When a light ray is incident on the media interface at an angle greater than the critical angle how do we know that the ray would be reflected?
Is there any mathematical proof?
Let me illustrate:
Let critical angle,a=45d then n2/n1=sin(45)=1/root(2)
if now i=60 then angle r must be 60d because of TIR but.....
sini/sinr=n2/n1 => sin60/sinr=1/root(2)=> r!=60.
Of course there is a mathematical proof - it is just Snellius' law:
sin(x1)/sin(x2) = n2/n1
if x1 is the incident beam and x2 is the refracted beam. n1 and n2 are the corresponding refractive indexes.
so, if n2 > n1 --> n2/n1 > 1, if furthermor n2 is large enough, it follows that:
n2/n1 * sin(x2) = sin(x1) > 1 which cannot hold for any real x1. Thus you have total reflection.
The reason why there is total reflection is destructive interference. When the angle is greater than the critical angle, there is no point at infinity in the second medium where the phases of the light wave received from every point on the surface are equal. Near the surface, there is still some field called the evanescent wave, because the interference is not totally destructive. That can be seen while deriving the Snellius law from the Huygens principle.
Your proposition is a proof by contradiction; because if we suppose that i > 45d for ex: i=60d we have sin60/sinr =1/root(2) and then sinr > 1, so it's impossible
we conclude that i must be i < 45d
Dear Avinash,
In your illustration, you applied the Snell's law for refraction. It failed because n2 > n1, as explained by Johannes Gruenwald above. At the incident angle equals to critical angle, the refracted angle is 90 degrees. When the incident angle is greater than the critical angle, there will be total internal reflection as described by Claude. In this case, we should apply the Snell's law for reflection: n2 . sin i = n2 . sin r (still in the same medium); therefore, we will get r = i = 60 degrees. ( r stands for reflection, not refraction).
I hope that will help.
If i is the angle of incidence and r is the angle of refraction at the interface separating the two media with r.i. n1, n2 (n1>n2), then for i greater than critical angle, sinr >1 (as shown by Johannes Gruenwald) that follows from Snell's law. This implies that cosr is purely imaginary. If we evaluate the amplitude reflection coefficient for the polarizations parallel and perpendicular to the interface (that are related to the cosr), then they come out to be complex. And their absolute value comes out to be 1. This proves the occurence of TIR.
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