# [Physics] When the angle of incidence is less than the critical angle, why is some light still reflected

reflectionrefractionvisible-light

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?

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.