# Change in Enlarger Head Height corresponds to Change in Exposure Time, but by how much?

Discussion in 'Darkroom Developing and Printing' started by Gregory W Blank, Jul 20, 2004.

1. ### Dan QuinnGuest

I read Mr. Kings remark as meaning double the area "not quite double"
the time. As the lens nears the film plane it's speed nears it's
maximum. A f5.6 lens is only that speed with an infinitely
large enlargement. Dan

Dan Quinn, Jul 23, 2004

2. ### f/256Guest

Again, I think the opposite is true, a f/5.6 lens is only that speed with
zero enlargement, in other words, with the film carrier at an infinitely
distance from the lens, that would place the baseboard one focal length from
the lens. It is the distance lens to baseboard what you have to divide by
the diameter of the aperture (entrance pupil) to find the f/stop of the
lens. An enlarger is a camera with the "object or subject" being
photographed inside of it and the "film" outside of the camera.

Guillermo

f/256, Jul 23, 2004

3. ### Dan QuinnGuest

I'm considerably less than 100% convinced. Dan

Dan Quinn, Aug 15, 2004
4. ### f/256Guest

You really gave it a good -4 weeks- thought, before deciding the % of
convincedness!!

The original msg included a description of why that is, but removed that
paragraph before sending the msg. It read something like this: Contrary to
your photographic camera, an enlarger is a camera with the subject/object of
the picture inside the "camera obscura" and the photographic emulsion
outside in the open, that is why, BTW, the need to situate the enlarger
inside a darkroom. In essence, an enlarger takes pictures of a piece of
film and project its image on the paper. It is the distance image plane to
lens (exit pupil) what constitute the focal length of the lens, in other
words, the distance lens to baseboard, and -similarly to any other regular
photo.camera- that distance is equal to the nominal focal length of the
lens only when the subject/object (in other words the piece of film being
enlarged) is at an infinite distance from the lens. The marked f/stops are
only that for when the lens is focused at infinity.

Guillermo

f/256, Aug 15, 2004
5. ### Mike KingGuest

When making large changes in magnification another variable is reciprocity
departure.

I usually try to vary light intensity rather than exposure time for this
reason.

A US\$20 Ilford EM-10 meter is better and easier to use to calculate exposure
than a calculator.

Mike King, Aug 15, 2004
6. ### Dan QuinnGuest

OK. The speed of the lens is the lens to baseboard focal length
divided by the diameter of the lens opening. A 50mm lens at f4 has
an opening of 12.5mm. If the baseboard is 500mm below the lens the
effective lens speed is f40. If the lens is stoped down to f8 the
effective speed becomes f80, at f11, f120. Is my arithmetic
correct? Dan

Dan Quinn, Aug 16, 2004
7. ### f/256Guest

Dan

Except for the obvious typo of f120 that should be f110, I believe you are
correct.

To prove it: if you see the answer to question 7 in the LENS FAQ
http://www.photo.net/learn/optics/lensFAQ, you will see that the factor by
which we have to multiply the nominal f-number to get the effective
f-number, is ( 1 + M ), where M is the magnification. In the same FAQ you
have that the Gaussian form of the lens equation is 1/Si + 1/So = 1/f
(question 4), and also M = Si/So (question 5). So we have:

Si=500mm
So= ?
f=50mm
1/Si + 1/So = 1/f
1/500 + 1/So = 1/50
1/So = 1/50 + 1/500
1/So = (10 + 1)/500
So = 500/9
M = Si/So
M = 500/(500/9)
M = 9
So the f-number bellows factor is ( M + 1) = ( 9 + 1) = 10
Therefore:
f/4 -> f/40
f/8 -> f/80
etc.

Admittedly a more complicated way to look at it. Used here just to up the

Guillermo

f/256, Aug 16, 2004
8. ### f/256Guest

A correction to the finding of So:

1/Si + 1/So = 1/f
1/500 + 1/So = 1/50