Question

Two wavelengths \( \lambda_1 \) and \( \lambda_2 \) are used in Young's double slit experiment. \( \lambda_1 = 450 \, \text{nm} \) and \( \lambda_2 = 650 \, \text{nm} \). The minimum order of fringe produced by \( \lambda_2 \), which overlaps with the fringe produced by \( \lambda_1 \), is \( n \). The value of \( n \) is _____.

Answer 1

Correct Answer: 9

Condition for Overlapping Fringes:
 In Young’s double slit experiment, the condition for overlapping fringes for different wavelengths is given by:
\[ n_2 \lambda_2 = n_1 \lambda_1 \]
where \( n_1 \) and \( n_2 \) are the fringe orders for wavelengths \( \lambda_1 \) and \( \lambda_2 \), respectively.

Determine the Ratio of Wavelengths:
 Given:
\[ \lambda_1 = 450 \, \text{nm}, \quad \lambda_2 = 650 \, \text{nm} \]
The ratio of the wavelengths is:
\[ \frac{\lambda_1}{\lambda_2} = \frac{450}{650} = \frac{9}{13} \]

Find the Minimum Order of Overlapping Fringes:
 For the fringes to overlap, \( n_2 \lambda_2 = n_1 \lambda_1 \).
Let \( n_1 = 13 \) and \( n_2 = 9 \) (the smallest integers that satisfy the ratio):
\[ n_2 = 9 \]

Conclusion:
The minimum order of fringe produced by \( \lambda_2 \) which overlaps with the fringe produced by \( \lambda_1 \) is \( n = 9 \).