Question

In Young's double slit experiment, the slits are 2mm apart and are illuminated by photons of two wavelengths $\lambda_1 = 12000 \mathring A $ and $\lambda_2 = 10000 \mathring A.$ At what minimum distance from the common central bright fringe on the screen 2m from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other ?

• A. 3 mm
• B. 8 mm
• C. 6 mm
• D. 4 mm

Answer 1

The Correct Option is C

According to question $ n_1 \lambda_1 = n_2 \lambda_2$
So $\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{10000}{12000} = \frac{5}{6}$
so minimum $n_1$ and $n_2$ are 5 and 6 respectively
$ X_{min} = \frac{n_1 \lambda_1 D}{d} = \frac{5(12000 \times 10^{-10})(2)}{2 \times 10^{-3}}$
$ = 6 \times 10^{-3} m = 6 mm $