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Specular reflection from moving mirrors

  1. May 25, 2012 #1
    Question:

    Does the law of specular reflection (angle in = angle out) for obliquely incident light rays still hold if the mirror is translating in a direction parallel to its normal direction?

    If not, what different effects attend said translation if

    a) the motion is toward or away from the stationary light source, and

    b) the mirror velocity is constant or varying?
     
  2. jcsd
  3. May 25, 2012 #2

    mfb

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    Staff: Mentor

    You can transform the system to get a stationary mirror. If the mirror is moving towards the light source, the transformation reduces the angle relative to the normal vector of the mirror plane. The outgoing angle is then the same as the incoming one. A backtransformation reduces the angle a second time.

    If the mirror is moving away, the angle is increasing.

    You can look at the velocity at a certain time for the reflection angle at this time. Effects from acceleration are negligible, unless you have really unrealistic values for that.
     
  4. May 27, 2012 #3
    Hey MFB!

    Thanks for your reply, but I'm not sure how to diagrammatically set-up or mathematically carry out the transformation you're talking about.

    Say the mirror is moving away from the light source (along the direction of it's normal with fixed velocity v) and the light ray is incident with angle "alpha" and speed c. If "beta" is the outgoing angle, what equation relates it to alpha?

    Let me explain what I think you mean, then if I have it right, I'd appreciate you letting me know. Thanks again for your help.

    1) break out the incoming light velocity c into its components parallel and perpendicular to the mirror surface using the product of c with the sine and cosine functions of angle alpha.

    2) reduce the magnitude of perpendicular component of the incoming light velocity by subtracting the speed of the mirror v, i.e. c x cos[alpha] - v.

    3) reflect the reduced perpendicular component and vectorally add the reversed perpendicular component to the original parallel component of the incoming ray to get the total reflected ray

    4) calculate "beta" angle as pi/2 minus the arc tangent of the ratio {[(c x cosine alpha) -v]/ [c x sine alpha]}

    5) since the magnitude of the reversed perpendicular component is in this instance less than the original perpendicular component, beta will have a larger value than alpha.

    6) if the mirror were moving towards the light the whole analysis would result in a beta less than alpha.

    Hope I'm right; thanks again, Eriel
     
  5. May 28, 2012 #4

    mfb

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    I think this should work as you described it.
    In addition, you get a frequency shift due to the moving mirror.
     
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