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From what I read, it sounds like when a light ray passes from one medium to the other, if the critical angle is reached so that the refractive ray is at 90 degree with the normal, the light ray does not enter the second medium but stays in the first. (not sure about this)

Since it's called total reflection, I assume there to be no refractive ray at the 90 angle. But this picture confuses me: enter image description here

The books says that all rays are reflected. But I am not sure the one ray at 90 counts as an reflective ray.?

My guess based on the pic, is:

  • either, the ray suddenly jumps to the position that agrees with the law of reflection, at the very moment it reaches 90degree.
  • or, stays at 90degree but will jump suddenly to that position, instead of going smoothly, if the incident ray gets any bigger.

To summarize, my questions are

  1. Which of my guess is correct? Or none of them is?
  2. What kind of ray (reflective or refractive) is the ray at 90degree?
  3. In which medium is that ray? (First or second?)
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  • $\begingroup$ All rays are partially reflected. Depending on the angle of incidence more or less intensity goes into the reflected ray and less or more into the refracted ray. At the critical angle and above it all intensity goes into the reflected ray. $\endgroup$ – LLlAMnYP Jul 31 '15 at 15:06
  • $\begingroup$ You're right, there is no 90 degree ray. That is the reference line which is used to measure the angle against. $\endgroup$ – WhatRoughBeast Jul 31 '15 at 16:01
  • $\begingroup$ @WhatRoughBeast, no I am not talking about the normal. It's the one on the horizontal. $\endgroup$ – most venerable sir Jul 31 '15 at 16:07
  • $\begingroup$ You need to show the other part of the figure - the part to the left of what you've shown. I believe it means that, if "normal" refraction, which is dealt with in the missing figure, produces a 90 degree (or more) ray, total internal reflection occurs. $\endgroup$ – WhatRoughBeast Jul 31 '15 at 16:19
  • $\begingroup$ @WhatRoughBeast, my question is in which direction does the ray being produced point if the angle of incidence becomes the critical angle? My books confuses me in that there are two rays, one at 90degree the other just like a normal reflective ray. $\endgroup$ – most venerable sir Jul 31 '15 at 16:26
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So at the boundary, what's really happening is: the frequency of light must remain the same (wave crests/troughs cannot exit any faster than they enter) and therefore the different speed of light changes the wavelength of the light.

Of a certain intensity of light, a proportion $T$ transmits into the new medium and a proportion $R$ exits. The barrier does not absorb energy, so $T + R = 1.$ It can only absorb momentum perpendicular to its surface, so there must be conservation of momentum in the other direction. Now for a light wave, the momentum is proportional to the inverse of the wavelength $\lambda$, so if $\theta$ is measured from the line perpendicular to the surface, momentum balance in this region means:$$\frac{1}{\lambda_1} ~ \sin \theta_1 = \frac{T}{\lambda_1}~\sin\theta_1 + \frac{R}{\lambda_2}~\sin\theta_2$$Conservation of momentum and energy therefore together give us Snell's law, since you can work out that if the frequency stays the same then $\lambda_1 / \lambda_2 = n_2 / n_1.$

You need another equation to successfully pin down the exact value of $R$ and see how it varies with respect to $\theta_1.$ This other equation is harder for me to explain, and it comes from the continuity of fields, which makes it polarization-dependent.

Typically light is a wave made of two fields, an electric and magnetic field, that oscillate perpendicular to each other, and perpendicular to the direction that the wave goes, too. So if light is going "up" then possibly the electric field is oscillating east-west while the magnetic field is oscillating north-south. The energy is contained in the square of the amplitude of these oscillations, and, if the wave is moving at a speed $v$, the magnetic field's amplitude is usually related to the electric field's amplitude by $E / v$. A full writeup of how exactly all this works is available here.

Here's one example. If the electric field points alongside the surface between the two media, then electric field continuity means that $E_i + E_r = E_t$. Defining the transmission amplitude $\tau = E_t / E_i$ and the reflection amplitude $\rho = E_r / E_i$, this means that $1 + \rho = \tau$. In general because it's the same medium, $R = \rho^2$ but $T \ne \tau^2$, so we need another equation. That's given by the magnetic field, $$\frac{1}{v_1} ~ (1 - \rho) ~ \cos\theta_1 = \frac 1 {v_2} ~\tau~\cos\theta_2, $$therefore $$ (1 - \rho) ~ n_1 ~ \cos\theta_1 = (1 + \rho) ~ n_2 ~ \cos\theta_2$$or once you work it all out, $$R = \left({n_1 \cos\theta_1 - n_2\cos\theta_2 \over n_1 \cos\theta_1 + n_2\cos\theta_2 }\right)^2.$$Again, that's only true for this one polarization, but let's look at this formula.

