0
$\begingroup$

So I came across the single slit interference in which a parallel beam of light is passed through narrow slit as shown in figure below. (I made only 3 rays in figure for simplicity there are actually infinite of them) enter image description here

I have a confusion regarding the derivation of the condition of minima or destructive interference(as mentioned in picture).

My textbook derived it by giving an example, say when dsin(theta) = 2(wavelength of monochromatic light) , here d is the slit width.

1) First divide the slit into four parts. (By making three imaginary lines which divides it into equal four parts having width d/4)

2) Then for every point in any of the parts there will be another one located at a distance d/4 the ray passing through these points will interfere destructively.

3) The whole slit can be divided into such pairs of rays and thus the whole summation (I don't know but I have a feeling that instead of this approach these relations can be derived using integral calculus) of rays interfere destructively. Thus at this angle satisfying the above relation, minima occurs.

The problem is that I can, instead of dividing the slit into four parts, divide slit into TWO PARTS and prove that Maxima occurs at this angle satisfying the relation above.

When slit is divided into two parts then for every point in any of the two rigions there will be another one seperated by distance d/2 and will interfere constructively! How's that possible!? What's wrong in dividing slit into 2 parts instead of 4?

Please tell me the error in my argument (this is making my head hurt) and if possible, is there's another way to get this condition using rigorous method (integral calculus)?

Any help will be appreciated.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

"When slit is divided into two parts then for every point in any of the two regions there will be another one separated by distance d/2 and will interfere constructively!"

This is true, but the light from these two points won't interfere constructively with the light from other corresponding points!

You can't say the same for destructive interference, because once light from two corresponding points has interfered destructively, it's gone!

Many students don't like the method of dividing up the slit into a finite number of parts. I recommend that you find out about the phasor (or vector) treatment: it's elegant and satisfying.

Improve this answer
$\endgroup$
3
  • $\begingroup$ @Phillip Wood You're absolutely correct! Thank you very much! So silly of me not to recognise this!!! $\endgroup$
    – Shivansh J
    Mar 22, 2019 at 6:11
  • $\begingroup$ So using calculus we can prove this right? $\endgroup$
    – Shivansh J
    Mar 22, 2019 at 6:12
  • $\begingroup$ You can certainly use calculus to add the (out-of-phase) contributions from infinitesimal portions of the slit. But there is a very nice graphical method for adding the contributions as vectors (phasors actually). It's easy to understand, though it helps if you've met the idea of phasors beforehand, for example in alternating current theory. $\endgroup$
    – Philip Wood
    Mar 22, 2019 at 8:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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