1/ Per request from @physicoyle, here is a discussion about this question about total internal reflection. The picture on the left is certainly wrong because it’s not showing the reflected beam, but the most-correct picture is probably a mix of the two! https://twitter.com/Hookean1/status/1297262225623851013?s=20

2/ First: when we draw a ray, what does that mean? A ray points along the direction of light propagation + is perpendicular to the optical wavefront. Rays are an approximation for when the beam and any object it encounters are much larger than the wavelength (“ray optics regime”)

3/ Now, say a light ray encounters an interface between a high-index material, and a low-index material. This ray experiences refraction (transmission at a different angle) and reflection (at the same angle), according to Snell’s law

(schematic from Griffiths EM)

4/ What’s the physical basis for this law? Here’s one way of looking at it: the wavelength in the material decreases by 1/n in material with index n. Snell’s law is the only way that the wave fronts can match up along the interface given the difference in wavelength

5/ Some caveats to the above picture: (1) there can be a lateral offset between the wavefronts; (2) in this diagram, I didn’t draw the reflected wavefronts which are also present

6/ From both Snell’s law and the picture of wavefronts needing to match up along the interface, it’s clear that when going from high index to low index, there is some critical angle beyond which no light can be refracted! Thus we get total internal reflection (TIR).

7/ TIR is a dramatic effect. Here’s an image from wiki of a turtle and its reflection, taken from inside water (n = 1.33), with light reflecting from the interface with air (n = 1). It’s not a perfect reflection because the interface is not perfectly smooth.

8/ Smooth surfaces between two transparent materials are more reflective than the very best metal mirrors. This is why light can bounce so many times within an optical fiber. Without total internal reflection, there would be no internet as we know it!

[img from hyperphysics]

9/ Here’s another example. This is a simulation of an optical emitter inside a block of diamond (2.4), and you can see only that most of the light is reflected from the diamond/air interface because of TIR (intensity is plotted). The critical angle between diamond and air is ~25°

10/ One more point: when light is incident beyond the critical angle, there still is some field on the other side of the interface! This is called the evanescent field, which skims the interface with the fields decaying exponentially from the interface, and carries no energy*

11/ You might think that because this field carries no energy away from the interface (typically), it’s not actually there. But it certainly is! In fact, if you put some object in the evanescent field, you can capture some of the energy and have it leak through. Compare:

... to the same case but with 2nd interface about a wavelength away. [gifs from: http://www.met.reading.ac.uk/clouds/maxwell/frustrated_total_internal_reflection.html]

12/ OK, but now you might notice something in the TIR example. You see the large reflection, and you see the little evanescent wave skimming the interface. But you also see some light refracting through the interface, even though you’re beyond the critical angle?!

13/ It turns out that the ray-optics picture where a ray is *exactly* perpendicular to a real, physical wavefront (an optical beam) is not actually possible. It only occurs when the beam is infinitely big...

14/ ... When a real light beam has finite size (like say from a laser), it is actually made up of many different angles. I give this example as a homework problem sometimes! Here is an excerpt from solutions to my problem set [hi, students!]

15/ So, now let’s go back to the question at hand. What happens when light is incident at *precisely* the critical angle? I’m claiming that it doesn’t really matter, for two reasons. https://twitter.com/Hookean1/status/1297262225623851013?s=20

16/ First, if you DO make the (nonphysical) assumption we have an infinitely big interface and an infinitely big beam, then as you change the angle and approach the critical angle, the reflectance gradually approaches 100% -- there is no discontinuity in reflectance vs angle

17/ [Note 1: There /is/ a discontinuity in the derivative of reflectance vs. angle

Note 2: There are two curves in the plot above because reflectance vs. angle is a bit different for different polarizations, though they converge for angles close to 0 and 90]

18/ And second, because the assumption of infinitely big interface and infinitely big beam is not physical, you can’t sit exactly at that point anyway! Any beam has finite size, and therefore is made up of many plane waves at slightly different incident angles...

19/ ... So if you try to sit at the critical angle, you will get a bit of refraction, a lot of reflection, and an evanescent wave at the interface...

20/20 Here’s a Gaussian beam (corresponding to angular range of 2°) coming in “exactly at” the critical angle (30°) and then at 32°, so that you’re sufficiently beyond the critical angle so that pretty much every component of the beam is totally reflected

OK I'm done!