[newarts]

Have you ever wondered exactly where your lens is in space? Like when they say the distance from the lens to the subject is 0.42m - is that from the glass -if so, which glass?

Perhaps you've heard of "principal planes" which are imaginary locations in space that act like they are the real location of a lens (looking from the front) - "Front Principal Plane" and "Back Principal Plane" (looking from the back of the lens. These locations are chosen to insure the thin lens optics equation work correctly*:

1/focal.length = 1/image.distance +1/object.distance where image.distance is between the image and the back principal plane, etc.

magnification = image.distance/object.distance

Formulas for compound lenses are often simple like this one for the focal length of stacked lenses of focal lengths f' and f" separated by a distance d.

f=f'f"/(f'+f"- d) where d is the distance between neighboring principal planes - one front and one back

So I wondered where these principal planes might be for a modern Internal Focus zoom lens like my Pentax DA 55-300mm 1:4-5.8. I focused the lens at infinity and added a small extension; some object was in focus for each zoom setting. I used the distance from the lens to the object and magnification measured from the photographs to locate the principal planes in space. Here's the results:



The principal planes when the zoom is adjusted to 300mm float freely in space completely outside the physical lens! The extended lens is less than 300mm long; and the back principal plane for a 300mm lens focused at infinity must lie one focal length in front of the image plane (sensor) - therefore the back principal plane must be outside the physical lens. The back of the virtual 200mm lens happens to coincide with the front end of the physical lens.

At an effective focal length of 100 mm front & back principal planes virtually overlap - somewhere in this focal length region the virtual lens may have a negative thickness ! Finally, the 55mm virtual lens has Principal planes that look normal.

I don't know of any practical purpose for this data, I figured it out as a matter of curiosity and to see if my experimental methods would work (they did.)

Comments are welcome especially if they can point out any practical value for such data - other than to increase our wonder that the theory and its practice should be so much like magic.

* This brilliant scheme to simplify the mathematics of optics was due to the great mathematician, Gauss (http://en.wikipedia.org/wiki/Carl_Friedrich_Gauss).

---

Last edited by newarts; 01-29-2012 at 03:09 PM.

[newarts]

.....
I don't know of any practical purpose for this {principal plane] data, I figured it out as a matter of curiosity and to see if my experimental methods would work (they did.)

Comments are welcome especially if they can point out any practical value for such data - other than to increase our wonder that the theory and its practice should be so much like magic.

Here's an application - calculating clear distance between physical lens and subject

Calculated macro working distance, wd=f(1+1/m), extends from the front principal plane of a lens, not from the physical lens. Hence knowing where the front principal plane is can be important for setting up macros.

Extending a lens with bellows or extension tubes etc does not change the location of the principal planes with respect to the physical lens so a principal plane in front of the physical lens adds to the working distance.

This is not the case when close-up lenses are used.

[newarts]

Here's a more useful diagram showing approximate principal dimensions of a Pentax K-X with a DA 55-300mm ED lens mounted; all dimensions in millimeters. The camera's sensor is about in alignment with the white mark near the mode dial and is the origin for the axial measurements shown.


These are suitable for estimating work only - they may be off by a few mm.

Here's an example calculation using front principal plane data;
What's the effective focal length of the combined lenses if a 200mm closeup lens is put on the above lens fully extended at 300mm?

f= f'f"/(f' + f" - d)

Where d is the distance between the rear principal plane of the closeup lens and the front principal plane of the primary lens. In this case, the closeup lens' rear Principal plane is behind the frn of the main lens by 411mm, hence d = - 411.

f = 300*200/(300+200+411) = 66mm.

This predicted focal length is consistent with predictions of magnification increase by lens extension; ie. change in mag = extension/(focal length.)