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Angle of Deflection of Laser Given Tilted Mirror

  1. May 4, 2015 #1
    Hi Everyone,

    A question in my latest Physics I lab wishes for a proof showing that if a laser beam is incident on a mirror that is then rotated an angle θ the beam is deflected an angle 2θ.

    I attempted to prove this geometrically below. In the diagram the angle γ, is the angle of deflection, and β is the incident angle. I was able to show that if the mirror was rotated θ degrees to the vertical, that β = θ. However,I haven't been able to show that γ = 2θ from the below diagram, as I am unable to determine the top right/bottom left angles of the parallelogram, or any additional angles of the triangles that exist in side of it.

    Any assistance or guidance would be very much appreciated.

    1. The problem statement, all variables and given/known data


    If a laser beam is incident on a mirror, and the mirror is rotated θ degrees, prove that the angle by which the beam is deflected is equal to 2θ.

    2. Relevant equations

    - Elementary geometric equations (e.g. sum of the interior angles of a triangle, properties of a parallelogram)

    3. The attempt at a solution

    YWlu2hN.jpg
     
  2. jcsd
  3. May 4, 2015 #2

    SammyS

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    How are the angles of incidence and reflection related?
     
  4. May 4, 2015 #3
    The angle of incidence is equal to the angle of reflection.
     
  5. May 4, 2015 #4

    SammyS

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    Did you use that anywhere?

    β = γ
     
  6. May 4, 2015 #5
    I see. That is what I initially thought, but then why is β ≠ θ when geometrically they seem to be equal when accounting for the isosceles triangle at the bottom of the diagram?
     
  7. May 4, 2015 #6

    SammyS

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    β does equal θ
     
  8. May 4, 2015 #7
    I see. Then given that β = γ = θ ≠ 2θ would that then mean that the angle of deflection is in fact not equal to twice the angle through which the mirror is rotated? Could it be a typo perhaps?
     
  9. May 4, 2015 #8

    SammyS

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    What are you calling the angle of deflection?
     
  10. May 4, 2015 #9
    γ
     
  11. May 4, 2015 #10

    SammyS

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    No.

    What is the angle between the incident ray and the reflected ray ?
     
  12. May 5, 2015 #11
    Uh, I see. This makes sense -- the reflected angle = β + γ. Given that θ = β = γ, the deflected angle = 2β = 2θ. Thank you very much for your help!
     
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