First off, observe that this is actually really well-defined for $\theta_2 = \pi/2$, which is when we know $\theta_1$ is critical: it limits to 1. For all $\theta_1$ less than this, we get a reflection coefficient R which smoothly increases from 0 to 1 before abruptly having a "kink" at 1 (because Snell's law can no longer be satisfied beneath that point and so it stays $R = 1$). And this means that $T$ goes to 0.

So, as the transmitted ray gets closer and closer to being "alongside the surface", it also gets weaker and weaker in intensity. By the time the transmitted ray lies purely "alongside" the surface it has smoothly transitioned into being nonexistent.

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  • $\begingroup$ I am in introductory course....... $\endgroup$ – most venerable sir Jul 31 '15 at 16:56
  • $\begingroup$ @user132522 Sorry, I know that I probably hit you with a bit too much information about where the laws come from etc. The final answer is in bold in the last paragraph. If you use a laser beam, the refracted light does not experience any sudden jumps in direction: it smoothly changes direction until it points more and more along the surface. As you get closer and closer to that direction, though, the light that is getting refracted is getting dimmer and dimmer, approaching 0 as you approach the critical angle. Once you hit that critical angle, refraction just can't happen. Does that help? $\endgroup$ – CR Drost Jul 31 '15 at 17:03
  • $\begingroup$ that was what I saw in a YouTube video. $\endgroup$ – most venerable sir Aug 1 '15 at 12:26
  • $\begingroup$ So by the time it's level with the surface, the ray disappear and an new reflective ray is formed? $\endgroup$ – most venerable sir Aug 1 '15 at 12:31
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    $\begingroup$ @user132522 Mostly. No "new reflective ray" is formed to my knowledge, although maybe you need an "evanescent wave" for some reason going along the surface. Rather, you start with a wave reflected from and a wave transmitted into the surface, and as the light from the transmitted wave is getting weaker and weaker, the light from the reflected wave is getting stronger and stronger. $\endgroup$ – CR Drost Aug 1 '15 at 19:36
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When light goes from a medium with a low index of refraction to an medium with a high index of refraction, at an angle that is less than 90 deg to the surface, it bends towards the normal (it becomes more vertical). The light paths are reversible, so when light goes from a high index medium into a low index medium, it bends away from the normal.

More specifically, if you are shining a laser pointer from the bottom of a swimming pool, straight up towards the surface of the water (perpendicular), the beam will not bend. As you increase the angle of incidence (measured from the normal), the beam exiting the water will bend away from the normal. At the critical angle, the beam exits the water surface parallel to the surface. Any angle greater than the critical angle causes the beam to reflect back into the pool, and no fraction of the light beam escapes into the atmosphere (you get total internal reflection).

Regarding the "jump" in the reflected ray, it isn't actually a jump. At all angles below the critical angle, part of the light ray is refracted and escapes into the atmosphere and part is reflected back into the pool. The amount of refracted light progressively diminishes as the angle of incidence increases (measured from the normal). At the critical angle, a small portion of light escapes the pool and a large portion is reflected.

If you want to see this phenomenon for yourself, and you can hold your breath for a fairly long time, find a calm swimming pool and quietly dive down in the deep end until you are a few feet underwater. Look up, and you will see light from the outside world, but only up to the critical angle. This will give you a view of the outside world that looks like a circle, and every image outside that circle will be reflections from inside the pool.

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  • $\begingroup$ What do I see when I am at the critical angle? The bottom of the pool? Or the wall if the ray is pointing horizontally? $\endgroup$ – most venerable sir Jul 31 '15 at 16:23
  • $\begingroup$ You see the very "edge" of what you will see from the outside world, which corresponds to the horizon that you would see if you lift your head above water. Of course, the view is a bit distorted due to refraction. At any greater angle, you see reflections from inside the pool, where the angle of incidence equals the angle of reflection. $\endgroup$ – David White Jul 31 '15 at 16:55
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See Chris Drost's excellent answer.

But I just wanted to point out the statistical mechanics perspective as this may give you an additional insight.

Index of refraction (squared) is proportional to the number of optical modes that can be supported in a material.

An occupied mode propagating from a high index material to a low index material at the boundary experiences an impedance mismatch. Moreover there is not an optical mode in the lower index material that can accept the mode from the higher index material. Thus the light is totally internally reflected.

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