How large a field do you want in focus? Benthos have a wide range of size according to Wikipedia.

Originally posted by cadmus How do I position the lenses to make a long tube? Or is it not possible?

It is possible with some restrictions on magnification and focal lengths involved for simple optics. You'll need a long lens for what I think you want to do.

It is important to avoid image transfer optics to decrease complexity. If so, the set-up is pretty simple. I assume you want the camera to be out of the water and the snout of the lens in the water. Here's the basic layout:

Calculations are complicated by the fact that when you look through a flat plate into water, things appear to be closer than they actually are by a factor of about 3/4

The optics involved impose the following limits* (for water with index of refraction n):

(water.depth+clearance)/((1+m)+n(1+m)/m) < f < water.depth/(n(1+m)/m))

The index of refraction for water is 4/3 so we'll use that from now on.

(water.depth+clearance)/((1+m)+4(1+m)/3m) < f < water.depth/(4/(1+m)/3m))

Example:

For water depth = 1 meter, camera clearance = .100meter, and magnification = 1,

(1+0.1)/(2+8/3) < f < 1/(8/3) or 236mm < f < 375mm

*ie. the focal length range where wet.lens/dry.camera 1:1 mag is possible.*
Dave

* I am no photographic optics expert but based on what I read about optics & underwater photography I believe the following is true:

The above is based on thin lens optics with the object side of the lens looking through a thin flat plate into a medium with index of refraction n. In this case the basic thin lens equation becomes:

1/f = 1/i + n/o : f & i & o are actual focal length, image and object distances; magnification = ni/o

1/f = (1/i)(1+ni/o)= (1/i)(1+m) or i = f(1+m)

1/f = (n/o)(1+(1/i)(o/n)) = (1/o)(n(1+m)/m) or o = f n(1+m)/m

Restrictions: Lens under water, camera body above water (see above schematic for nomenclature).

lens.to.object.distance = f (n(1+m)/m) < water.depth

*f < water.depth/(n(1+m)/m)*
image.to.lens.distance + lens.to.object.distance = f ((1+m) + n(1+m)/m) > water.depth + clearance

*f > (water.depth+clearance)/((1+m)+n(1+m)/m)*
Solving for minimum magnification (when the lens just touches the water) one finds that for fixed focal length:

m.minimum = 1/(water.depth/(nf) - 1)

*as focal length decreases the minimum magnification decreases*
Solving for maximum magnification is more difficult because some squared terms are involved. Here's the result:

m.max = (b+SQRT(b^2-4n))/2

*where b=(water.depth+clearance)/f -(1+n)*
DISCLAIMER: the above has not been verified by anyone, but I think it is a reasonable thin-lens result. It is based on thin lens optics with the object side of the lens looking through a close thin flat plate into a medium with index of refraction n. In this case one can show the approximate basic thin lens equation becomes:

1/f = 1/i + n/o : f & i & o are actual focal length, image and object distances; magnification = ni/o

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Last edited by newarts; 02-22-2011 at 11:21 AM.
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