Question

A beam of light consisting of two wavelengths, $800$nm and $600$nm is used to obtain the interference fringes in a Young's double slit experiment on a screen place $1.4$m away. If the two slits are separated by $0.28$mm, Calculate the least distance from the central bright maximum where the bright fringes of the two wavelengths coincidence.

Answer 1

Here,

$D=1.4m$

$d=0.28mm=0.28\times {10}^{-3}$

${\lambda }_1=800\times {10}^{-9}m$

${\lambda }_2=600\times {10}^{-9}m$

Let y be the common distance of the bright fringes by the both wavelength.

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

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

$\Rightarrow n_1\times 800\times {10}^{-9}=n_2\times 600\times {10}^{-9}$

$\Rightarrow 4n_1=3n_2$

$n_1\ne 0,n_2\ne 0$

For y to be minimum and since n are integers, $n_1=3,n_2=4$

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

$\Rightarrow y=\frac{3\times 800\times {10}^{-9}\times 1.4}{0.28\times {10}^{-3}}$

$\Rightarrow y=1.2\times {10}^{-2}m$

$\Rightarrow y=1.2cm$