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Physics3, angle of refraction, beam displacement

  1. Feb 26, 2009 #1
    hi everyone, i've got another question. i feel good about my methods, but i'm getting an incorrect answer.

    1. The problem statement, all variables and given/known data
    A ray of light impinges at a 60° angle of incidence on a glass pane of thickness 6 mm and index of refraction 1.50. The light is reflected by a mirror that touches the back of the pane (Fig. 35-37)(see attachment). By how much is the beam displaced compared with the return path it would have if the pane were absent?

    2. Relevant equations
    n_1/n_2 = (sin theta_2) / (sin theta_1)
    opposite/adjacent= tan theta
    c^2= a^2 + b^2

    3. The attempt at a solution

    (descriptions may be terrible w/o diagrams/letters, as I am without a drawing program just now)

    so I use the first equation above to get the angle of refraction, I'll call theta_1. I've used the above tangent equation to get the bottom-length=opposite-side of the right triangle formed by the ray, the the mirror, and the vertical line to the left of the ray inside the mirror. twice this bottom length is the distance between where the ray enters the top of the glass and where the ray leaves the top of the glass, the distance between the two vertical dashed lines, a distance which I'll call x.

    i need to find the distance between where the beam would 'naturally' strike the mirror and the point beneath where it leaves the glass (the right dashed line). i'll call this distance y.

    To do this I need to consider a third larger right triangle formed by the left vertical dashed line, the line of the ray's 'natural' path to the mirror, and the mirror. the height of this triangle is given. its upper angle has to be 60 degrees (just like the given angle), by the geometry of two lines crossing having equal angles kitty-corner. So using the tangent formula, I can get the length of the bottom of this triangle, = the distance y, along the mirror.

    Subtracting this longer distance y from the shorter distance x gives the distance between where the ray would 'naturally' strike the mirror and the point beneath where it leaves the top of the glass.

    for this small right triangle, I know the height (given), and the base is y-x, so by Pythagoras, I calculate the length of the hypoteneuse.

    in my understanding, the length of this hypoteneuse is the displacement of the beam and a would-be beam that never got refracted.

    i tried an identical practice problem and got it right, but my answer to the homework question is wrong (somehow). any tips or suggestions would be helpful, and if there's any questions to clarify the problem, please do ask and I'll provide further details.

    Again, sorry about the difficulty in description. eventually I'll figure out how to draw on OS X. (no paint brush!)

    thanks much,

    Attached Files:

  2. jcsd
  3. Feb 26, 2009 #2


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    Sounds good! So 35.3 degrees for the angle of refraction and the ray comes out 8.5 mm away from the point of incidence on the glass?
  4. Feb 26, 2009 #3
    i agree with both of those figures, but the displacement that this leads to for me, 6.29577 mm, is apparently incorrect. what I want is just the distance between the two parallel lines/beams, right?... because that's what i thought i've calculated.

    i don't want to doubt the professor/answer, but on this problem i'm running out of ideas for how I could possibly do this differently.
  5. Feb 26, 2009 #4


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    I have the beam - without the glass - emerging 20.8 mm over.
    tan(60) = x/6 where 2x is the horizontal distance the beam travels.
    This is at the top surface of where the glass would be.
    The displacement is 20.8 - 8.5 = 12.3 mm
  6. Feb 26, 2009 #5
    ok, I just tried that, and my answer is incorrect. (using my numbers, it came to be 12.2993 mm) also, significant digits are ignored, I just have to be within 1% of the right answer.

    but good thinking. because now I see that the displacement of the beam that you're describing and the displacement I described are two different things. to me, it's not clear in the problem which they actually want. a simple little thing, like an x inscribed somewhere on the diagram, would have been so helpful in determining what they want.

    be that as it may, i tried that and it didn't work, so.... i'm still not sure what i'm doing wrong.
  7. Feb 26, 2009 #6


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    I don't understand what you found equal to 12.2993. Maybe I made a mistake - often do.

    I found the horizontal displacement of the beam, but perhaps we are supposed to find the distance between the two parallel beams. That would be the horizontal answer times cos(60).
  8. Feb 26, 2009 #7
    yes, my 12.2993 and your 12.3 are the same, the horiz. displacement.

    what i found earlier, the 6.29577 mm is (what I think is ) the distance between the two parallel beams. well, taking the horiz answer 12.2993 times cos 60 = 6.14966, which is a new number and potential answer. but before i try it out, i want to understand what it means.

    which triangle in the diagram are you using for this horiz * cos(60) ?... i know that cos is adj/hyp of a triangle, and the horizontal is the 12.2993 mm between the two vertical dashed beams, but... i can't figure out which triangle you're using.
  9. Feb 26, 2009 #8


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    Two parallel beams emerge at 60 degrees from the glass surface.
    Draw a line perpendicular to the beams from one to the other, with one end on the glass surface.
  10. Feb 26, 2009 #9
    well that answer that you proposed is correct. i delineated the triangle you were talking about, but I haven't convinced myself that it's a right triangle. the concept i couldn't get is how you were treating the horizontal displacement as the hypoteneuse, since all the distances i'd been finding out were horizontal or vertical lines, bottoms and sides of triangles.

    for me, there's just an errant unknown angle that has to be 30 degrees so that that particular triangle can be right, and i haven't yet convinced myself that it's 30 degrees by other angles that I know. geometry is a bit rusty, i suppose. but since it's right, eventually i'll figure out why it has to be 30 degrees.

    thanks for your help!
  11. Feb 26, 2009 #10


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    The "displacement between two rays" should be the perpendicular distance between them.

    Most welcome - I enjoyed the problem!
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