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Angle of Incidence with partial reflection and partial refraction

  1. Mar 31, 2010 #1
    1. The problem statement, all variables and given/known data
    A beam of light in air strikes a piece of glass (n=1.51) and is partially reflected and partially refracted. Find the angle of incidence if the angle of reflection is twice the angle of refraction.


    2. Relevant equations
    n1sintheta1 = n2sintheta2


    3. The attempt at a solution
    The angle of reflection equals the angle of incidence, which is twice the angle of refraction.
    Therefore, I set up the equation as:
    1 * sin (2theta) = 1.51 * sin (theta)
    But when I try to solve for theta I get that they cancel out (sin 2theta / sin theta) just equals 2? I think I'm doing the trig wrong but I don't know why?
     
  2. jcsd
  3. Mar 31, 2010 #2
    Yeah you are doing the trig wrong. You cannot just take the 2 out of the sin function and cancel.

    Use this trig identity: sin(2x) = 2 sin x cos x
     
  4. Mar 31, 2010 #3
    yep, i'm awful at trig so that isn't surprising. haha.

    i'm still not getting it right though: if i use that identity, i get:

    1 * 2 sin theta cos theta = 1.51 * sin theta

    If I divide the left by sin theta, the sin theta cancels out (right???), leaving me:
    2 cos theta = 1.51
    which gives me a theta of 40.97 deg, which is wrong...

    Not sure what I'm doing wrong now...
     
  5. Mar 31, 2010 #4
    You need to double that angle, they ask for the angle of incidence not the angle of refraction.
     
  6. Mar 31, 2010 #5
    right. i'm an idiot. thanks for your help!
     
  7. Mar 31, 2010 #6
    Yep no problem.
     
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