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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?

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  • $\begingroup$ A material always have emmision + absorption + transmission + reflection = 1 $\endgroup$ – Creepy Creature Apr 19 '17 at 8:21
  • $\begingroup$ That is not always the case. There is a special case in which no light is reflected. $\endgroup$ – Lelouch Apr 19 '17 at 15:28
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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.

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  • $\begingroup$ Also, even for the case of total reflection, there are non-zero electric and magnetic fields that exist in medium 2, but they die off quickly (i.e. evanescent) and do not carry any energy. $\endgroup$ – LedHead Apr 19 '17 at 7:28
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In the general case, there are 3 components, namely, the incident light, refracted light and reflected light. However, when the light is plane polarised along the plane of incidence,one can show using wave theory of light that if $E_I,E_R,E_T$ denotes the amplitudes of incident,reflected and refracted electric fields, then the following holds:

$E_R = \frac{(\alpha - \beta)}{(\alpha + \beta)}.E_I$, $E_T = \frac{2}{(\alpha + \beta)}E_I$

where $\beta = \frac{\mu _1 . n_2}{\mu _2.n_1}$ ($\mu _i$ denote the permeability of the 2 media and $n_i$ denote the refractive indices.)

and $\alpha = \frac{cos (T)}{cos(I)}$ ($T$ denotes the angle of refraction/transmission and $I$ denotes the incident angle).

Neglect the mathematics if you don't want them, but just notice that for the case $\alpha = \beta$, $E_R$ becomes zero. So there is no reflected ray. This happens for a particular angle of incidence given by:

$tan(I) = n_2/n_1$. This is known as Brewster's Angle. At this angle, plane polarised light(w.r.t the plane of incidence) is transmitted 100% into the second medium. This can be shown using rigorous maths, and you can read that up from Griffith's Electrodynamics as mentioned in the other answer by Renan.

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