A common way for deriving the expression for the minima in a single slit diffraction pattern involves dividing the slit into even number of zones and pairing up the wavelets so that they cancel out for a particular angle.

Dividing the slit into two zones and applying the condition for destructive interference gives

$$\frac{a}{2} \sin{\theta} = \frac\lambda 2$$

And generalizing this, we get

$$a \sin\theta = m \lambda$$

Now this is my question:

The expression $\frac{a}{2} \sin{\theta} = \frac\lambda 2$ says that for some angle $\theta$ , the wavelets from the two zones will interfere destructively, producing the first minima. Now, $\frac{a}{2} \sin{\theta}$ is simply the path difference between the wavelets, and if I equate the path difference to, say, $\lambda$ , i.e, the condition for constructive interference , I will get

$$a \sin\theta = 2 \lambda$$

which should give the angular position of a maxima. But this contradicts with the position for minima, given by $a \sin\theta = m \lambda$ , when $m = 2$ .

So what's wrong with my reasoning ?


1 Answer 1


When you equate the path difference between any two corresponding wavelets from both the segments to $\lambda$, you surely do get,

$$a\sin \theta=2\lambda$$

But this condition only implies that two corresponding wavelets will interfere constructively. It does not imply that all the wavelets will interfere constructively. The resultant of any one pair of corresponding wavelets will not interfere constructively with the resultant of any other pair of corresponding wavelets. So the electric field vectors will interfere like this.

enter image description here

So all the resultant wavelets will be separated by a constant phase difference which would not be $0$. So your phasor will look something like this.

enter image description here

Courtesy:- https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_%28OpenStax%29/Map%3A_University_Physics_III_-_Optics_and_Modern_Physics_%28OpenStax%29/04%3A_Diffraction/4.03%3A_Intensity_in_Single-Slit_Diffraction

And this phase difference will be such that all the phasors will add up precisely to give a phasor having $0$ magnitude. So it would be, in fact, a destructive interference.

Now the reason why you never faced such dilemma in the case of destructive interference was because you were adding zero vectors which were out of phase. So the resultant was also zero. Here you are adding non zero vectors which are out of phase, so they add up to zero.

Hope this helps!!

  • $\begingroup$ So even though a pair of wavelets will constructively interfere, the resultant of all such constructive interferences will be zero, due to them being in different directions ? $\endgroup$ Commented Dec 31, 2019 at 6:52
  • $\begingroup$ @Gokul Exactly!! $\endgroup$
    – user243267
    Commented Dec 31, 2019 at 7:00
  • 1
    $\begingroup$ Ah right, I get it now. Thanks for answering ! $\endgroup$ Commented Dec 31, 2019 at 7:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.