Question

In Young's double slit experiment, the slits are $2mm$ apart and are illuminated by photons of twowavelengths ${\lambda }_1=12000A$ and ${\lambda }_2=10000A$ . 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. $4mm$
• B. $3mm$
• C. $8mm$
• D. $6mm$

Answer 1

Answer: (D)

Let $n_1$ bright fringe of ${\lambda }_1$ coincides with $n_2$ bright
fringe of ${\lambda }_2$ . Then $\frac{n_1{\lambda }_{1}D}{d}=\frac{n_2{\lambda }_{2}D}{d}$ or $n_1{\lambda }_1=n_2{\lambda }_2$
$\frac{n_1}{n_2}=\frac{{\lambda }_2}{{\lambda }_1}=\frac{10000}{12000}=\frac{5}{6}$
Let x be given distance $\therefore x=\frac{n_1{\lambda }_{1}D}{d}$
Here, $n_1=5,D=2m,d=2mm=2\times 10^{-3}m$
${\lambda }_1=12000A=12000\times 10^{-10}m=12\times 10^{-7}m$
$x=\frac{5\times 12\times 10^{-7}m\times 2m}{2\times 10^{-3}m}=6\times 10^{-3}m=6mm$