
#1
Nov2209, 06:30 AM

P: 651

a waveguide consists of two mirrors separated by distance d, facing each other, the light ray that reflects between them forms angle (theta) with one surface. there is a reflection phase shift of pi.
for the light to continue to the end, there must be constructive interference between the first wavefront and the 3rd wavefront(the twice reflected ray). the equation given is 2dsin(theta) = mλ anyone knows why this is so? i don't understand what is the path difference to start with 



#2
Nov2309, 12:07 AM

P: 87

Ok so I know what happens but im going to try to explain it.....
So lets take this in terms that it is a laser. Then lets say that this laser consists of two waves. They are both in sinc which means that if you were to put them in a horizontal straight line and drew a straight vertical line then you would see that all the crests and troughs would line up perfectly. so this laser is going along and then it is encountered by the mirror. If you notice the top wave will hit first before the second. That means that the first wave will then be bouncing back will the other one will be still heading for the mirror. Once the second one hits the mirror, the first one has moved some distance 'x' on the rebound. If this distance coresponds to 1 wavelength of the laser frequency then nothing will happen, but if it is 1/2 of its wavelength then you will notice if you again put them on a horizontal line you will notice that when you draw a straight line the crests will match up with the troughs, and therefore canceling themselves out. and you will then not see a laser. Now i think this happens in like double slit experiments and such and also in crystal diffraction but mirrors have something different. But the way i described it should answer your question on why the angle is what it is Hopefully this helps Sincerely, FoxCommander 



#3
Nov2309, 04:11 AM

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PF Gold
P: 11,341

The initial model you describe is not clear. Is it a 'long' path between the two mirrors, with many reflections on the way? Is there only one 'mode' of propagation (i.e. are all the angles identical or is there a range of angles)?
You seem to be bordering onto a discussion of optical fibre propagation. What is the actual context of the post? Can you include a diagram? 



#4
Nov2309, 08:19 AM

P: 651

waveguideerm. ok... but i still don't know what is the path difference in this question 



#5
Nov2309, 08:22 AM

P: 651

i don't really understand what he's doing... hehex, 



#6
Nov2309, 09:13 AM

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P: 11,341

He seems to have only included one ray and you can't have a path difference between one thing. Unless there is light exiting along b cdashed. In which case it needs to be in phase with the light finally exiting out of the end.
He could be meaning a partsilvered mirror, where the portions of each exiting ray need to be in phase to get constructive interference in any particular direction. That's what his sums seem to imply. At other angles, the light will not be at a maximum coming out. It's a bit like the Newton's rings calculation and similar things. Give him a hard time and make him explain himself!! 



#7
Nov2309, 10:03 AM

P: 651

er does it have anything to do with the requirement i stated in the first post?




#8
Nov2309, 06:04 PM

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P: 11,341

I reread the OP and it would appear that, for light to continue to the end, you would actually want DEstructive interference between each ray of light light emerging through the sides and the next so that no energy would be lost and all the energy would get to the end. That would be when the path difference between each emerging ray was a half wave. That sort of makes sense now  at least to me.
:) 



#9
Nov2409, 02:05 AM

P: 651

??? oh my . i am lost lolx. i still don't really understand :(
i think i will post the original question. maybe i didn't explain it well in the 1st thread does it make sense ? 



#10
Nov2409, 04:50 AM

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P: 11,341

I have a problem with it,
If, as he says, you need constructive interference for light to emerge and if the reflections are perfect and if the Energy has to go somewhere (which it must) then it would have to be reflected back the way it came. (It can't be absorbed anywhere within the waveguide because there is no loss mechanism). The entrance to the guide would then appear as a perfect reflector at certain angles. To get interference, you normally need two or more waves in the same place  as in the two slits experiment or when light passes through a thin film with partial reflections. I can't identify two paths here where light can actually interfere. Apart from the impressive "[12marks]" at the bottom, the question looks wrong. You need to ask your guy where and how the interference takes place. Else you can assume he is right and, looking at the geometry of the diagram and work out the value of angle where the waves on the extreme left would be in step with the waves parallel with them on the extreme right. The distance bc would need to be a whole number of wavelengths. Thinking again about it, I suppose that you could imagine a whole bunch of 'rays' entering the waveguide gap over its whole width (the laser beam will have a finite width) and, because it is all coherent, then bits of one part of the beam could interfere with bits of another part of the beam unless the condition in the question is met. That may be what it's all about. Laser light does behave strangely because of its coherence  hence the speckles you see when it hits an object. Unlike a microwave waveguide, which is made as small as possible, an optical guide is pretty wide in terms of wavelength so what I know of normal waveguide didn't include this problem. Go and do the calculation I suggest  you will get a range of possible angles which satisfy the condition: an n sin(theta) or an n cos(theta) should be in there somewhere! Good luck. 



