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

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 $n = 9$.