by Din » Sun Dec 10, 2017 8:05 pm
I think the important thing to understand is that the modulation is dependent on the beam ratio and the exosure, not on the absolute intensity of either beam.
Looking at the modulation equation again, but this time, inserting the beam ratio, and changing I(1) and I(2) to I(o) and I(r), we have:
I(t) = I(o) + I(r) + 2√{I(o)*I(r)}*cos(φ)
If we now insert the beam ratio, k = I(o)/I(r), we get
I(t) = k*I(r) + I(r) + 2√{k*I²(r)}cos(φ)
The modulation amplitude is now 2√{k*I²(r), which is now dependent on √k, the square root of the ratio. This will reduce the modulation, unless k=1. Of course, other factors also affect this, such as the linear portion of the H & D curve, where you have to work in the linear part of the curve. But, a combination of ratio and exposure determines the modulation.
Now, remember that the beam is Gaussian. So, if the beam is collimated, ie a flat wavefront, then any point on the beam profile at some height r from the beam axis has an intensity given by
I = I(0)*exp(-ar²)
That is, the beam intensity drops off, as the square exponential of the height above the plate centre, or beam centre (assuming the beam is centred on the plate) . However, if you have a divergent beam, then the Gaussian spread is along the arc of the circle of the divergent wavefront. So, now, if the point of divergence of the beam from the plate centre is d,the intensity along the arc of the wavefront, at some angle θ, relative to a line drawn from the centre of divergence to the plate centre, is
I = I(0)*exp(-ad²θ²)
That is, the Gaussian is spread along an arc of the divergent waveform. Notice, the intensity is dependent on θ², so it falls off much more rapidly, as you mentioned.
Thus, what we have to consider, in terms of the ratio - which is what determines the modulation - the spread of the Gaussian linearly (collimated beam), as a ratio of the spread of the Gaussian along the arc of a circle, for a collimated/divergent recording. Alternatively, the ratio depends on the divergence angles of the two beams for a divergence/divergence recording. As the intensity of a divergent beam at any position of the plate is dependent on the square of the divergence angle, and so falls off rapidly, if you have very different divergences, then the ratio falls off much more quickly, as you approach the edge of the plate. This may lead to vignetting.
I think the important thing to understand is that the modulation is dependent on the beam ratio and the exosure, not on the absolute intensity of either beam.
Looking at the modulation equation again, but this time, inserting the beam ratio, and changing I(1) and I(2) to I(o) and I(r), we have:
I(t) = I(o) + I(r) + 2√{I(o)*I(r)}*cos(φ)
If we now insert the beam ratio, k = I(o)/I(r), we get
I(t) = k*I(r) + I(r) + 2√{k*I²(r)}cos(φ)
The modulation amplitude is now 2√{k*I²(r), which is now dependent on √k, the square root of the ratio. This will reduce the modulation, unless k=1. Of course, other factors also affect this, such as the linear portion of the H & D curve, where you have to work in the linear part of the curve. But, a combination of ratio and exposure determines the modulation.
Now, remember that the beam is Gaussian. So, if the beam is collimated, ie a flat wavefront, then any point on the beam profile at some height r from the beam axis has an intensity given by
I = I(0)*exp(-ar²)
That is, the beam intensity drops off, as the square exponential of the height above the plate centre, or beam centre (assuming the beam is centred on the plate) . However, if you have a divergent beam, then the Gaussian spread is along the arc of the circle of the divergent wavefront. So, now, if the point of divergence of the beam from the plate centre is d,the intensity along the arc of the wavefront, at some angle θ, relative to a line drawn from the centre of divergence to the plate centre, is
I = I(0)*exp(-ad²θ²)
That is, the Gaussian is spread along an arc of the divergent waveform. Notice, the intensity is dependent on θ², so it falls off much more rapidly, as you mentioned.
Thus, what we have to consider, in terms of the ratio - which is what determines the modulation - the spread of the Gaussian linearly (collimated beam), as a ratio of the spread of the Gaussian along the arc of a circle, for a collimated/divergent recording. Alternatively, the ratio depends on the divergence angles of the two beams for a divergence/divergence recording. As the intensity of a divergent beam at any position of the plate is dependent on the [i]square[/i] of the divergence angle, and so falls off rapidly, if you have very different divergences, then the ratio falls off much more quickly, as you approach the edge of the plate. This may lead to vignetting.