Question

A double-slit of separation $1.5mm$ is illuminated by white light (between $4000$ and $8000\mathring{A}$). On a screen $120cm$ away colored interference pattern is formed. If a pinhole is made on this screen at a distance of $3.0mm$ from the central white fringe, some wavelength will be absent in the transmitted light. Find the second longest wavelength (in $\mathring{A}$) which will be absent in the transmitted light.

Answer 1

The distance of $n^{th}$ dark fringe from the center of central achromatic fringe is given by,

$x_n=(2n+1)\frac{\lambda }{2}.\frac{D}{2d}$

Those wavelengths will be absent whose dark fringe fall on the hole.

In the given problem, $x=3mm=0.3cm,D=120cm,2d=1.5mm=0.15cm$

$\therefore \lambda =\frac{4d{c}_n}{D(2n+1)}=\frac{0.30\times 2\times 0.15}{120(2n+1)}=\frac{3}{4000(2n+1)}cm=\frac{75000}{(2n+1)}$Å

This is much large than the given range.

So we will Substitute n from 6,7,8,... we get

${\lambda }_6=\frac{75000}{11}\AA =6818\AA$, this will be the first longest absent wavelength

${\lambda }_7=\frac{75000}{13}\AA =5769\AA$, is the required second longest absent wavelength.