Question

In Young’s double slit experiment, the slits are 2 mm
apart and are illuminated by photons of two
wavelengths λ1 = 12000 Å and λ2 = 10000 Å. At
what minimum distance from the common central
bright fringe on the screen 2 m from the slit will a
bright fringe from one interference pattern coincide
with a bright fringe from the other?

Answer 1

6mm if this is correct I 'll give full explanation

Answer 2

The minimum distance from the common central is 6 mm.

Solution:

The minimum distance is attained by the following method,

Given: D=2 m  

d=2 m m=$2 \times 10^{-3} m$

As we know that the $n \lambda= constant$

$\begin{array}{l}{n_{1} \lambda_{1}=n_{2} \lambda_{2}} \\ \\{\frac{n_{1}}{n_{2}}=\frac{\lambda_{2}}{\lambda_{1}}} \\ \\{\frac{n_{1}}{n_{2}}=\frac{10000}{12000}} \\ \\{\frac{n_{1}}{n_{2}}=\frac{5}{6}}\end{array}$

The $5^{t h}$ fringes of the YDSE and the $6^{t h}$ fringe of the YDSE will happen to coincide with each other.

Therefore, the minimum distance assumed to be x will be given by $x=\frac{n_{1} \lambda_{1} D}{d}$

$\begin{array}{l}{x=\frac{\left(5 \times 12000 \times 10^{-10} \times 2\right)}{2 \times 10^{-3}}} \\ \\{x=60 \times 10^{-4}} \\ \\ \bold{{x=6 \mathrm{mm}}\end{array}}$