Physics Help Forum Proof of total internal reflection

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 Oct 25th 2009, 09:35 PM #1 Junior Member   Join Date: Feb 2009 Posts: 15 Proof of total internal reflection I have to prove that when a ray in the air (n=1) is penetrating in a long solid where n=1.5, there will always be a total internal reflection no matter what the orientation of the ray incident is. First, I believe by "orientation" they mean the angle, correct me if it could be something else. I tried to check what happens when I set the ray incident angle to 0 degrees and 90 degrees. (extreme values, so theoretically everything in between works too) For: (LaTeX doesn't seem to work at the moment) 1*sin 90 = 1.5*sin A I have A = 41.81 degrees and for: 1*sin 0 = 1.5*sin A I have A = 0 degrees Now I want to find the critical angle inside the solid that tells me if a ray will do a total internal reflection or not: 1.5 * sin B = 1 * sin 90 I have B = 41.81 degrees (of course since it is the same equation as the first) So since my refracted angle is always in-between 0 and 41.81 degrees, we can say that it will always be a total internal reflection. This is my "solution" but I believe it is flawed (I have serious doubts about the method I used to find the critical angle) Any help would be appreciated! Thanks a lot in advance.
 Oct 26th 2009, 12:06 AM #2 Physics Team   Join Date: Feb 2009 Posts: 1,425 The method used is OK, but there is a point you have missed. For the angle of incidence of 90, you have the angle of refraction as 41.81. But this is not the angle with which the ray is incident on the side of the solid. That angle will be 90 - 41.81 = 48.19. It will be clearer if you draw a ray diagram. However since this is greater than the critical angle, T I Reflection occurs anyway. Actually 90 degree incidence may cause the ray to just go past and not refract at all. So what we are really interested in is an angle infinitesimally smaller than 90. This will make the angle of refraction slightly less than 41.81 and the next angle of incidence greater than 48.19 which is definitely more than the critical angle. Hence our previous argument holds .
 Oct 26th 2009, 12:57 PM #3 Junior Member   Join Date: Feb 2009 Posts: 15 So would it be safer to say (except for the straight-through case) that the angle of refraction will always be in-between (90-41.81) and (90-0), so between 48.19 and 90, which is always greater than the critical angle of 41.81? Here is a quick mockup: Is it something like that? (3 rays are on it)
 Oct 26th 2009, 10:12 PM #4 Physics Team   Join Date: Feb 2009 Posts: 1,425 Yes that is the case, except that the solid is supposed to be very long. So long in fact that the blue ray would strike the upper side , not the opposite side. In fact i was puzzled about the condition "long solid" as that seemed unnecessary. But your diagram made things clear. Thanks. And how have you made this diagram? I am generally lost on stuff like this. Last edited by physicsquest; Oct 26th 2009 at 10:19 PM.
 Oct 27th 2009, 04:21 AM #5 Junior Member   Join Date: Feb 2009 Posts: 15 Oh I see what you mean. Yeah I just reproduced the diagram from the manual (of course the rays weren't on it :P), but I can understand the unpracticality of printing an extremely long solid on a page :P I made the diagram quickly using Photoshop (which is supposed to be used for photo editing haha) Thanks a lot for your help!

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# how can we prove internal reflection

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