Holograms

I'm old enough to remember when holograms first appeared commercially.  I remember being amazed and trying to look behind to see if it wasn't just a 3-dimensional object masquerading as a 2-dimensional object (masquerading as a 3-dimensional object).  It must have looked like when you show a chimp a mirror and it tries to reach behind it to touch its reflection.  (I guess the same would happen to a human if it saw a mirror for the first time as an adult.)

I did one year of physics at university before switching to maths.  In general I didn't like the practical sessions because they were always at the end of the day, and very long and tiring.  However, on one occasion we got to make our own holograms, and that really did impress me.  IIRC mine was of a 2 pence piece.

What I found really interesting, was finding out how they worked.  I don't think we were taught this as such - it was up to us whether we wanted to go off and find out for ourselves.  Anyway, the diagram above explains it to the best of my abilities.  Here's the key

Manufacture

a.    A point on the surface of the object being recorded.

b.    Convex mirrors.  These reflect the beam and spread it out.  An alternative option is a standard mirror and a concave lens.

c.    Half silvered mirror.  This lets half the beam pass through and reflects the other half.

d.    A laser light source.  This is generates monochromatic (single wavelength) light.  It is also coherent light, i.e. the photons are all in phase.  I don't think the coherence is strictly necessary since each individual photon takes all paths and interferes only with itself.

e.    A very high resolution photographic plate.  For the hologram to work this needs to be able to resolve significantly below the wavelength of the light used.  For green light that's around 500nm.

f.    The reference beam reflected by one of the convex mirrors towards the photographic plate.  Note that this light hits the plate more or less perpendicularly.  Also, since the mirror is relatively far from the plate the distance travelled by the reference beam does not vary very fast as you go from one end of the plate to the other.

g.    The beam reflected from point a.  This is much closer to the plate and therefore the distance travelled by this beam varies much faster as you go from one end of the plate to the other.

h.    Antinodes.  These are points where interference between the reference and reflected beams interfere constructively.  These will correspond to dark spots or lines on the developed plate.  Note that light from point a. must travel almost exactly one wavelength more to get one antinode than it must to get to a neighbouring antinode.  (Because the path length change in the reference beam is negligible.)

Viewing

i.   An ordinary light source, such as a incandescent bulb, or sunlight

j.   Light reflected off the plate.   This light is monochromatic because the hologram is made from a material that absorbs all but one frequency.  In the picture I've shown the reflected light to be green and the laser as red to distinguish the two, but in reality they need to be the same colour.  The light is always reflected in the same direction because the microscopic dots or lines on the plate act as a diffraction grating.  Note also, that the direction matches the direction of the light that was originally reflected from the point on the object.  Only in this direction does the light from each gap reinforce rather than cancel.  In a slightly different direction light from the first gap cancels with light from gap n+1, light from the 2nd gap cancels with light from gap n+2, and so on.

k.    The viewer sees a virtual point a. behind the plate.

Multi-point objects

It's not obvious that if there are multiple points a, the viewer still sees a virtual object exactly like the one used to make the hologram.  Let's see if we can demonstrate this general case.  Firstly note that if $\mathbf{E}(\mathbf{x},t)$ and $\mathbf{B}(\mathbf{x},t)$ solve Maxwell's equations, then so do the following:

$$
\begin{align}
\mathbf{E'}(\mathbf{x},t) &= \mathbf{E}(\mathbf{x},-t) \\
\mathbf{B'}(\mathbf{x},t) &= -\mathbf{B}(\mathbf{x},-t) \\
\end{align}
$$

If we let $E$ and $B$ represent the electric and magnetic fields present during the creation of the hologram then $E'$ and $B'$ are exactly the same, except that they evolve backwards in time, and the direction of $B$ is reversed.  This means that instead of light from two sources (the reference beam and the reflected beam) landing simultaneously on the plate, light starts on the plate and forms two beams.  (The first is a plane wave emanating from the plate and the second focuses light towards the object.)

It's important to note that the final hologram should have dark regions where the light did not fall rather than where it did.  To put it another way, if the plate goes dark where the photons land then the plate is a negative not the hologram itself.  This wasn't important when we were considering a single point a on the object, but it is important now.

Imagine replacing the plate with the final hologram, removing the laser and the mirrors, and placing an ordinary spot light at b.  Turn it on for just a picosecond so that we have a plane wave just 0.3mm thick flying towards the hologram.  The plane wave will pass the dark spots on the surface of the hologram and hit the surface below.  Immediately after that the wave will

• only contain photons of the same wavelength as those used to create the hologram - since the others will have been absorbed by the surface
• have the same intensity distribution across the surface as the plane wave used to create the hologram, because the dark areas will have absorbed light where the intensity was low originally
  
• be moving in the opposite direction to the light used to create the image since the light has been reflected

This means that, if $\mathbf{E}$ and $\mathbf{B}$ represent the fields at creation, then just after reflection at the surface of the hologram the fields must match $\mathbf{E'}$ and $\mathbf{B'}$.  And the only way to extend this to a solution of Maxwell's equations is to assume they continue to match $\mathbf{E'}$ and $\mathbf{B'}$ at all later times too.(*)

The result is an inverted virtual image in front of the surface at a.  However, we actually place the spot light and the ink on the opposite side of the reflective surface.  This means that the virtual image is still at a, but is behind the surface of the hologram and not inverted.

FOOTNOTES

• (*) A couple of caveats here: Firstly we should note that the intensity may not be the same so the solution is actually just proportional to $\mathbf{E'}$ and $\mathbf{B'}$. Secondly, the light used to create the hologram was coherent - i.e. all the photons were in phase - whereas the light reflected from the hologram from an ordinary spot light is not in phase.  However, this doesn't matter too much because we can apply the same reasoning to individual photons and add the results.  The only difference this makes is that instead of calculating the intensity of light we are calculating the probability of detecting a photon at each location in space.