I think you may be describing an idealised reflection hologram with on axis beams from both sides. In this case, there is a standing wave whose time-averaged amplitude at every position is recorded. Since, in a standing wave, the amplitude at different points in space varies, the recorded intensity at any specific point varies as a function of the point. I remember a demonstration by my school Level Physics teacher where he took a pipe and drilled holes along the top. He then made an additional hole to attach to a source of gas. He attached speakers to either end of the pipe and lit the holes at the top. Since the pressure of the gas inside was a function of the pressure due to the sound wave and he had two speakers at either end, the gas inside the pipe was a pressure standing wave. You could see that the flame was at different heights along the pipe.PinkysBrain wrote:They generally capture the effect of phase as the time averaged amplitude of the standing wave caused by the interference of the reference and object wave, but they don't generally capture phase proper because that amplitude is also dependent on the amplitude of the object wave proper ... which is unknown in display holography (obviously not necessarily so in interferometry). Amplitude and phase get all mushed up in recording, and only a single one is generally modulated in playback ... it's a miracle holography works as well as it does
By the way, even if you have a perfect system as above, with two beams coming in from opposite sides of the plate, you cannot get a pure standing wave, the wave will creep forwards and so the "fringes" will blur. The reason is that pure standing waves will occur if, and only if, the amplitude of both waves is exactly equal. If there is any dissimilarity between the waves, there will be a small traveling wave component. When launching into a medium, the absorption of that medium will reduce the amplitude of one wave slightly. This causes a dissimilarity in amplitudes and so causes the waves to travel. This is why you cannot have a single beam Denisyuk brighter than an H1/H2 system, even if you matched the beam ratios perfectly.
In a transmission hologram, or any hologram where the waves are not counter-propagating, the phase is captured via the complex amplitude in the exponential. The general light field due to an oscillating point source, oscillating at w, is given by
E = E' expi(wt-kr)
If you had a lot of these points at r_1, r_2...r_n, you'd add them all up (due to the Principle of Superposition) and get
E = E' {expi(wt- kr_1) + expi(wt-kr_2) .. + expi(wt-kr_n} = E' exp(iwt)*expi(kr_1 + kr_2 +...+kr_n) = E' exp(iwt)* exp(iphi(r))
The phi(r) term captures all the phase variations of all the light from all the various point sources. So, the generalised "object wave" is
O = O' exp(i*phi(x,y)
The energy carried by a wave cannot be negative. So, to force it to be positive, we take the square of the E field:
W=E^2 = (E')^2 sin^2 (wt-kr)
Since the variation of this term is extremely rapid and cannot be tracked, we instead state that the average energy delivered by a lightfield is an average over many cycles (remembering that a cycle, for light, is about a tenth of a picosecond or so). The average of the above function varies wildly for a few cycles, but then settles down to half the maximum value of E'. Since the "few cycles" is only a picosecond, we then simply state that the energy delivered by a lightwave is half the maximum value of the sin function, or half the amplitude. So, the energy of object wave W_O(which is what converts the silver halide to silver) is, using complex notation:
W_O proportional to (1/2)(O(x,y))^2 proportional to (1/2) {O(x,y)O*(x,y)} (where O* is the complex conjugate of O)
If you took a photograph, this is all you'd get because all phase information is lost due to the presence of the complex amplitude. But, in a hologram you have a reference wave which has a fixed phase
R = R' exp(wt-kr)
So, at the hologram plate, the total lightfield is:
H = (R + O)(R* + O*) = RR* + OO* + RO* + OR* = |R| + |O| + RO* + OR*
Now, RO* = R'O'exp(ikr - ikphi) and OR* = R'O'exp(ikphi-ikr), their sum is R'O'cos(phi)
And so, on the plate, is an overall increase in exposure (due to the |R| + |O| term) which is what I call the 'dc' level and a term in O and a term in O*. The O term is an exact copy of the object wave, phase and all. The O* term is the conjugate term, phase and all. This O* term is usually called the 'pseudoscopic term", though the actual light field is not always pseudoscopic - especially in an H1/H2 setup.
