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The distance between the incident point and the refracted point

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  1. May 19, 2016 #1
    1. The problem statement, all variables and given/known data
    http://s32.postimg.org/4urcaq0xx/image.png
    The incidence angle of a light ray over a tile of glass with dy faces parallel is pi/3 rad. Find the distance between the incidence point and the refraction one of the ray with the tile, if the distance between these rays is 0.7 cm.

    2. Relevant equations
    sina/sinb=n2/n1

    3. The attempt at a solution
    sin pi/3/sinb=1.5/1 sinb=1/3
    tga=sqrt (2)/2 x=0.7*sqrt(2)/2=0.5
     
  2. jcsd
  3. May 20, 2016 #2

    BvU

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    Hi Zade,
    Your post is a bit hard to follow, since you don't list the meanings of the symbols you use - and I don't see any in the drawing (which suggests 60 degrees is halfway between 0 and 90 degrees ?). Furthermore $$\sin{\pi\over 3}\ne {1\over 2}$$

    You don't really ask a question and end your post with '=0.5'. Do you think the distance the exercise asks for is 0.5 cm ?
     
  4. May 20, 2016 #3

    Merlin3189

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    Perhaps it would help if you marked on your diagram (or on a diagram); the 0.7cm and the distance to be calculated.

    You don't mention any refractive indices: are you assuming air=1 (I assume it's air) and glass = 1.5 ?
     
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