# Why is the light reflected at the same angle from mirror?

From the school physics I know that the material objects bounce from the plane surface at the same angle, losing some kinetic energy. In the same school I was taught that the light (and waves in general) obeys this principle too.

Obviously, in the case of light the plane surface should be a perfect mirror. But I can't understand how should this work from the quantum point of view. Let's assume that our mirror consists of a single silver atom.

Why should the electrons of this atom re-emit consumed photons at some specific angle?

• A single atom won't re-emit light at the same angle. It's far too small for that and will scatter light in a wide angle distribution. One would need a large number of atoms on the surface of a macroscopic mirror to reproduce the angle law of classical optics. – CuriousOne Dec 29 '14 at 6:55
• @CuriousOne: then how much is enough? and why that much will work as a mirror and less won't? – abyss.7 Dec 29 '14 at 6:57
• A macroscopic mirror would have to have a physical size of many wavelengths and the thickness of the metal layer would have to be close to one wavelength. Say we need approx. 100um square of surface area and 100nm thickness to make something resembling a classical mirror. That's a volume of $(10^{-4}m)^2*10^{-7}m=10^{-15}m^3$. Density of silver is $10.5*10^3kg/m^3$, so the mirror mass is $10.5*10^{-12}kg$. Atomic weight of silver is $107au=107*1.66*10^{-27}kg=1.14*10^{-25}kg$, which gives us approx. $9.2*10^{13}$ silver atoms to make a reasonable macroscopic mirror. – CuriousOne Dec 29 '14 at 7:12
• And if you make the mirror smaller than that, you will get significant diffraction and the simple geometric theory of reflection won't apply. – CuriousOne Dec 29 '14 at 7:14
• Related: physics.stackexchange.com/q/32483/2451 and links therein. – Qmechanic Dec 29 '14 at 9:07

That's a good question. Without realising it you have stumbled across the Huygens-Fresnel principle.

The starting point it that a single silver atom is far smaller than the wavelength of light, so any scattering from it will be isotropic i.e. it will scatter the light equally in all directions.

But suppose we have two silver atoms side by side. Each atom will scatter isotropically, so in effect we have two closely spaced emitters of light and the system behaves like a Young's slits setup. Now the light isn't simply isotropically scattered, but instead it's scattered into preferred directions. (I'm oversimplifying because two atoms would be too closely spaced to act as Young's slits, but bear with me.)

Now add lots of atoms in a row, and you get something like a diffraction grating. Add lots more to make a 2D surface, then add more layers of silver atoms below, and you're building up a system where the overall light scattering is the sum of individual scattering from huge numbers of individual silver atoms. This is basically the Huygen's construction, and if you do the sums for a surface you can show that the overall scattering is only non-zero when the angle of reflection is equal to the angle of incidence. Any optics textbook should have the calculation, or a quick Google found an example here.

• en.wikipedia.org/wiki/… – Leos Ondra Dec 29 '14 at 10:01
• Rayleigh scattering isn't isotropic. – Rob Jeffries Dec 29 '14 at 10:19
• A single atom scattering isotropically violates momentum conservation. – Mark Eichenlaub Jun 3 '15 at 8:08
• @John Rennie Great answer. Could you please sort this out: The link actually does not show any calculations or is it the Huygens principal that you referred as calculation ? – Zam Jun 12 at 9:42

The alternative "textbook" electromagnetism answer is to use the boundary condition for the electric field either side of an interface - this is that the component of electric field parallel to the interface (i.e. perpendicular to the normal to the surface) must be the same immediately either side of the interface.

If we have an incoming wave travelling in the $xy$ plane, incident upon an interface plane defined by $y=0$, with amplitude $E_i$ at incidence angle $\theta_i$ that produces a reflected wave at angle $\theta_r$ and a transmitted wave at $\theta_t$ to the normal, then you can write down the following equation (where I assume the electric field is $z$-polarised with its direction also parallel to the interface plane). $$E_i \exp(i(k_ix\sin\theta_i - \omega_i t )) + E_r \exp(i(k_r x\sin\theta_r - \omega_r t)) = E_t \exp(i(k_t x\sin\theta_t - \omega_t t)),$$ where this must be true for all values of $x$ and $t$.

The only way this can happen is if the arguments of the exponentials are all equal to each other for all values of $x$ and $t$ and this yields $\omega_i=\omega_r=\omega_t$ (frequency unchanged for the reflected and transmitted components) and $k_i \sin\theta_i = k_t\sin \theta_t$ (Snell's law); and finally that $k_i \sin\theta_i = k_r \sin\theta_r$ and since they travel in the same medium with the same frequency, speed and wavelength, then $k_i = k_r$ and hence the desired result $\theta_i = \theta_r$.

This explains nothing from the atomic/quantum point of view, but as explained in one of the comments, it is only when you have vast numbers of atoms that this classical behaviour emerges at all.

In many materials, the outer electrons are not confined to an orbital around a single atom. The orbitals spread out, extending across many atoms. Many electrons overlap. The Pauli exclusion principal says overlapping electrons cannot all have the same state. The orbitals have a range of slightly different energies and momenta. This forms a band.

Sometimes the electrons fill all available states in the band. These materials are insulators. For every electron with a momentum to the right, there is another to the left. There is no net transport of charge.

Sometimes the band is half full. This is a a conduction band. There are empty states at energies just above the most energetic electron. This material is a metal. Metals are good conductors because an electric field can easily excite an electron to a state with a momentum in the direction of the field.

To make a mirror, metal is polished to a smooth surface. Usually the surface is a plane. This defines the boundary condition for the electrons.

Typically a mirror is used at a distance from the light source. The light can be reasonably approximated by a plane wave. The approximation is not perfect, and the light is often not monochromatic.

Light is comprised of photons. If light is a monochromatic plane wave, the wave function of each photon is a plane wave. If not, the wave function is a superposition of monochromatic plane waves with various directions and frequencies. We can understand what happens by considering a monochromatic plane wave photon.

When a photon strikes a mirror, it will be absorbed by one electron. That electron will be excited, gaining energy and momentum. Later, it will drop back to the previous state, emitting a photon.

Neither the photon nor the electron have a well defined position. Both can be decomposed into states with well defined energies and momenta. The momentum of the photon can be resolved into components parallel and perpendicular to the plane of the mirror.

The momentum of the emitted photon will have the same parallel component, but a perpendicular component in the opposite direction. I can't give you a quantum mechanical reason for that. As one of the comments indicated, the theory of how light and electrons interact is quantum chromodynamics. Perhaps someone familiar with it can add to this?

## protected by Qmechanic♦Dec 31 '14 at 14:00

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).