# Complete Destructive Interference?

1. Dec 19, 2004

### jfrusciante

A teacher of mine showed this problem to me and could not see a solution just as i cannot either.

A single laser beam is sent on a path as in the diagram enclosed as an attachment.

If you look in the diagram at both detectors, the beams will be totally out of phase. The path difference is half a wavelngth and the total difference in relflections between the two is an even integer so the beams are 180 degrees out of phase. Therefore no light should be observed at each detector.

How is this possible as this would surely result in complete destructive interference and violate the conservation of energy?

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2. Dec 19, 2004

### dextercioby

By the looks of that diagram,there is no problem with the conservation of energy.Half of it will do to one detector and the other half at the other detector.As for destructive interference,i'll think about it.

Daniel.

3. Dec 19, 2004

### jfrusciante

I think both detectors will have destructive.

Here is why...look at this break down for each mirror:

Mirror 1 (beam splitter): half goes straight on half is refelcted, thus 180 out of phase

Mirror 2: Beam has travelled extra 1/4 of a wavelength compared to straight on beam. It is reflected again and thus now 450 degrees out.

Mirror 3: Both beams travel same horizontal distance but the second beam is reflected again and thus another 180 degrees. The total difference is now 630 degrees.

Mirror 4 (second beam splitter):
The second beam now travels an extra quarter of a wavelength and carries straight on at the beam splitter. The beam that has not been reflected yet partly reflects up and puts that 180 degrees out. Therefore the total phase difference at top detector is 900 degrees...which is just 180 degrees. Therefore destructive interference.

At horizontal detector the second beam is reflected again and the first beam carries on, so we have 630+90+180=900 degrees.
Therefore 180 degrees out of phase.
Therefore destructive intereference.
Destructive interference at both detectors.

That may have been quite hard to follow as i did that rather quickly but i am pretty sure that argument is correct.

4. Dec 19, 2004

### Gonzolo

You would see circular fringes of alternating maxima and minima. With the path difference you have chosen, the center of the spot would be black.

The path difference is on the order of a wavelength, so you have to consider related effects, such as linewidth.

5. Dec 19, 2004

### jfrusciante

I don't think i really understand what you mean . What is linewidth?
Thanks.

6. Dec 19, 2004

### Gonzolo

The model which you are aware of, that may give 0 intensity, is that of a plane wave, which is crucial for understanding wave phenomena, but does not actually exist.

A real laser, as well as a single photon, has a middle wavelength plus an incertainty. Imagine a bell-shaped curve. Its width is the linewidth. For a He-Ne laser : 632 nm plus or minus a few nm.

Also, in a real laser, the beam naturally diverges a tiny bit. You may want to see this as all photons not coming out in the exact same direction (although a single photon also diverges), so that the path length of the beam has its own uncertainty. This path length uncertainty and slight divergence will cause the circular fringes, perhaps more than the linewidth actually.

Last edited by a moderator: Dec 19, 2004
7. Dec 19, 2004

### pervect

Staff Emeritus
I think your phase assumption of 180 degrees is wrong for the thin film half silvered mirrors. However, I'm not positive. It's also possible that the thin film mirrors are lossy. Eiither possibility gets rid of the cosnervation of energy problem. But my memory is that the half silvered beamsplitters give two beams 90 degrees out of phase without loss, (but I couldn't find a reference to check my fallible memory). But here's the argument that it must be one or the other.

The only lossless beamsplitter is a polarizer. A beamsplitter that is not a polarizer is lossy.

Draw a vector diagram of the E fields (assuming plane waves).

If you split an E field of 1, pointing north, into two parts, .5 north and .5 north, you will have two beams with an intensity of E^2 = .25,, for a total power of 1/2 the original beam.

The only way to conserve energy is to split the E field into a pair of orthogonally polarized waves - for example, .707 northwest and .707 northeast.. Then when you add up the intensites (E^2) you get 1.