Question

A parallel beam of light of $500nm$ falls on a narrow slit and the resulting diffraction pattern is observed on a screen $1m$ away. It is observed that the first minimum is at a distance of $2.5mm$ 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 is, $x_n=\frac{nD\lambda }{a}$
Where, $D=$ distance of the slit from the screen
$\lambda =$ wavelength of the light
$a=$ width of the slit
For first minimum, $n=1$
Thus, $2.5\times {10}^{-3}=\frac{\left(1\right)\left(1\right)(500\times {10}^{-9})}{a}$$m$
$a=2\times {10}^{-4}$$m$
$a=0.2$$mm$