Ray of light travels from an optically denser medium

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When light transitions from an optically denser medium to a rarer one at the critical angle, the refracted ray grazes the interface, demonstrating the principle of reversibility of light. Snell's law applies in both directions, confirming this reversibility. However, the formation of an evanescent wave is not feasible with ordinary light sources, as it is non-propagating and cannot be directed to achieve the necessary conditions at the interface. The critical angle calculation shows that while light can exit at 90 degrees when moving from a denser to a rarer medium, a vertical beam will still exit vertically rather than at the critical angle. This highlights the complexities of light behavior at boundaries between different media.
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When a ray of light travels from an optically denser medium to a rarer one such that the angle of incidence equals the critical angle, then the refracted ray grazes the interface.

Now, will the principle of reversibility of light work in this case? If yes, please explain.
 
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Does Snell's law depend on direction?
 
Yes, it is reversible.

But the corresponding evansecent wave is somthing that you cannot form
with an ordinary source of light. If you could, you could launch a planewave
into the denser medium at the critical angle.
 
Antiphon said:
Yes, it is reversible.

But the corresponding evansecent wave is somthing that you cannot form
with an ordinary source of light.
why is that?
 
loop quantum gravity said:
why is that?

The evanescent wave is not space-propagating. So there is no way to
form a planewave of (space-propagating) light which could be directed at the interface from
the low-index side and result in the right surface fields needed to launch
the planewave at the critical angle into the dense medium.
 
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but surely it doenst correspond with snell's law.

if we have n1=1 and n2=1.33
then the critical angle from a light medium (n1) to a denser medium (n2) is
a_cr=arcsin(n2/n1)=arcsin(1.33) and needless to remind you the limits of sin function.
 
But you are not "running it in reverse". As light goes from a lighter medium to a heavier (higher speed of light in the medium to lower), the "critical angle" is the incoming angle at which the outgoing angle is 90 degrees. That does not mean that, going the other way, a vertical beam will leave at the "critical angle". A vertical beam will leave vertically.
 
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