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Loss in Mirror Reflection

  1. Nov 9, 2005 #1
    I have two mirrors at the same height on opposite walls facing each other in my bathroom.
    I took a laser pointer and put it in between the mirrors and directed the beam at one, such to where the reflected beam hit the other mirror, and back on the first and so on.
    Anyway, I noticed probably 16 or more multiple reflections of that laser beam.

    Of course, each time the beam is relected a loss occurs. I understand that.
    My question is how many times can that beam be reflected between the mirrors before becoming too faint for me to see.
    Perhaps to put it in another way, what is the loss percentage of "standard glass mirrors"? Is it "generally" 1% or much higher? I realise that that depends on the specific materials used for mirror construction, but is there a "generalization"?
    Also, are there "mirrors out there that have a reflective loss percentage lower than 1%?
     
  2. jcsd
  3. Nov 9, 2005 #2

    Claude Bile

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    Can't say what the reflctivity of an ordinary mirror is off the top of my head, I would imagine it would be at least 90%. High quality AR coated laser mirrors can have reflectivities yp to 99.9999% (Yes, that is a quoted percentage).

    Claude.
     
  4. Nov 9, 2005 #3
    I think it depends on how smooth the surface of the mirror is. If the mirror is perfectly smooth, in theory there should be no loss in the intensity of the beam no matter how many times it is reflected. If the mirror has imperfections, then only part of the light will be reflected and the other part will be sent away in a random orientation (due to "diffuse reflection" I think it is called) because it will be deflected away by any imperfect surfaces on the glass. This at least would account for the "loss" in intensity after being reflected off the glass multiple times (each time losing a little bit of the incident light due to diffuse reflection).

    Course I could easily be confused about this so please correct me if my thinking is in error. As to whether or not you can make a "perfectly smooth" mirror that will transmit light without any loss I have no idea if one has actually been made, but I assume the closer you get to that the more reflections it will take before the beam becomes too faint to see (disclaimer: my knowledge on this matter is purely acadamic, and might have zero basis in reality :redface:)
     
    Last edited: Nov 9, 2005
  5. Nov 9, 2005 #4

    ZapperZ

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    In principle, even if the surface is perfectly smooth, you will still lose light. This is because the metallic surface (the typical mirror is a metal) has conduction electrons that are responsible for the reflection process. As they move and react to the light's electric field, they lose a fraction of energy due to the non-zero resistivity.

    Zz.
     
  6. Nov 9, 2005 #5
    So in reality there would be a small amount of heat loss on each reflection? Is that where the energy loss is going?

    I should think that a perfectly reflecting mirror would be able to make a laser of unimaginable power (capabable of performing nuclear fusion for reactions), and the fact that such a laser hasn't been discovered yet supports what you are saying, I just want to be clear on how to interpret the effect of resistivity.
     
  7. Nov 9, 2005 #6

    Danger

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    Another loss factor is the air between the mirrors. There'll always be scattering and absorbtion/emission losses. Before that takes effect, your beam will be off the edge of one of the mirrors anyway because you have to fire the laser in at an angle to prevent the pointer itself from being in the beam path. (Or can that be remedied by angling one of the mirrors? I don't have time to think that through right now.)
     
  8. Nov 9, 2005 #7

    ZapperZ

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    When light hits a metallic surface, the conduction electrons actualy ABSORBS the photons, causing a transition from one conduction band to another (there's a whole spiel here that I'm skipping about the transition is actually between two points between two different band seperated by a reciprocal lattice vector). But this doesn't last and the conduction electrons will drop back to the original band and re-emit this photon, preserving the transverse momentum but with a 180 phase difference. Classically, the conduction electron gains energy from the oscillating E-field in the photon, and re-radiate this field. But since the electrons have negative charge, the field shift phase by 180 deg.

    So if one prefers, the light being reflected is not the same light that was incident onto the mirror.

    Now, since the oscillating conduction electrons is in a material with a finite resistivity, it will always lose energy via ohmic heating. This is unavoidable, no matter how smooth the surface is. This is why copper waveguides are lossy.

    Now, one can greatly diminish this by using a superconducting surface. Unfortunately, one runs into the same problem. While superconductors have zero DC resisitivity, it does not have a zero AC resistivity. Since the E field in light is an oscillating field, it implies that these losses are unavoidable, even under the perfect condition.

    Zz.
     
  9. Nov 9, 2005 #8
    Ok. So, If the mirror becomes electrically charged(below surface plasma disharge), does that affect reflectivity?
    In like manner, If one were to apply a high frequency AC potential to a metallic mirror, would photon iteraction be affected?
     
  10. Nov 9, 2005 #9
    Thanks Zapper, that reply was satisfying.
     
  11. Nov 10, 2005 #10

    Mk

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  12. Nov 10, 2005 #11
    heh, just the other day I was wondering why in a completely dark room with 2 mirrors, why does light in 2 mirrors stop reflecting the second you turn the light off, shouldnt there be a delay? I know the speed of light is just over 186,000 miles per second. But I wonder why would it dissipate so fast in a completely dark room.
     
  13. Nov 10, 2005 #12

    Mk

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    The thread I just posted elaborates on that.

    This is assuming in a vacuum.
     
    Last edited: Nov 10, 2005
  14. Nov 10, 2005 #13

    Claude Bile

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    A perfectly reflecting mirror alone would not accomplish this, it would only serve to reduce the threshold of the laser and increase the cavity lifetime. The increased cavity lifetime will allow you to obtain higher energy pulses, but that is still subject to a number of other considerations such as gain medium, pump power etc. and would not necessarily mean a higher average power, though the peak power would certainly be increased.

    Claude.
     
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