In a single slit diffraction pattern, the distance between first minima on the right and first minima on the left of central maximum is `4mm`. The screeon on which the pattern is displaced is `2m` from the slit and wavelength of light used is `6000Å`. Calculate width of the slit and width of the central maximum.

Explanation

Given: `lambda=6000Å`, `D=2m`
Consider that `X_(1)` and `X_(2)` be the distance between first minima on the right and left respectively of central maximum.
`X_(1)+X_(2)=4mm`
`X_(1)+X_(2)=(2lambdaD)/(d)`
`4xx10^(-3)=(2xx6xx10^(-7)xx2)/(d)`
`d=6xx10^(-4)m`
`d=0.6mm`
`:. ` Width of central maximum,
`W=X_(1)+X_(2)`
`=4xx10^(-3)m`
`=4mm`.