The discussion revolves around the implications of having a slit separation (d) in Young's double slit experiment that is less than the wavelength (λ) of light. Participants explore whether interference fringes can still exist under these conditions and the validity of the double slit formula.
Participants do not reach a consensus on whether fringes can exist when d < λ. Multiple competing views are presented regarding the implications of this condition on interference patterns.
Participants note that the discussion involves assumptions about the nature of the slits and the simplifications inherent in the double slit formula, particularly regarding the treatment of slits as omnidirectional sources.
This discussion may be of interest to those studying wave optics, interference phenomena, or the applications of the double slit experiment in various contexts.
mehul mahajan said:I'm finding problem understanding your question, what i know is that the path difference = d sinθ
and for constructive interference d sinθ =n λ
where n is an integer.
terryds said:In Young's double slit formula, the equation is
d sin \theta = m \lambda
Which means that
sin \theta = m \lambda / d
But, what if d < \lamda so the sine value becomes greater than 1?
Will there be any interference?
sophiecentaur said:If d is small, there will be no total cancellation but interference will still cause a drop in the amplitude as the angle increases.
It is valid because it only describes the positions of zeros and there is no zero for closely spaced slits.terryds said:So, does the formula still valid for d < lambda?
If the slits are too close together, the phase difference between the waves from the two sources can't be as great as π and so you can't have complete cancellation at 90°. Nevertheless, if there is a reflex angle between the two E vectors, the resulting amplitude will be noticeably lower than where they add and you will get a minimum but the zero is, as you say, at an invalid angle for the formula. I tried a brief search for a suitable diagram of the patterns of two omnidirectional sources on Google but I could find nothing. It's the sort of picture you get in books on antenna theory.terryds said:Why will there be no total cancellation?
sophiecentaur said:It is valid because it only describes the positions of zeros and there is no zero for closely spaced slits.
The formula is based on a simplified situation with infinitely narrow slits (omnidirectional individual sources)-more like what you get with two co-phased radio frequency dipoles than two slits. If the sources are just greater than λ apart, there will only be a single zero, which will be at nearly 90° off beam. As the spacing gets narrower still, this zero will pass through the 90° direction and disappear. In any case, the beam (the diffraction you are referring to??) pattern of each 'real' single slit will have a zero at 90° or greater. Two dipoles will work fine, though, as a model because they are truly omnidirectional, individually.If the slits are too close together, the phase difference between the waves from the two sources can't be as great as π and so you can't have complete cancellation at 90°. Nevertheless, if there is a reflex angle between the two E vectors, the resulting amplitude will be noticeably lower than where they add and you will get a minimum but the zero is, as you say, at an invalid angle for the formula. I tried a brief search for a suitable diagram of the patterns of two omnidirectional sources on Google but I could find nothing. It's the sort of picture you get in books on antenna theory.
I got to know this stuff from antenna theory, rather than optics and it's actually much more approachable (imo).
If there fringes then there would have to be zeros, wouldn't there? This is just one very broad maximum.terryds said:Will there be any fringes then?