Question

For a single slit of width 'a' the first minimum of the interference pattern of a monochromatic light of wavelength e occurs at an angle of `(lambda)/(a)`. At the same angle of `(lambda)/(a)`, we get a maximum for two narrow slits separated by distance 'a'. Explain.

Answer 1

The path difference two secondary wavelengths is given by mn = `a sin theta.` Since `theta` is very small `sin theta=theta`. So for the first order diffraction `n=t`, then angle is `(lambda)/(a).`
Now we know that `theta` must be very small `theta=0` (nearly) because of which the diffraction pattern is minimum.
Now for interference case, for two interfering waves of intensity `I_(1) and I_(2)` we must have two slits separated by a distance. We have the resultant intensity, `I=I_(1)+I_(2)+2sqrt(I_(1)I_(2))cos theta`
Since, `theta=0` (nearly) corresponding to angle `lambda/a` so `cos theta=1`(nearly)
So,`" "I=I_(1)+I_(2)+2sqrt(I_(1)I_(2))`
We see the resultant intensity is sum of the two intensities, so there is a maxima corresponding to the angle `lambda/a`.
This is why, at the same angle of ea, we get a maximum for the narrow slits separated by a distance 'a'.