Question

In Young's double slit experiment the distance between two slits is $2mm$ and screen is at a distance of $120cm$ from the plane of slits. The smallest distance from the central maxima where the bright fringe due to light of wavelength $6500A^o$ and $5200A^o$ would coincide?

• A. $0.156cm$
• B. $0.186cm$
• C. $0.486cm$
• D. $0.456cm$

Answer 1

The correct option is A $0.156cm$

$n^{th}$ maxima is formed at the distance:

$y_n=\frac{n\lambda D}{d}$

At the locations where the maximas of the two wavelengths coincide,

$y_{n1,\lambda 1}=y_{n2,\lambda 2}$

$\frac{n_1{\lambda }_{1}D}{d}=\frac{n_2{\lambda }_{2}D}{d}$

$\frac{n_1}{n_2}=\frac{{\lambda }_2}{{\lambda }_1}=\frac{5}{4}$

$⟹n_2=4k;k\in Z^{+}\because n_1\in Z^{+}$

Smallest distance would correspond to $n_2=4$

$\therefore y_{4,{\lambda }_2}=\frac{4\times 6500\times {10}^{-10}\times 1.2}{2\times {10}^{-3}}$

$=15.6\times {10}^{-4}m$

$=0.156cm$