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Old May 20th 2008, 02:55 PM   #1
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Reflection after having reached the critical angle?

Is this equal to the angle of incidence or is it equal to angle of incidence-the critical angle. Judging by the simple laws of reflection I'm guessing that it's the first one, in fact I'm very sure. Some confirmation?
Thanks, Rob
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Old May 20th 2008, 05:01 PM   #2
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Originally Posted by rsmith View Post
Is this equal to the angle of incidence or is it equal to angle of incidence-the critical angle. Judging by the simple laws of reflection I'm guessing that it's the first one, in fact I'm very sure. Some confirmation?
Thanks, Rob
I'm not sure what you mean. Below is a nice diagram which I've shamelessly copied from here.

The incident ray reflects. Thus the angle of incidence is equal to the angle of reflection. The reflection angle is the critical angle by definition.

-Dan
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Reflection after having reached the critical angle?-u14l3b2.gif  
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Old May 21st 2008, 02:53 AM   #3
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Yeah that diagram helps. It just seemed strange that once the critical angle had been reached, that the TIR ray would suddenly jump to equal the incident ray. What I thought might happen is that once the critical angle had been reached, every degree past that critical angle would equal the IR ray if you get what I mean. It's kind of logical I suppose, but it doesn't obey the laws of reflection. Just thought I'd check. Thanks, Rob
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Old Nov 5th 2008, 03:28 AM   #4
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It is equal to angle of incidence when light is travelling from denser to lighter medium
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Old Dec 8th 2009, 06:31 AM   #5
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hi there , i think you are CORRECT
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Old Dec 14th 2009, 02:06 PM   #6
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Originally Posted by rsmith View Post
Yeah that diagram helps. It just seemed strange that once the critical angle had been reached, that the TIR ray would suddenly jump to equal the incident ray. What I thought might happen is that once the critical angle had been reached, every degree past that critical angle would equal the IR ray if you get what I mean. It's kind of logical I suppose, but it doesn't obey the laws of reflection. Just thought I'd check. Thanks, Rob
There is no jump.
Some of the light is reflected at any angle (and the rest is transmitted/refracted). The amount that is transmitted (refracted) decreases as the angle of incidence approaches the critical angle and becomes zero at this specific value.
So what happens is that for incidence at an angle equal or larger than the critical angle there is only reflection and no refraction (that's why is called Total internal reflection).
The reflection part of the ray goes according to the usual law of reflection for any angle, below or above the critical angle.
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Old Jan 16th 2010, 03:31 PM   #7
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Smile perfect

Originally Posted by mircea View Post
There is no jump.
Originally Posted by mircea View Post
Some of the light is reflected at any angle (and the rest is transmitted/refracted). The amount that is transmitted (refracted) decreases as the angle of incidence approaches the critical angle and becomes zero at this specific value.
So what happens is that for incidence at an angle equal or larger than the critical angle there is only reflection and no refraction (that's why is called Total internal reflection).
The reflection part of the ray goes according to the usual law of reflection for any angle, below or above the critical angle.


This is the best answer I've seen for describing the critical angle. Any rays of light that strike a surface at angle that is greater than the critical angle are completely reflected back in the direction of travel at the same angle they struck the material at. Interestingly, if the incident ray is precisely at the critical angle, the refracted ray is tangent to the boundary at the point of incidence.

The actual formula for the critical angle is:




where n2 is the refractive index of the less optically dense medium, and n1 is the refractive index of the more optically dense medium.

Many Smiles,
Craig
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