WHY are reflections stronger at an angle acording to the Fresnel equation? When looking straight at a glass surface you can mostly see through it, but if you look at it at an angle it becomes more and more like a mirror surface. This is described by the Fresnel equation, or Schlick's approximation that is often used in computer graphics so simulate this.
But why is this happening? The material property of the surface is not changing, so the surface is still reflecting just as much light, so it only "looks like" more light is being reflected.
What is happening, why does it look like this?
 A: The best intuitive explanation I can come up with is that the Fresnel equations, which perfectly describe the reflexion of light from a plane, optical quality interface, are an expression of the requirement that the field vectors should be continuous across the interface. This requirement is itself an expression of the absence of either static or flowing nett electric charge on the interface.
Because the interface causes a "jump" in the transmitted field vectors, there must be a reflected wave to add just the right electromagnetic field at the interface to compensate for this jump and thus uphold continuity of the field vectors.
The bigger the incidence angle, the sharper the change in direction of the field between the incident and transmitted field, and therefore bigger the jump in the field vectors that must be compensated by the reflected wave to uphold continuity of the field vectors.
When the field propagates at a glancing angle to the interface, the field is almost all reflected. This is because the reflected field's direction is almost the same as that of the incident field, whilst the transmitted field's is radically different. Field vector continuity is thus upheld by the mixture comprising mostly reflected field and very little transmitted field.
