1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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)


    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted