Question

A parallel beam of light of 600 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1.2 m away. It is observed that the first minimum is at a distance of 3mm from centre of the screen. Calculate the width of the slit.

Answer 1

The distance of the $n^{th}$ minimum from the centre of the screen is, $x_n=\frac{nD\lambda }{a}$
where, D = distance of slit from screen
$\lambda$= wavelength of the light
a = width of the slit
For first minimum, n = 1
$x_n=3\times {10}^3$m
$\lambda =600\times {10}^{-9}$m
D = 1.2 m
$3\times {10}^{-3}=\frac{1\times 1.2\times 600\times {10}^{-9}}{a}$
$a=\frac{1\times 1.2\times 600\times {10}^{-9}}{3\times {10}^{-3}}$
$=\frac{1.2\times 6\times {10}^{-4}}{3}$
$=2.4\times {10}^{-4}$m
$=0.24\times {10}^{-3}$m
$a=0.24$mm