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.

Answer

`0.1 mm`

Explanation

Here, `lambda = 600 mm = 6 xx 10^(-7)m, D = 0.8 m`
`n = 2, x = 9.6 mm = 9.6 xx 10^(-3)m, a = ?`
For 2nd order minimum, `a sin theta = n lambda`
`a((x)/(D)) = 2 lambda, a = (2 lambda D)/(x) = (2 xx 6 xx 10^(-7) xx 0.8)/(9.6 xx 10^(-3))`
`= 10^(-4) m = 0.1 mm`