For producing a fraunhofer diffraction fringe, a screen is placed 2m away from a single narrow slit. If the width of slit is 0.2 mm, it is found that first minimum lies 5 mm on either side of the central maximum. Find the wavelength of the incident light.

Explanation

We know, if the distance between n th minimum and central maximum be `x_(n)`, then
`x_(n) = (nDlambda)/(a)`
or , `0.5 = (1xx 200 xx lambda)/(0.02)` [ here, n = 1 ,`x_(n)` = 5 cm = 0.5 cm,
D = 2 m = 200 cm, a = 0.2 mm = 0.02 cm ]
`therefore " " lambda = (0.5 xx 0.02)/(1 xx 200) = 5000 xx 10^(-8) ` cm = 5000Å