Ruby double-pulse fun

Present your work.
Loic74
Posts: 51
Joined: Sat Aug 10, 2019 9:09 pm
Location: France

Ruby double-pulse fun

Post by Loic74 »

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
lobaz
Posts: 280
Joined: Mon Jan 12, 2015 6:08 am
Location: Pilsen, Czech Republic

Re: Ruby double-pulse fun

Post by lobaz »

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.
Loic74
Posts: 51
Joined: Sat Aug 10, 2019 9:09 pm
Location: France

Re: Ruby double-pulse fun

Post by Loic74 »

Thanks Iobaz,
I will investigate and report back my findings
Din
Posts: 402
Joined: Thu Mar 12, 2015 4:47 pm

Re: Ruby double-pulse fun

Post by Din »

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.
Petr, It's because lateral magnification is 10 times transverse magnification:

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).
Post Reply