Originally posted by RBellavance ....

The working distance with the DCR-250, regardless of the lens it is attached to, is 11cm (it is a +9 diopter, working distance = 1/9 meters). That is the distance from the DCR-250 itself to the subject.

For example, the Vivitar 105/2.5's minimum focusing distance is 351mm. That is measured from the sensor, and translates to a working distance of 13.8 cm at 1:1 (the lens measures 17cm, and the registration distance is 4.3cm. 35.1 - 17 - 4.3 = 13.8). In this case, the gain in magnification is insignificant........

I'm sorry, but your assumption is incorrect. A close-up lens is like any other lens; it does not have a fixed working distance, it has a focal length. In the case of the Raynox DCR 250 (an 8 diopter lens) this focal length is 125mm. See:

DCR-250 Super Macro conversion lens for D-SLR camera
Here's what's actually going on when a close-up lens is added to a camera lens, macro or not (some simple algebra is required):

(1) When a lens of focal length f' is placed next to a lens with focal length f" the combination has a focal length given by:

*f=f'f"/(f'+f") if there's a distance d between the lenses the formula is f=f'f"/(f'+f"-d)*
The equation can be re-arranged to yield

f=f'/(1+(f'-d)/f") = f'/(1+m_close-up) where

**m_closeup = (f'-d)/f"** - we'll use this later

Putting a close-up lens close to a camera lens always results in a "new" lens with a shorter focal length. If the distance to the sensor doesn't change, the image gets larger.

A lens projects an image on a sensor according to the following rule:

(2) when a lens is positioned at a distance from a sensor, the magnification, m, focal length, f, and distance, s, from the sensor are related as:

**s=f(1+m)** - when the lens is focused at infinity, f=s, ie the lens is one focal length from the sensor, and m=0
Now when a close-up lens f" is added to a camera lens of focal length f' already focused at infinity, the distance from the sensor doesn't change but the focal length does:

s=f'=f(1+m)

Substituting equation (1) for the new focal length f results in:

s=f'=f'f"(1+m)/(f'+f"-d)=f'(1+m)/(1+m_close-up)

solving for m results in m_close-up=(f'-d)/f".

* the magnification due to the close-up lens is (f'-d)/f"*__ when the primary lens is focused at infinity__.

Now let's start with the same f' lens, this time positioned such that the magnification due to extension is m_extension. The distance from the sensor is now:

s_extended=f'(1+m_extension)

Now we add a close-up lens f" to the original lens; the distance from the sensor doesn't change, so

f'(1+m_extension) = f(1+m)

Substituting for the new focal length due to the close-up lens gives

f'(1+m_extension)=f'f"(1+m)/(f"+f"-d) = f'(1+m)/(1+(f'-d)/f") = f'(1+m)/(1+m_close-up)

Solving for total magnification m yields a useful, simple result

**m=(1+m_extension)(1+m_close-up)-1**
I am sorry about the algebra, but it is needed to explain what's going on.

To understand what happens with working distance (distance from lens to subject) or closest focus distance (distance from sensor to subject) use:

working_distance = f(1+1/m)

closest_focus_distance = f(1+m)(1+m)/m

and make the appropriate substitution for f as modified by the addition of a close-up lens.

Dave