Question

A parallel beam of light of 450 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1.5 m away. It is observed that the first minimum is at a distance of 3 mm from the 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, $x_n=\frac{nD\lambda }{a}$
Where, D $\to$ distance of slit from screen.
$\lambda \to$ wavelength of the light.
a $\to$ width of the slit
For first minimum, n = 1,
$x_n=3\times {10}^{-3}m$
$D=1.5$ m
$\lambda =450\times {10}^{-9}$ m
$3\times {10}^{-3}=\frac{1\times \left(1.5\right)\times (450\times {10}^{-9})}{a}$
$a=\frac{1\times \left(1.5\right)\times (450\times {10}^{-9})}{3\times {10}^{-3}}$
$a=0.225$mm