Light of wavelength 6000Å is incident on a slit of width 0.30 mm. The screen is placed 2 m from the slit. Find (a) the position of the first dark fringe and (b). The width of the central bright fringe.

Explanation

The two dark fringes that flank the central bright fringe correspond to `n=pm1` in Eq. (i) hence we find that
`sintheta=pmlambda/a=pm(5.80xx10^(-7)m)/(0.300xx10^(-3)m)=pm1.93xx10^(-3)`
From the triangle in figure note that `tan theta=y_(1)//D` Because `theta` is very small, we can use the approximation `sinthetaapproxtantheta` thus `sin theta approxy_(1)//D` Therefore, the positions of the first minima measured from th central axis are given by
`y_(1)approxD sin theta=pmDlambda/a=pm3.87xx10^(-3)m`
The positive and negative signs correspond to the dark fringes on either side of the central bright fringe. Hence, the width of the central bright fringe is equal to
`2|y_(1)|=7.74xx10^(-3)m=7.74mm` Note that this value is much greater than the width of the slit. However, as the slit corresponding to smaller values of `theta` in fact for large values of a, the various maxima and minima are so closely spaced that only a large central bright area resembling the geometric image of the slit is observed.