A light Interference doubt

jaumzaum

I'm really confused about the way the light can suffer interference. I'll try to explain the way I think all this occurs and the question I have if this explanation was correct. I would like you guys to correct me if anything I say is wrong and to try to explain me the final question.

Light is a set of photons that oscilates creating an eletric and a magnetic field around them. When 2 photons are too close their eletric/magnetic fields can interact generating the interference. But this would need the 2 photons to be really close, doesn't? And this interference would occur only while the 2 photons were together. Like in the picture.



Now look at the picture



Consider all light rays to be almost perpendicular to the surface. So the difference of luminous path would be 2d and, like all interference book says:

If 2d = (2n+1)/2 λ - constructive interference
If 2d = n λ - destructive interference

This is due to the phase inversion in B.

OK, BUT WHERE THE HELL DOES THE INTERFERENCE OCCURS?
B isn't in the same place than D, so the interference can't occur ON the surface. Does it occur in the human eye? How? Because I don't thinks the 2 rays will actually hits each other exactly IN the human eye.

the_emi_guy

When thinking about interference of light, do not think about photons. Think about waves.
Light waves interfere with each other in the same manner as other types of waves. Go fill your bathtub and experiment with disturbing the smooth surface with different objects in different locations. Look at the interference patterns that occur.
Rays BE and DF are out of phase, so their amplitudes subtract. One is "pulling" while the the other is "pushing". Like two water waves encountering each other but one pushing water up while the other pushing the water down.

jaumzaum

I understand this push and pull interference. But the waves should be close (in other words, superposing) each other to interact, right? BE and DF aren't superposing each other. How can they interact? How can theey interfere?

Cthugha

What makes you so sure these are not superposing? There are at least three ways to answer your question. I do not know which is best for your level of understanding, so I will try all three of them.

1) In the kind of drawing you show, you see straight light rays. If light were really only located on a small spatial scale, it would suffer badly from diffraction and spread quickly. That is why books typically tell that they assume plane waves for the kind of image you show. That means you have lots of such parallel beams next to each other. Therefore, there is also a parallel beam to A-B which ends at D and overlaps perfectly with the CD-Beam.

2) The lateral extent of any wave cannot be made arbitrarily small. The lower bound is roughly the wavelength of the light involved. In the picture you have shown, yon need a total path difference of lambda or 0.5 lambda for constructive or destructive interference. As most of the difference in path is covered by the vertical thickness of the thin film, the horizontal offset is significantly smaller than that and therefore typically much smaller than a wavelength. As the lateral extent of the wave is typically larger than the wavelength of the light, the two beams will indeed overlap pretty well.

3) The do-it-yourself-approach. Read up on the Huygens-Fresnel principle, which states that every point of the light wave serves as a source of a spherical wave, grab a pair of compasses and draw the whole interference scenario yourself. This will not help you at the moment, but is very instructive in the long run.

jaumzaum

Thanks for the answers Cthugha.

The light I've considered in the second image was a light source, not only a light beam (I've draw n a single ray just to make it easier to understand).
I've got your point now. Yes, there are many other rays can interact and "superpose" the AB ray. Part of my question is answered. BUt I still have some doubts.

Look at the following image (third image)



In the second image we had that a ray parallel to AB that passes through D would superpose COMPLETELY DF. But that's because the surface was plane.

At third image we have that as the surface is not plane, the green ray cannot superpose completely the red one. They only intersect at B. The blue ray, eather. That way, the interference occurs only in a single point.

Now let's suppose an absurd thing for a moment (I know it's absurd but it is important for my understanding). If there was only the three beams drawed in the picture (no more light beam). For we to see the interference , should we put the eye exactly in A?

Cthugha

I suppose from a technical point of view, it would be much easier to put a screen there and watch the screen, but in principle yes. The interference will take place only at the positions where beams which are somewhat phase shifted with respect to each other actually have some spatial overlap.

jaumzaum

Thanks Cthugha.

A final and last question.

For we to predict (mesure) where the interference will occur, we have that the light beams (emmited by the light source) have to be in phase immediatelly before the hit the surface. How can this be possible? Is the sun light an example of that? Would the interference occur if the rays were not in phase?

[]'s

João

Cthugha

If the rays are not in phase initially, that basically just introduces an overall offset. The constructive interference occurs if two different waves are in phase at some point. If you now have an initial phase difference, the additional phase difference you need to achieve for interference by having the two beams take different paths is now not 2d = n λ anymore, but a bit shorter or longer.

For interference patterns to occur, it is more important that two beams have some fixed phase relationship. They need not be in phase, but the phase difference should not change over time. Light from the sun does not have this property as it is composed of many wavelengths and the phase changes randomly (due to the random nature of the light emission process) on a timescale of the order of femtoseconds.

Khashishi

If we measure the light pattern on a screen, we are viewing the interference at the screen. Any interference that occurs on the way to the screen doesn't matter, since we are not looking at that point. Different light sources don't interact with each other as they travel through space. They just pass through each other.

If you cross the beams of two lasers, the waves will overlap in the crossed region, but two beams will pass out the crossed region as if nothing happened.

jaumzaum

But if we put a screen in the crossed region, we would see interference right?

jaumzaum

Thanks Cthugha. The light I've mentioned was not composed by 2 beams only, but infinite beams that have a randomly phase difference (like the common-light we see in our houses, that is reflected many times). Would this cause a interference pattern?

Cthugha

No, if you have infinite beams at random phases, every possible phase difference is realized and therefore also every possible scenario from constructive to destructive interference (and partial interference in between). In summary all of these contributions cancel out and there is no interference at all.

sophiecentaur

The picture suggests the 'Newtons Rings ' phenomenon. This is visible with 'ordinary' light sources (the old slide projectors were very susceptible). It's only the first one or two fringes that were visible, though. The point is that all the "infinite beams" will interfere with themselves over a small range of angles, because there is a small degree of coherence. Obviously, with a laser source, the pattern will be much stronger and appear over a wider area of any image.
Light from all over a room will never give you visible fringes but once you start to collimate the beam (in a projector, for instance or just light from one direction), the effect starts to appear. The first thing you tend to see is coloured fringes as RG and B cancel at different angles.

jaumzaum

Thanks sophiecentar, but no one has answered my question yet. Would kashishish light cause interference in the crossed region?

sophiecentaur

What is that? I googled and got nothing relevant except your post. I am not sure what you mean by "crossed region" either. It would be best if you googled thin film interference and newtons rings for yourself. You will see plenty of diagrams which should help you understand this better. I think you may be making wrong assumptions about which of the light beams you will actually see in your diagram. This could account for your confusion.

I can say that light arriving from all directions that started off as incoherent won't usually produce visible interference patterns. However, if you look at the colours of oil films on puddles, you are seeing an interference effect and light is arriving from all directions (the sky). The interference is very much diluted in this case because of the large amounts of other light. Birds' feathers and butterflies' wings have vivid colours because of interference (not pigments) and, again, the light arrives from all over the place. In the case of so-called interference filters, the films are very thin and the 'fringes' are very wide - consisting of no more than one minimum, which will be a minimum for only one narrow range of wavelengths. The colours you see are 'complementary' colours and not spectral colours (e.g. bright magenta, where yellow has been eliminated)

jaumzaum

Kashishi is the guy up there, sorry

he posted:

I want to know if you put a screen at the crossed region of the 2 lasers, would you see interference?

Khashishi

yeah, you will see interference in the crossed region.