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Why does water reflect light? What is actually happening when light is reflected by water? We know why metals reflect light; water, however, is not metal, but it still reflects light and we can see our image reflected on calm water.

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  • $\begingroup$ Everything reflects some EM waves. And you might be interested in the actual "why" of metals reflecting light, especially those metals with color. $\endgroup$ – Carl Witthoft Oct 4 '15 at 17:05
  • $\begingroup$ @CarlWitthoft you mean like complexes? $\endgroup$ – TanMath Oct 4 '15 at 18:53
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The most fundamental answer is that water reflects light because the wave impedance of water is different than the one of air and the electric and magnetic field must be continuous everywhere in space.

The important thing to note is that the wave impedance is the fixed ratio of the electric and magnetic field amplitude of the light wave and that the electric field and magnetic field must be continuous, i.e. cannot change their value discontinuously at material boundaries.

This is a direct consequence from Maxwells Equations, the fundamental equations describing the propagation of light.

If you think about the two requirements, you may think that this is contradictory and cannot be fulfilled simultaneously under all cases, e.g. when a light beam hits a boundary where two materials with different wave impedance meet. This is true only at the first sight and is the reason why there is a third beam.

In a simplified view, the wave impedance of the reflected beam has a negative sign so that for the three beams (incoming beam, transmitted beam and reflected beam) the ratio of E and H field is the wave impedance of the relevant material and also E and H are continuous at the boundary. If you think about it you will notice that this is only possible with three beams and not two.

From this fundamental principle (taking into account polarization of E and H) all other laws dealing with reflection follow. Especially the Snell law with the refractive index. The fundamental principle predicts the amount of reflection as well as transmission and also the direction. Also it predicts the Fresnel equations which specify the reflection and transmission for different polarization (which cannot be derived with arguments involving only the refractive index).

For metals the explanation also holds, the wave impedance is ideally zero for metals so that there is only a reflected beam.

For a list of effects why the wave impedance is different for different materials see John Fistere's answer.

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  • $\begingroup$ +1 Good answer. Might be good to refer to the Fresnel equations as the quantitative description of these ideas. $\endgroup$ – WetSavannaAnimal Oct 4 '15 at 8:08
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    $\begingroup$ This is an answer, but not a good one. Invoking impedance does not answer the question, and it's certain not the "fundamental principle" as you call it. It replaces the question with another question: what is impedance. There's really no physical content in this answer. $\endgroup$ – garyp Oct 15 '15 at 17:23
  • $\begingroup$ This answers the question "why does anything reflect light?" It seems the question is intended to be "why does water reflect images?" $\endgroup$ – Richard H Downey Nov 12 '15 at 19:56
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Whether a material is reflective or not, and to what extent, is dependent upon the atomic structure of the material. When photons hit a surface, they interact with the atoms of that surface, usually raising the energy level of the electrons. When the electrons re-radiate the energy, the structure of the material determines whether it is released as heat in the material, re-radiated diffusely, or re-radiated at the same angle of impingement. The difference appears as dark material, light matte material, or mirror-like, respectively. The result is influenced by the atoms on both sides of the interface.

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    $\begingroup$ "usually raising the energy level of the electrons" Er ... mostly no for visible light. If they did (a) we would see absorption behavior in that particular band and (b) we would see the subsequent re-radiation in that band. but the visible band is situation neatly in a deep valley in the absorption spectrum of water. Light certainly does interact with the atoms but interaction is better described classically. $\endgroup$ – dmckee Oct 4 '15 at 16:24
  • $\begingroup$ THANX FOR UR ANSWER $\endgroup$ – ffahim Oct 6 '15 at 4:13
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EM wave consists of electric and magnetic component. Electric field is responsible for reflection. When EM wave hits the water surface electric component turn molecules of water (H2O is a dipole) Next moment water molecules want to go back in initial position. So they turn back and thus so many molecules at the same time re-emit new EM wave in the opposite direction

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It's simply because water is much flatter and smoother than most surfaces. You see reflections in water but not, say, sand, for the same reason you see your reflection in a polished piece of steel but not a rough-sanded piece of steel. All materials reflect light to some extent, but a rough surface scatters the reflected rays in all directions, so reflected images are blurred beyond recognition. On the other hand, with a very smooth surface, all the reflected light rays stay arranged in the same way they were arranged before hitting the surface (except for being flipped into a mirror image, of course).

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