#11
Nov2409, 06:02 AM

P: 651

oh so bc is my path difference?
it looks kina profound to me i only learnt stuff like double slit and single slit where the path difference is kind of clear but in this question, i don't understand why he even extends the bc to bc'. so to start of, i presume the path difference is bc? so it has to be an integer of lamda for constructive interference? 



#12
Nov2409, 01:30 PM

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P: 11,341

bc isn't the distance you're interested in. That isn't a path difference, in fact (which was what I couldn't get my head round at the start) this diagram only shows you one ray going through. What I think you are after is the difference in path length for two rays, side by side, so that they will both have waves that are in step. I think this will happen if you make sure that the angle is such that the wavefront going across at b is a whole number of wavelengths from the wavefront going across at c. Under those conditions, I think any wave starting at the beginning of the waveguide, over the whole width, will be in step with any other  thus avoiding destructive interference. You could try drawing the diagram with two rays (different coloured pens, perhaps) and you'll see that they will only be in step when 'my' condition is met.
Looking at the hand drawn diagram, I think the extended line is on the way to doing the calculation I suggest but he certainly hasn't made it clear to a 'humble student' what he wants you to do. Typical teacher (I should know  I am one!) Carry on . . . . . 



#13
Nov2409, 08:29 PM

P: 651

huh? i don't understand.
does it have anything to do with the pi reflectance every time the ray hits one of the mirrors? and what exactly is my path difference lolx! 



#14
Nov2509, 01:47 AM

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P: 11,341

The difference between the distance that two rays will travel down the guide. Your diag just shows one but they enter the guide over its full width. Any ray could destroy another unless the criterion I suggested is met. If the path difference is n lambda then there is no destruction.
There are more ways of killing a cat than drowning it. Just work out the angle as I said and you can get the answer. 



#15
Nov2509, 06:25 AM

P: 651

ok, so dsin(theta) = m lamda. so where did the 2 appear from? AD is my path difference right? does the 2 in my lecturer's answer come out due to the pi reflectance? also how do i account for the 2 times of pi reflectance? apparently my lecturer did account for it in his answer when he  2pi. but is his answer some form of general solution or something ? 



#16
Nov2509, 09:24 AM

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P: 11,341

There are two reflections more for one ray than for the other  so 2pi is in there. But, as it is a whole cycle, it doesn't matter. What matters is the physical difference in path length for the two rays.
The path difference of interest corresponds to the difference in length between the distance to the first of the red reflections and the distance to the third of the blue reflections and, indeed, to the third of the red reflection, too.. If that is n lambda, then there won't be any destruction. The factor of two is due to the fact that the beam travels along two diagonals, each one being d sin theta . The waves / rays are totally in phase when they enter the guide and they will stay that way at the end if the total path for each is always n lambda. This only happens for the correct angles (there may be more than one solution if the guide is wide enough). 



#17
Nov2509, 09:38 AM

P: 651

so the path difference is AD?
or did you mean the path difference is the horizontal distance between the 1st red reflection point to the 3rd blue reflection point? oh so there will be a path difference for the 1st reflections and a path difference for the 2nd reflections so total is times 2? 



#18
Nov2509, 11:22 AM

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P: 11,341

The path difference refers to the paths followed by the waves. That's where the d sin theta bit comes from. The point is that light enters the guide over its whole width. When it comes out, it is essential that its actual path (not the horizontal distance) inside is the same for all the light  or an integral number of wavelengths different. So, if the route via an extra double bounce is n lambda different from the route involving no extra bounces then this condition is met. If it's not n lambda different then you can get cancellation of some rays by others.
Look, you can't have a difference unless you are comparing two paths (not reflections)  isn't that obvious? It's what happens on the actual way out of the guide. Some rays will come out with two more bounces than others and will be mistimed but the mis timing doesn't matter if its a whole number of cycles. Two of those diagonal lines represent this extra distance traveled by those rays on the extra two bounces. Have you seen the Newton's rings and 'oil film' stuff in textbooks? It's the same thing at work; two different routes through the system need to be n lambda different for a maximum of constructive interference. I have told you more than enough by now for you to work out what it's all about, I'm sure. Go back and look at other examples and explanations of interference and diffraction if it still isn't clear and start from scratch. 


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