Light of wavelength 6000 o A is incident on a slit of width 0.30 mm. The screen is placed 2 m from the slit. find the position of the first dark fringe

Explanation

The first fringe is on either side of the central bright fringe.
here `n pm 1, D=2m, lamda=6000 dot A=6 times 10^-7 m`
`therefore sin theta= x/D`
`impliesa=0.30mm=3 times10^-4`
`impliesa sintheta= n lamda`
`implies (ax)/D=n lamda`
`x= (n lamdaD)/a`
`impliesx = pm[(1 times6 times10^-7 times2)/ (3 times10^-4)]= pm4 times10^-3m`
The positive and negative signs corresponds to the dark fringes on either side of the central bright fringe.