A parallel beam of light of 500 nm falls on a narrow slit

cbse
waveoptics

#1

A parallel beam of light of 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minima is at a distance of 2.5 mm from the centre of the screen. Calculate the width of the slit.


#2

The distance of the nth minima from the centre of the screen is given by

$x_{ n }$ = nDλ/a

Where D = distance of slit form screen = 1m
λ = wavelength of the light = 500nm
= 500 x ${{10}^{-9}}$ m,
n=1
$x_{ n }$ = 2.5 mm = 2.5 x ${{10}^{-3}}$ m,

and a = width of the slit first minima=?
puting these values we get

2.5 x ${{10}^{-3}}$ = 500 x ${{10}^{-9}}$/a
====> a = 2 x ${{10}^{-4}}$ m = 0.2 m