}{\lambda_2 - \lambda_1} \]

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