The question arose when my physics teacher taught me geometrical optics and told that the phenomenon of both refraction and $reflection$ occur by change in medium. First of all, I'm not sure about the above and would like a bit clarity on this. And the main doubt emerges when I make the scenario given in the picture below:

enter image description here

Sorry for my bad drawing, but as you can see, my question is that what happens when I make the ray incident at (a)almost equal to critical angle (i limiting to $i_c$) and (b) at exactly $i_c$ . For the (b) part I saw this (Does light reflect if incident at exactly the critical angle?) question. But the above concept my sir taught kinda contradicts for part (a). I have given the options which should occur in the picture, (a) and (b), with (a) being wha my sir said and (b) being the other option.

If (a) is occurring, which should occur according to my teacher, then it contradicts principle of reversibility, or actually common sense, since the reverse ray won't even touch the surface, and if (b) occurs then it contradicts what my teacher taught me.

So is the teacher wrong? Where lies the fallacy?


3 Answers 3


In general, reflection and refraction happen when light passes from one medium to another. You can see this if you see your own reflection in a window. Now, as a light ray approaches the critical angle, not only does the refracted ray get closer to the surface, but the amount of light transmitted gets less and less. At the critical angle, the refracted ray would be along the surface, but the amount of light is zero.

The graph below shows how much light is reflected at various angles of incidence. The graph on the right shows your situation of a ray of light moving from a dense medium to a lighter medium (glass to air, for example). At the critical angle, 100% of light is reflected, leaving none to propagate along the surface.

from http://en.wikipedia.org/wiki/Fresnel_equations

  • $\begingroup$ So you are saying that both the options (a) and (b) are not correct. The light will simply 100% reflect. $\endgroup$
    – Rohinb97
    Sep 28, 2014 at 14:20
  • $\begingroup$ @Rohinb97 Yes, it will 100% reflect. $\endgroup$
    – Mark H
    Sep 28, 2014 at 14:39

Let me go a little further than @mark-h's answer:

The behavior of light at an interface is described by the electro-magnetic field solution to the Helmhotlz equation. It gives the reflected and transmitted electric and magnetic components as a function of the refractive indices of the incident and exiting media.

From those solutions we can derive the Fresnel transmission and reflection coefficients (cf. figures in @Mark-H s answer) as a function of the angle of incidence. They describe the amplitude (and phase) of the reflected and transmitted beams.

There are 2 consequences you might be interested in:

  1. Light is never be 100% transmitted. There is always some amount of reflection, even when the incident angle is below the critical angle. For air to glass interface, about 4% of the light is reflected at normal incidence (just look at your reflection in a window).
  2. Above the critical angle, the phenomenon is called Total Internal Reflection (TIR). 100% of the light is reflected. No light transmitted means that there is no viable propagating solution to the Helmholtz equation in the exiting medium. BUT, there is still an EM field that goes through the interface. It is just not propagating, and is called evanescent. It exists only within a few microns from the interface and decays exponentially (If you are familiar with partial differential equations, the solution has an imaginary propagation constant).

There are other consequences at the microscopic scale, like the Goos-Hanchen effect: the reflected wave is slightly shifted parallel to the surface.

The evanescent wave phenomenon is widely used in photonics: waveguide coupling, fiber optics transmission, ellipsometry, etc. It is basically responsible for making your microprocessor work and your internet reach your home :)

  • $\begingroup$ Nice answer. I didn't know all that. You've perfectly explained as to why all of this occurs. But I have a doubt. If there exists a small layer of EM field(although decaying) then the actual ray, which is supposed to be refracted at the critical angle, has to follow option (a), which reverts back to my question. Although it's intensity is 0, but it was supposed to go like this, and was going to violate the principle of reversibility. Please explain and clarify this. $\endgroup$
    – Rohinb97
    Oct 2, 2014 at 7:53
  • $\begingroup$ Thank you! The best way to think about this is that there is not 1 ray, but two. One refracted and one reflected, and the incident energy is distributed to them according to the Fresnel coefficient. As you can see in the graphs, the reflection coefficient tend to 1 before the critical angle. Just flip them upside-down and you get the transmittion coefficients, that are near 0 before the critical angle... $\endgroup$
    – Anael
    Oct 3, 2014 at 2:10
  • $\begingroup$ I'm not quite sure what the difference is between your a) and b). But as for any problem in physics when you consider the special case of a singular point (where the behavior is not smooth / the derivative is not continuous), you have to interpolate from the limits. It also never happens since it is a discreet event and you can never get 1 perfect infinitely thin, non divergent ray (what does it mean in terms of EM wave?) that will hit a perfectly plane surface (nanometer smooth?) at exactly the critical angle (local variations of the refractive index?)... $\endgroup$
    – Anael
    Oct 3, 2014 at 2:20

If the light ray is normal to the surface, the maximum amount of light is transmitted.

As the light ray bends, as in your part (b), a percentage of the light will be transmitted (refracted) and the remaining will be reflected (at the incidence angle).

Very near the critical angle $\theta_c - d\theta$, likewise, some of the light will be transmitted (refracted almost tangent to the surface) and some will be reflected. At this point you can imagine the majority of the light would be reflected and only a small percentage of light is refracted.

At the critical angle $\theta_c$, although the refracted angle is tangent to the surface, the intensity is actually 0, and all the light is being reflected.

Past the critical angle $\theta_c + d\theta$, all the light is reflected.

Likewise, in the reverse case, because the optical density of the medium is reversed, there is no $\theta_c$. At $\theta < \frac\pi2$ some of the light gets transmitted(refracted) and some gets reflected. At $\theta = \frac\pi2$ all the light gets reflected at the angle $\frac\pi2$ so the resulting conclusion is that the light simply goes in a straight line.

  • $\begingroup$ What reverse case are you talking about? And I'm not talking about incident angle $\pi/2$. I didn't catch the last para of your answer. $\endgroup$
    – Rohinb97
    Sep 27, 2014 at 23:00
  • $\begingroup$ "If the light ray is normal to the surface, 100% of the light is transmitted." This is false. Just look at glass. $\endgroup$ Oct 1, 2014 at 21:28
  • $\begingroup$ changed@TylerHG $\endgroup$
    – t.c
    Oct 2, 2014 at 1:54

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