A screen is placed 50cm from a single slit, which is illuminated with 6000Å light. If distance between the first and third minima in the diffraction pattern is 3.0 mm, what is the width of the slit?

Explanation

In case of diffraction at single slit, the position of minima is given by `a sin theta = nlambda`. Where d is the aperture size and for small `theta` :
`sin theta= theta= (y//D)`
`:. a((y)/(D))= n lambda, i.e., y= (D)/(a) (n lambda)`
So that, `y_(3)-y_(1)= (D)/(a)(3lambda-lambda)= (D)/(a)(2lambda)` and hence, `a= (0.50xx(2xx6xx10^(-7)))/(3xx10^(-3))= 2xx10^(-4)m`
`= 0.2mm`