B Can Fringes Exist with Sine Value Greater Than 1 in Young's Double Slit Formula?

[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?

 

[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.

 

[terryds]

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.

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

 

[sophiecentaur]

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?

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

 

[terryds]

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

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

 

[tade]

you won't see the fringes no more

 

[sophiecentaur]

So, does the formula still valid for d < lambda?

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.

Why will there be no total cancellation?

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).

 

[terryds]

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).

Will there be any fringes then?

 

[sophiecentaur]

Will there be any fringes then?

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.