B Young's Double Slit Formula

1. Mar 29, 2017

terryds

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?

2. Mar 29, 2017

mehul mahajan

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.

3. Mar 29, 2017

terryds

So, what happens if the distance between slits is much smaller than the wavelength?

4. Mar 29, 2017

sophiecentaur

If d is small, there will be no total cancellation but interference will still cause a drop in the amplitude as the angle increases.

5. Mar 29, 2017

terryds

So, does the formula still valid for d < lambda?
Why will there be no total cancellation? I mean, you mean the cancellation caused by the difraction? Or, what? Please explain it.

Thanks

6. Mar 29, 2017

tade

you won't see the fringes no more

7. Mar 29, 2017

sophiecentaur

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).

8. Mar 29, 2017

terryds

Will there be any fringes then?

9. Mar 29, 2017

sophiecentaur

If there fringes then there would have to be zeros, wouldn't there? This is just one very broad maximum.
We're talking a slit separation of only 600nm or so.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted