A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observe on screen 1 m away. It is observed that the first minimum is at a distance of `2.5 mm` from the centre of the screen. Find the width of the slit.

Answer

`0.2 mm`

Explanation

Here, `lambda = 500nm = 5 xx 10^(-7)m, n = 1, a = ?`
`D = 1m, x = 2.5 mm = 2.5 xx 10^(-3)m`
For diffraction minimum
`a sin theta = n lambda`
`a((x)/(D)) = n lambda`
`a = (n lambda D)/(x) = (1 xx 5 xx 10^(-7) xx 1)/(2.5 xx 10^(-3))`
`= 2 xx 10^(-4)m = 0.2 mm`