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At what frequency does the rotating mirror need to turn?

  1. May 30, 2013 #1
    A rotating mirror with 16 sides was used to measure the tine it took light to travel 3.5km to a concave mirror and back. At what frequency did the rotating mirror need to turn to make 1/16th of a rotation in the time it took light to travel to 3.5km and back again?



    im having a bit of trouble with the second part. is this correct?


    d=3.5x10^3m x 2
    =7x10^3m
    c=speed of light
    =3 x 10^8 m/s
    c=d/t
    t=c/d
    t=3x10^8m/s/ 7x10^3m
    =4.28x10^4 s

    t= 1/8 T
    T= t/(1/8)
    =4.28x10^4s/(1/8)
    =3.42x10^5 s


    T=1/f
    f= 1/T
    =1/(3.42x10^5)
    =2.92x10^-6s
     
  2. jcsd
  3. May 30, 2013 #2

    gneill

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    Staff: Mentor

    Does that time for the round trip of the light seem reasonable to you? It's nearly half a day!
     
  4. May 30, 2013 #3
    ok so is it rather. t= d/c?
     
  5. May 30, 2013 #4
    so is it rather t= d/c or t=dc?
     
  6. May 30, 2013 #5

    gneill

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    Staff: Mentor

    What does units analysis tell you? What are the units of distance, speed?
     
  7. May 30, 2013 #6
    time= speed of sound/distance or
    s= m/s / m
    dont the the meters cancel out? giving seconds?
     
  8. May 30, 2013 #7

    gneill

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    Staff: Mentor

    Do the seconds end up in the numerator or the denominator? And what's with the "speed of sound"? There's no sound involved here. You just need to invoke the base units: m,kg,s.

    If you want to end up with units of time, then time better end up in the numerator :wink:
     
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