Me playing with my JK HLS2 ruby on an old Agfa 8E75 plate.
Next I wish to transfer it while reducing the image size with a lens between H1/H2.
From what I read this greatly reduces the perceived depth of field?
https://www.youtube.com/watch?v=TR09kAzSfm8
Ruby double-pulse fun
Re: Ruby double-pulse fun
First of all, if you make the image smaller 10x in XY (lateral) directions, it gets 100x smaller in the Z (depth) direction. Instead of a smaller sphere you get mostly flat lens shape.
I am not sure what may be the main reason for apparent depth of field reduction. First, if the large original is recrded with laser of limited coherence length (background is not recorded), that limit gets much closer in the smaller H2. Second, aberrations of lenses used to make the image smaller might affect distant parts more than closer parts, but I am just guessing.
I am not sure what may be the main reason for apparent depth of field reduction. First, if the large original is recrded with laser of limited coherence length (background is not recorded), that limit gets much closer in the smaller H2. Second, aberrations of lenses used to make the image smaller might affect distant parts more than closer parts, but I am just guessing.
Re: Ruby double-pulse fun
Thanks Iobaz,
I will investigate and report back my findings
I will investigate and report back my findings
Re: Ruby double-pulse fun
Petr, It's because lateral magnification is 10 times transverse magnification:lobaz wrote: ↑Tue Oct 06, 2020 1:14 pm First of all, if you make the image smaller 10x in XY (lateral) directions, it gets 100x smaller in the Z (depth) direction. Instead of a smaller sphere you get mostly flat lens shape.
I am not sure what may be the main reason for apparent depth of field reduction. First, if the large original is recrded with laser of limited coherence length (background is not recorded), that limit gets much closer in the smaller H2. Second, aberrations of lenses used to make the image smaller might affect distant parts more than closer parts, but I am just guessing.
M(l) = m(t)²
You can prove this by using the lens equation
1/u + 1/v = 1/f
where u = object distance, v = image distance and f = focal length.
The longitudinal magnification is then dv/du and the lateral magnification is -(v/u).