#### Question

For a single slit of width "a", the first minimum of the interference pattern of a monochromatic light of wavelength λ occurs at an angle of λa. At the same angle of λa, we get a maximum for two narrow slits separated by a distance "a". Explain.

#### Solution

The path difference between two secondary wavelets is given by nλ = asin*θ*. Since, θ is very small sin*θ* = *θ*. So, for the first order diffraction n = 1, the angle is λ/a. Now we know that θ must be very small θ = 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_1I_2)costheta`

Since, *θ* = 0 (nearly) corresponding to angle λ/a so cos*θ* = 1 (nearly)

So,

`I=I_1+I_2+2sqrt(I_1I_2)costheta`

`=>I=I_1+I_2+2sqrt(I_1I_2)cos(0)`

`=>I=I_1+I_2+2sqrt(I_1I_2)`

We see the resultant intensity is sum of the two intensities, so there is a maxima corresponding to the angle λ/a.

This is why, at the same angle of `lambda/a`we get a maximum for two narrow slits separated by a distance "a".