Question

In Young's double slit experiment using monochromatic light of wavelength $600$nm, $5^{th}$ bright fringe is at a distance of $0.48$mm from the centre of the pattern. If the screen is at a distance of $80$cm from the plane of the two slits, calculate:
(i) Distance between the two slits.
(ii) Fringe width, i.e., fringe separation.

Answer 1

wavelength of light $\lambda =600nm=6\times {10}^{-7}m$

position of $5th$ bright fringe $y_5=0.48mm=0.48\times {10}^{-3}m$

Distance between slits and the screen $D=0.8m$

(i) : Position of $5th$ bright fring is given by $y_n=\frac{n\lambda D}{d}$

$\therefore$ $0.48\times {10}^{-3}=\frac{5(6\times {10}^{-7})\left(0.8\right)}{d}$

$⟹$ $d=5\times {10}^{-3}m=5mm$

(ii) : Fringe width is given by $\beta =\frac{\lambda D}{d}$

$\therefore$ $\beta =\frac{(6\times {10}^{-7})\left(0.8\right)}{0.48\times {10}^{-3}}={10}^{-3}m=1mm$