When the angle of incidence is less than the critical angle, why is some light still reflected? When the angle of incidence is greater than the critical all light is reflected, why is not all light transmitted when it's less?
[Physics] When the angle of incidence is less than the critical angle, why is some light still reflected
reflectionrefractionvisible-light
Related Solutions
When one is exactly at the critical angle, the light behaves in a way that may be interpreted as "something in between" refraction and reflection: it continues in a direction that is tangent to the boundary of the mediums.
When the angle is smaller than the critical angle, we get refraction. At the critical angle, $\theta_2$ of the refraction becomes 90 degrees, so we get the tangent propagation. At angles larger than the critical ones, there is a discontinuity: the equation for $\theta_2$ (arcsine of something) has no solutions which is why we get a total internal reflection.
There is nothing about these facts that would contradict reversibility or time-reversal symmetry of the laws of physics. If we time reverse the behavior at the critical angle, it indeed looks like the light must "randomly pick" a moment at which it enters the medium with the higher refraction index and there isn't any unique way to pick the preferred moment.
But that's not a problem because the probability that the direction of light is "exactly" tangent to the boundary is zero. In a real-world situation, the light will be a superposition of beams with angles $\theta_2=\pi/2-\epsilon$ for various small values of $\epsilon$, and for any nonzero $\epsilon$, the light will know very well when it hits the boundary. So your problem only occurs at a negligible, "measure zero" portion of the situation, so it is at most a "measure zero" problem. When one adds the appropriate degree of realism and specifies the precise angles and deviations from the "idealized model", the problem goes away.
Imagine the speed of light to be $1$ meter per second and the speed of light in the medium with a high refractive index to be $\frac{1}{2}$ meters per second.
If you have a single peak of a wave in the slower medium, that peak must move forwards at speed $\frac{1}{2}$, no matter what angle it's facing. In the faster medium, that peak must move forwards at speed $1$, no matter what angle it's facing.
The critical angle comes into play when you consider where the peak of the wave is on the boundary between two mediums. If $\theta$ is the angle between the wave direction and the surface normal and $v$ is the speed of the wave, this point travels at a speed $\csc(\theta) v$. This makes sense: if $\theta=\pi/2$, the point at the boundary is just the wave speed. If $\theta=0$, the wave passes instantly and so the question isn't really defined (because there is no point where the peak of the wave is on the boundary).
We demand $\csc(\theta_1) v_1=\csc(\theta_2) v_2$. That is, there should be a single point on the boundary where the peaks of both waves meet. The velocity of the point can be expressed in two ways, and both must be equal.
Unfortunately for the faster medium, if you have a full wave, the point on the boundary can never move slower than $v_2$, in this case, $1$ meter per second. But we're trying to send in a wave whose boundary point can move as slow as $\frac{1}{2}$ meter per second. There is absolutely no way any wave in the faster medium can satisfy that.
The result of this is an evanescent wave, where something is "transmitted" but decays exponentially (so that nothing is truly transmitted over long distances). You can't see that very well in optical light, but you can in microwaves. Take for example this Sixtysymbols video. Around three minutes in, two microwave prisms get pushed together. The reading starts to increase slowly before the two prisms are mushed up right next to each other because there is an evanescent wave "escaping" the prism but transmitting nothing over long distance. If the evanescent wave hits another prism, there is some actual transmission.
Best Answer
Light is an electromagnetic wave, and as such is governed by the Maxwell equations. These equations (considering only the linear mediums 1 and 2) give certain boundary conditions: $$ (\text{i})\ \epsilon_1E^{\perp}_1=\epsilon_2E^{\perp}_2\ \ ;\ \ (\text{ii})\ \bf{E}^{\parallel}_1=\bf{E}^{\parallel}_2 $$ $$ (\text{iii})\ B^{\perp}_1=B^{\perp}_2\ \ ;\ \ (\text{iv})\ \frac{1}{\mu_1}\bf{B}^{\parallel}_1=\frac{1}{\mu_2}\bf{B}^{\parallel}_2 $$
These conditions determine how the light behaves at the surface, and note that they imply the total E-parallel component in medium 1 is equal to the E-parallel component in medium 2. Mathematically,
$$ \bf{E}^{\parallel}_1=\bf{E}^{\parallel}_{\text{incident}}+\bf{E}^{\parallel}_{reflected}=\bf{E}^{\parallel}_{transmitted} $$
This happens independently of the angle of incidence, hence there must be a reflected wave in order to have a refraction. You can calculate the amplitude for each of the components and check that they obey it. I suggest reading chapter 9 of Griffiths' "Introduction to Electrodynamics" for a better understanding.