In Young's double slit experiment, the slits are 2mm apart are illuminated by photons of two wavelengths $I_1=12000\mathring{A}$ and $I_2=10000\mathring{A}$. 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?

The correct option is B 6 mm
Now, from the question we can infer that

$n_1{\lambda }_1=n_2{\lambda }_2$

so,

$\frac{n_1}{n_2}=\frac{{\lambda }_2}{{\lambda }_1}$

or

$\frac{n_1}{n_2}=\frac{10000A}{12000A}$

thus,we have

$\frac{n_1}{n_2}=\frac{5}{6}$

5th and 6th fringes will coincide respectively.

the minimum distance is given as

$X_{min}=\frac{n_1{\lambda }_{1}D}{d}$

here,

$n_1=5$

D = 2m

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

so,

$X_{min}=\frac{[5\times 12000\times {10}^{-10}\times 2]}{2\times {10}^{-3}}$

thus, we get

$X_{min}$ = 6mm

hence B option is the correct answer