# Index of refraction optics problem

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

#### Attachments

Edit: Ah I misread the question (didn't realise 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 [Broken]

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 [Broken]

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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