In Young's double slit experiment the 7th maximum with wavelength `lambda_(1)`is at a distance `d_(1)`, and that with wavelength `lambda_(2)` is at distance `d_(2)`. Then `d_(1)//d_(2)` is

Options

A. `lambda_(1)/lambda_(2)`
B. `lambda_(2)/sqrtlambda_(1)`
C. `lambda_(1)^(2)//lambda_(2)^(2)`
D. `lambda_(2)^(2)//lambda_(2)^(1)`

Answer

To solve the problem, we need to find the ratio of the distances \( d_1 \) and \( d_2 \) for the 7th maximum in Young's double slit experiment for two different wavelengths \( \lambda_1 \) and \( \lambda_2 \).

### Step-by-Step Solution:

1. **Understanding the Setup**: In Young's double slit experiment, the distance between the slits is \( d \), and the distance from the slits to the screen is \( D \). The position of the \( n \)-th maximum on the screen is given by the formula:
\[ y_n = \frac{n \lambda D}{d} \]
where \( y_n \) is the distance from the central maximum to the \( n \)-th maximum.

2. **Position of the 7th Maximum for Wavelength \( \lambda_1 \)**: For the first wavelength \( \lambda_1 \), the position of the 7th maximum (\( n = 7 \)) is:
\[ d_1 = \frac{7 \lambda_1 D}{d} \]

3. **Position of the 7th Maximum for Wavelength \( \lambda_2 \)**: For the second wavelength \( \lambda_2 \), the position of the 7th maximum is:
\[ d_2 = \frac{7 \lambda_2 D}{d} \]

4. **Finding the Ratio \( \frac{d_1}{d_2} \)**: To find the ratio of the distances \( d_1 \) and \( d_2 \), we can write:
\[ \frac{d_1}{d_2} = \frac{\frac{7 \lambda_1 D}{d}}{\frac{7 \lambda_2 D}{d}} = \frac{\lambda_1}{\lambda_2} \]

5. **Final Result**: Therefore, the ratio of the distances \( d_1 \) and \( d_2 \) is:
\[ \frac{d_1}{d_2} = \frac{\lambda_1}{\lambda_2} \]