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 0.50 mm and the separation between the slits is 0.12 cm. The refractive index of mica and polystyrene are 1.58 and 1.55 respectively for the light of wavelength 590 nm which is used in the experiment. The interference is observed on a screen a distance of 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 that, $t_1=t_2=t=0.5mm=0.5\times {10}^{-3}m$
${\mu }_m=1.58and{\mu }_p=1.55$
$\lambda =590nm=590\times {10}^{-9}m,$
$d=0.12cm=12\times {10}^{-4}m$
$D=1m$

(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 the strips are fitted, the optical path changes by
$△x=({\mu }_m-1)t-({\mu }_p-1)t$
$=({\mu }_m-{\mu }_p)t$
$=(1.58-1.55)\times \left(0.5\right)\left({10}^{-3}\right)$
$=\left(0.015\right)\times {10}^{-3}m$

So, Number of fringes shift = $\frac{\left(0.015\right)\times {10}^{-3}}{590\times {10}^{-9}}=25.43$
$\Rightarrow$ There are 25 fringes and 0.43th of a fringe.
$\Rightarrow$ There are 13 bright fringes and 12 dark fringes and 0.43th of a dark fringe. So, position of first maximum on both sides will be given by,
$\therefore x=\left(0.43\right)\times 4.91\times {10}^{-4}=0.021cm.$
$x^{\prime }=(1-0.43)\times 4.91\times {10}^{-4}=0.028cm$
$[Since,fringewidth=4.91\times {10}^{-4}m]$