Question

In Young’s double slit experiment, two wavelengths $\lambda_1 = 780 \, nm$ and $\lambda_2 = 520 \, nm$ are used to obtain interference fringes. If the $n^{th}$ bright band due to $\lambda_1$ coincides with $(n + 1)^{th}$ bright band due to $\lambda_2$, then the value of n is

• A. 4
• B. 3
• C. 2
• D. 6

Answer 1

Answer: (C)

$x_{n} = x_{n+1} $
$^{n }\beta_{1} = ^{\left(n+1\right)}\beta_{2} $
$ \frac{n\lambda_{1}D}{d} =\left(n+1\right) \frac{\lambda_{2}D}{d} $
$ n \lambda_{1} = \left(n+1\right)\lambda_{2} $
$ \frac{n+1}{n} =\frac{\lambda_{1}}{\lambda_{2}} = \frac{780}{520} $
$52n + 52 = 78\, n$
$26 \,n = 52$
$n = \frac{52}{26} = 2 $