Index of refraction optics problem

AI Thread Summary
A light ray strikes a glass pane at a 60-degree angle of incidence, reflecting off a mirror behind it. The index of refraction of the glass is 1.54, and the pane is 5mm thick. The user calculated the angle of refraction as 34.2 degrees and determined the length of the ray within the glass. They suggest using trigonometry and the Pythagorean theorem to find the displacement of the beam compared to its path without the pane. The discussion emphasizes the complexity of the problem but provides a pathway for solving it.
brad sue
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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
 

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Edit: Ah I misread the question (didn't realize there was a mirror).
 
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See my image, where the capital letters are points and 'd' is the displacement.

http://img89.imageshack.us/img89/8789/optics9xq.gif

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

Now I'm pretty sure \angle FCE 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.
 
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Beam me down said:
See my image, where the capital letters are points and 'd' is the displacement.

http://img89.imageshack.us/img89/8789/optics9xq.gif

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

Now I'm pretty sure \angle FCE 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.
Make more sense now. Thank you I will try to go from here.
B.
 
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