A parallel beam of light of `600 nm` falls on a narrow slit and the resulting diffraction pattern is observed on a screen `.12 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.

Explanation

The distance of the nth minimum from the centre of the screen is `x_(n)=(n D x)/(a)`
Where D - distance
`lambda` - wave length
a - width
For first minium, n = 1
Thus, `2.5 xx 10^(-3)=(1(1)(500 xx 10^(-9)))/(a)a= 2 xx 10^(-4)m = 0.2 mm`