Lens formulas for enlarging

[lbi115l]

Hello,

 

I was wondering whether the lens formulas are the same for enlarging

as photographing.

 

Specifically, I have an 8x10 enlarger, and want to find out the

optimum focal length with a certain maximum bellows and a maximum

distance to the paper (i.e. object distance and image distance). I'm

planning on using it in a number of formats, so not just for 8x10.

 

Thanks.

[michael_briggs2]

<p>The optical formulae relating focal length, magnification, object and image distance don't care whether the situation is a camera or an enlarger. The formulae are given in the first section of the Photo.Net Lens Tutorial: <a href="http://www.photo.net/learn/optics/lensTutorial">http://www.photo.net/learn/optics/lensTutorial</a>.</p>

 

<p>One constraint is to select a focal length that the manufacturer says covers 8x10. This is typically 240 mm or higher. Then you can use the formula with the dimensions of your enlarger to determine the maximum magnification possible. You may find that the resulting magnification isn't as large as you want for smaller formats. This one reason why most people enlarging several formats own several lenses.</p>

[tim_curry]

A good rule of thumb is to use a slightly larger focal length than is "normal" for a given format, if possible. For example, I use a 75mm lens for enlarging 35mm film, although a 50mm is considered normal in this application. I use a 135mm Rodenstock lens as a "normal" lens for 4x5 enlargements, but I use a 190mm lens for bigger enlargements.

 

With that 8x10 beast, there will be a point at which you run out of bellows draw which will limit your lens focal length. I would start by looking at your maximum bellows draw and then work backwards from that point. A 600mm lens might be just dandy, but not if you run out of bellows before focusing is possible.

[silent1]

An additional tidbit, related to bellows draw: to make a 1:1, the lens will be twice its focal length from the negative, and the same distance from the paper. So, if you have a "normal" 8x10 lens of about 13" (aka 330 mm), you'll have around two feet of bellows extension (allowing for a couple inches in the rigid frame below the negative), and the lens only a little over two feet from the paper. Any larger print will use less bellows and put the head higher.<p>

 

So, if you have a 13" lens, you won't possibly need more than about two feet of extension -- after all, if you want to print your 8x10 negatives to 8x10, you'll just contact them -- unless, for some perverse reason, you want to print <i>smaller</i> than the negative. Wallet size from a 4x5, anyone? B)

[michael_briggs2]

<p>Measure the maximum distance between negative and photo paper possible with the enlarger: call this "D".

Using the notation of the Photo.Net Lens Tutorial, <a

href="http://www.photo.net/learn/optics/lensTutorial">http://www.photo.net/learn/optics/lensTutorial</a>,

D = So + Si. Further define R = D / f, where f is the focal length of the enlarging lens in mm. Manipulating the optical equations in the Lens Tutorial, one finds the maximum magnification M possible is

M = [R - 2 + square_root (R*R - 4*R)] / 2.</p>

 

 

<p>Using a lens with a shorter focal length will allow a larger magnification. With 8x10, focal lengths around 240 mm are popular to keep the enlarger size down. For most formats the standard focal length is equal to the format diagonal, which would be 300 mm for 8x10. If the enlarger is larger enough or if large magnfications aren't wanted, 300 mm would be better than 240 mm.

</p>

 

<p>An example: D = 1600 mm (1.6 meters = 5.2 feet) and f = 300 mm. The equation gives M = 3. From the other optical equations, So = 400 mm (this is the distance from lens to film, i.e., the bellows extension) and Si = 1200 mm.</p>

 

<p>The maximum bellows extension limits the minimum enlarging magnification possible, not the maximum magnification. As the lens is moved farther from the film, the magnification and print size decreases.</p>