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, $n_{1}=5,D=2m,d=2mm=2\times 10^{- 3}m$
$\lambda _{1}=12000A=12000\times 10^{- 10}m=12\times 10^{- 7}m$
$x=\frac{5 \times 12 \times 10^{- 7} m \times 2 m}{2 \times 10^{- 3} m}=6\times 10^{- 3}m=6mm$