In Young's double slit experiment, the 8th maximum with wavelength `lambda_1` is at a distance `d_1` from the central maximum and the 6th maximum with a wavelength `lambda_2` is at a distance `d_2`. Then `(d_1//d_2)` is equal to

Options

A. (a) `4/3(lambda_2/lambda_1)`
B. (b) `4/3(lambda_1/lambda_2)`
C. (c) `3/4(lambda_2/lambda_1)`
D. (d) `3/4(lambda_1/lambda_2)`

Explanation

To solve the problem, we need to find the ratio \( \frac{d_1}{d_2} \) where \( d_1 \) is the distance of the 8th maximum for wavelength \( \lambda_1 \) and \( d_2 \) is the distance of the 6th maximum for wavelength \( \lambda_2 \).

### Step-by-Step Solution:

1. **Understanding the Young's Double Slit Experiment:**
In Young's double slit experiment, the position of the maxima on the screen is given by the formula:
\[ d_n = \frac{n \lambda D}{d} \]
where:
- \( d_n \) is the distance of the nth maximum from the central maximum,
- \( n \) is the order of the maximum,
- \( \lambda \) is the wavelength of the light used,
- \( D \) is the distance from the slits to the screen,
- \( d \) is the distance between the two slits.

2. **Finding the Distance for the 8th Maximum with Wavelength \( \lambda_1 \):**
For the 8th maximum (\( n_1 = 8 \)) with wavelength \( \lambda_1 \):
\[ d_1 = \frac{8 \lambda_1 D}{d} \]

3. **Finding the Distance for the 6th Maximum with Wavelength \( \lambda_2 \):**
For the 6th maximum (\( n_2 = 6 \)) with wavelength \( \lambda_2 \):
\[ d_2 = \frac{6 \lambda_2 D}{d} \]

4. **Finding the Ratio \( \frac{d_1}{d_2} \):**
Now, we can find the ratio of \( d_1 \) to \( d_2 \):
\[ \frac{d_1}{d_2} = \frac{\frac{8 \lambda_1 D}{d}}{\frac{6 \lambda_2 D}{d}} = \frac{8 \lambda_1}{6 \lambda_2} \]
Simplifying this gives:
\[ \frac{d_1}{d_2} = \frac{4 \lambda_1}{3 \lambda_2} \]

### Final Result:
Thus, the ratio \( \frac{d_1}{d_2} \) is:
\[ \frac{d_1}{d_2} = \frac{4 \lambda_1}{3 \lambda_2} \]