In Young's double slit experiment, the `10^(th)` maximum of wavelength `lambda_(1)` is at a distance `y_(1)` from its central maximum and the `5^(th)` maximum of wavelength `lambda_(2)` is at a distance `y_(2)` from its central maximum. The ratio `y_(1)//y_(2)` will be

Options

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

Explanation

To solve the problem, we will use the formula for the position of the bright fringes in Young's double slit experiment. The position of the nth bright fringe is given by the formula:

\[ y_n = \frac{n \lambda D}{d} \]

where:
- \(y_n\) is the distance of the nth bright fringe from the central maximum,
- \(n\) is the order of the fringe,
- \(\lambda\) 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.

### Step-by-step Solution:

1. **Identify the given information:**
- For wavelength \(\lambda_1\), we are interested in the 10th maximum, so \(n_1 = 10\).
- For wavelength \(\lambda_2\), we are interested in the 5th maximum, so \(n_2 = 5\).

2. **Write the equations for \(y_1\) and \(y_2\):**
- For the 10th maximum of wavelength \(\lambda_1\):
\[ y_1 = \frac{10 \lambda_1 D}{d} \]
- For the 5th maximum of wavelength \(\lambda_2\):
\[ y_2 = \frac{5 \lambda_2 D}{d} \]

3. **Set up the ratio \(\frac{y_1}{y_2}\):**
\[ \frac{y_1}{y_2} = \frac{\frac{10 \lambda_1 D}{d}}{\frac{5 \lambda_2 D}{d}} \]

4. **Simplify the ratio:**
- The \(D\) and \(d\) terms cancel out:
\[ \frac{y_1}{y_2} = \frac{10 \lambda_1}{5 \lambda_2} \]
- Simplifying further gives:
\[ \frac{y_1}{y_2} = \frac{10}{5} \cdot \frac{\lambda_1}{\lambda_2} = 2 \cdot \frac{\lambda_1}{\lambda_2} \]

5. **Final result:**
\[ \frac{y_1}{y_2} = \frac{2 \lambda_1}{\lambda_2} \]

### Conclusion:
The ratio \(\frac{y_1}{y_2}\) is \(\frac{2 \lambda_1}{\lambda_2}\).

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