Does light reflect if incident exactly at critical angle ?

[Murtuza Tipu]

A lot of textbooks and exam boards claim that light incident at exactly the critical angle is transmitted along the media boundary (i.e. at right-angles to the normal), but this seems to violate the principle of reversibility in classical physics. How would a photon or ray traveling in the reverse direction "know" when to enter the higher refracting medium? It can't know, so I conclude that such light is simply reflected?

Is this correct?

 

[AlephZero]

Unless your textbooks and exam boards don't believe in the existence of atoms, a boundary which is a geometrically perfect plane surface doesn't exist.

The general idea of what "light incident at exactly the critical angle is transmitted along the media boundary" means is clear enough, but don't confuse a simple mathematical model with reality.

 

[Murtuza Tipu]

can you explain me what exactly it is

 

[Claude Bile]

It's a good question, but one that classical optics has covered I think.

Ray approximations are useful, but don't forget that light is ultimately a wave; and in the wave picture, the plane-wave components are completely non-localized.

In other words, light incident upon a surface with some critical angle, will be incident at that angle on ALL points on the surface. Thus there is no need for the time-reversed wave to "decide" a position from which to refract back out into space.

Claude.

 

[Redbelly98]

A lot of textbooks and exam boards claim that light incident at exactly the critical angle is transmitted along the media boundary (i.e. at right-angles to the normal), but this seems to violate the principle of reversibility in classical physics. How would a photon or ray traveling in the reverse direction "know" when to enter the higher refracting medium? It can't know, so I conclude that such light is simply reflected?

Is this correct?

Yes and no.

No because, at the critical angle, the reflectivity is 100% and transmissivity is 0%. So there is no light transmitted along the surface, it is completely reflected.

And yes, because to reverse the situation, you would have a wave coming from the direction of the reflected beam, which is also at the critical angle.

You can look into the math by looking at the expressions for reflection and transmission, referred to as r and t at this link:
http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/freseq.html

At the critical angle, the transmitted angle θ_t is 90°, so you can work out what happens to r and t (at website linked above) in that case.