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Index of refraction optics problem

  1. Apr 15, 2006 #1
    Hey , I need some indication to solve this problem:

    A ray light impinges at a 60 degres angle of incidence on a glass pane of thickness 5mm and index of refraction of 1.54.
    the ligth is reflected by a mirror that touches the back of the pane. By how much is the beam displaced compared with the return path it would have if the pane were absent?

    I found the angle x --normal with the refracted ray. x= 34.2 degres.
    I also found the length of the ray entering in the glass (up to the mirror)--L= 5.10-3\ cos(x)=6.0 10-3m

    From here what can I do to find the displacement?
    I attach a picture of the problem.

    Thank you

    Attached Files:

  2. jcsd
  3. Apr 15, 2006 #2
    Edit: Ah I misread the question (didn't realise there was a mirror).
    Last edited: Apr 16, 2006
  4. Apr 16, 2006 #3
    See my image, where the capital letters are points and 'd' is the displacement.

    http://img89.imageshack.us/img89/8789/optics9xq.gif [Broken]

    In the situation [tex]\angle EBF = 90 - 60 = 30[/tex]. Which indicates that [tex]\angle BEF = 180 - 30 - ( 90 + 30 ) = 30[/tex] as points BEF form a triangle (which have a maximum of 180 degress, so from that subtract [tex]\angle EBF[/tex] then [tex]\angle BFE[/tex]).

    Now I'm pretty sure [tex]\angle FCE[/tex] forms a right angle. If they do you then get a whole series of right angle triangles and using trignomentry and pythagoras's theroem enable you to work out "d".

    This is probably a cumbersome way of going about it, but its the only way I can see at the moment.
    Last edited by a moderator: May 2, 2017
  5. Apr 18, 2006 #4
    Make more sense now. Thank you I will try to go from here.
    Last edited by a moderator: May 2, 2017
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