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Speed of a moving laser due to refraction

  1. Apr 29, 2013 #1
    1. The problem statement, all variables and given/known data
    A transparent cube with index of refraction n = 1.6 and side length a = 4.0 cm is free to rotate about an axis passing through its center. A laser beam is aimed at the cube. After passing through the cube the beam hits a screen on the other side forming a spot. The spot is displaced a distance y from the straight line path. The cube is then rotated at a constant angular velocity ω = 10 rad/s. As the cube rotates the spot moves.

    What is the speed of the spot at the instant that y = 0? (Answer 15 cm/s)

    uFnPftE.jpg

    2. Relevant equations
    Snell's Law: n1*sin([itex]\theta[/itex]1)=n2*sin([itex]\theta[/itex]2)


    3. The attempt at a solution
    I think I have to find y in terms of theta(from the center of the cube to the midpoint of edge of the cube), time t, side length a, and n. Then I could find dy/dt when y = 0. However I am unsure how to start or how I would use the angular velocity. Does the refracted laser inside the cube always hit the midpoint of the side like in the picture?
     
  2. jcsd
  3. Apr 29, 2013 #2

    rl.bhat

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    In the figure, y is the lateral shift. Write down the expression for the lateral shift in terms of refractive index and the angle of incidence. Then find dy/dt to find the velocity.
     
  4. Apr 29, 2013 #3
    I have this expression for the lateral shift.
    y = a*[sinθ√(1-(sin2θ/n2) - [itex]\frac{1}{n}[/itex]cosθsinθ]/[√1-(sin2θ/n2)]

    I'm not sure it's correct, it seems overly complicated.

    So I assume the incident angle moves at the angular velocity ω=10rad/s, how do I use it with the expression for y? I know ω=dθ/dt, but all the trig in the expression is confusing me.
     
  5. Apr 29, 2013 #4

    rl.bhat

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    Further simplification gives you
    y = a(sinθ - remaining term)
    Now the velocity is given by dy/dt = dy/dθ*dθ/dt
    Find the derivative of y with respective θ. At y = 0, what is the value of θ? Put this value in the above derivative and find the velocity.
     
  6. Apr 30, 2013 #5
    The method makes sense, but I am still not getting the right answer.

    if y = a[sinθ - other term]

    then dy/dθ = a[cosθ - (derivative of other term w/ sines and cosines)]

    and I assume the incident angle would be 90° because it would be a straight line through the cube.

    I plug in 90° in dy/dθ, the sines all = 1 and the cosines = 0 so it's fairly straight forward to calculate.

    I want dy/dt which is dy/dθ * dθ/dt so I would multiply dy/dθ by 10rad/s to get the final answer.

    In the end I get 32cm/sec.
     
  7. Apr 30, 2013 #6

    rl.bhat

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    When y = 0, the angle of incidence of the light is zero.
    Now write down the derivative and put the angle equal to zero.
     
  8. Apr 30, 2013 #7
    Finally got it, thanks so much.
     
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