Question

A mica strip and a polystyrene strip are fitted on the two slits of a double slit apparatus. The thickness of the strips is $0.50mm$ and the separation between the slits is $0.12cm$. The refractive index of mica and polystyrene are $1.58$ and $1.55$ respectively for the light of wavelength $590nm$ which is used in the experiment. The interference is observed on a screen a distance one meter away. (a) What would be the fringe-width? (b) At what distance from the centre will the first maximum be located?

Answer 1

Given: $t_1=t_2=0.5mm=0.5\times {10}^{-3}m$

${\mu }_m=1.58$

${\mu }_p=1.55$

$\lambda =590nm=590\times {10}^{-9}m$

$d=0.12m=12\times {10}^{-4}m$

$D=1m$

(a) Fringe width = $\frac{D\lambda }{d}=\frac{1\times 590\times {10}^{-9}}{12\times {10}^{-4}}=4.91\times {10}^{-4}m$

(b) When both strips are fitted, the optical path changes by

$\Delta x=({\mu }_m-1)t_1-({\mu }_p-1)t_2=({\mu }_m-{\mu }_p)t(1.58-1.55)\times 0.5\times {10}^{-3}=0.015\times {10}^{-3}m$

So, Number of fringes shift =$\frac{0.015\times {10}^{-3}}{590\times \times {10}^{-3}}=25.43$

There are 25 fringes and 0.43${}^{th}$ of a fringe.

There are 13 bright fringes and 12 dark fringes and 0.43${}^{th}$ of a dark fringe. So, position of the first maximum on both sides will be given by

$\therefore x=0.43\times 4.91\times {10}^{-4}=0.021cm$

$x^{\prime }=(1-0.43)\times 4.91\times {10}^{-4}=0.028cm$ (since, fringe width=$4.91\times {10}^{-4}m$)