# How do focal length, minimum focus distance, and magnification interact on a lens?

I have a question about zoom lenses that I can't find the answer. It's about these 3 concepts:

• Maximum Magnification (MM)
• Minimum Focusing Distance (MFD)
• Longest Focal Length (let's say LFL :) )

On its own, each is a understandable concept, but I am wondering how the three interact. I mean, do you have the MFD of the whole lens at the LFL? If yes, then obviously it will produce the MM there. If no, will it still have the MM with the LFL at its closest distance?

I don't know if I explained it well, but if you don't understand or have any suggestion to improve the question, feel free to give comments.

• It's really not clear what is your question and I don't even see a question mark.
– Itai
Commented Sep 9, 2016 at 17:21

I suppose it might be theoretically possible for a zoom lens design to have a shorter MFD at a shorter focal length, but I've never came across such a lens.

Most zoom lenses have an MFD that is either the same or very near the same for all focal lengths. This necessarily means the magnification will increase as the focal length increases given that the MFD is the same.

A lens with a retrofocus type of design such as is used for wide angle lenses would be my guess as to the most likely candidate to be the exception if some other element of the lens' design, such as internal focus and/or internal zoom, prevented movement of the lens' elements to allow for the same MFD at longer focal lengths as the MFD allowed at a shorter focal length.

The vast majority of telephoto zoom lenses have relatively small MM: somewhere in the neighborhood of between 0.15X and 0.20X. Some may approach 0.30X, but they are the exception more than the rule. They accomplish this MM at the longest focal length of which the lens is capable.

Most zoom lenses in the normal range of about 24-70mm have MM that ranges from around 0.20X to 0.30X. There are a few exceptions of zoom lenses with significantly higher MM. The Canon EF 24-70mm f/4 IS has a MFD of around 8 inches which gives it an MM of 0.70X. This magnification is only achievable when the zoom ring is moved past 70mm to the Macro position, and infinity focus is lost at this setting. When set to 70mm the MM is a fairly pedestrian MM of 0.21X at about 15 inches MFD. This is similar to most other 24-70mm lenses made by Canon, Nikon, Tamron and Sigma.

For true 1:1 macro capability, the available lenses are almost exclusively prime lenses. This allows lensmakers to avoid both the complexity of a zoom lens and the extension needed by a macro lens to focus on short distances to be combined in the same lens design.

The Canon MP-E 65mm 1-5X Macro is a one-of-a kind lens that some might define as a zoom lens, since the field of view changes by a factor of 500% from 1X to 5X, that is only capable of a single focus distance (e.g. the MFD) at each zoom position. Since focal length is defined by how much a lens bends collimated light to focus it at a particular distance, and since the MP-E 65mm 1-5X can't focus collimated light at the registration distance for which it is designed, in a sense it has no truly definable focal length in the conventional sense.

• Thank you. The reason I asked is because the manufacturer usually mentions only the MFD and MM but does not make it very clear that at which focal length (or zoom position instead) that they happen. So it is safe to imply that the two happen at the same time at the longest zoom position. Right? Commented Sep 12, 2016 at 0:10
• That's a fairly safe assumption. I've never noticed a zoom lens that didn't achieve its MM at the longest focal length. Keep in mind that the vast majority of what most consider to be true macro lenses capable of 1:1 reproduction (1.0X MM) are not zoom lenses but instead are single focal length prime lenses. Commented Sep 12, 2016 at 0:15
• I have a Panasonic FZ100 with effective focal length ranging from 25mm to 600mm. The minimum focus starts at 0.99 feet at minimum focal length and extends to 6.57 feet at maximum focal length. Commented Oct 9, 2016 at 22:48
• With a 4.5-108mm lens for a 6x4.5mm sensor there are a lot of design considerations that are different from larger interchangeable lens cameras. The distance of the lens in front of the sensor is significantly further as a ratio to the sensor diagonal than is the case with most cameras with large sensors. That makes the lens retrofocal at most, if not all, of the focal lengths of which it is capable. Even then, the magnification at 108mm and a focus distance of 6.57 feet will be significantly higher than the magnification at 0.99 feet with a 4.5mm field of view. Commented Oct 10, 2016 at 15:21

As you know, the zoom sports a variable focal lengths. It is the focal length that conveys the power of the lens. We can classify lenses as to their power. A “normal” lens has a focal length about equal to the diagonal measure of the format frame. The typical 35mm camera will be fitted with a “normal” 50mm focal length. Such a lash-up delivers an image one-tenth life size written as X0.1 or 1:10. In other words, a 10 inch tall object will image 1 inch in height. A wide-angle is about 70% of “normal” and shorter. A telephoto is about 200% of “normal” and longer.

When we image far distant objects, the lens-to-film (sensor) distance is equal to the focal length. Indeed, this is how the focal length is defined. When we focus on nearby objects, the lens-to-sensor distance must be extended. We focus by racking the lens forward, extending its distance from the sensor. At life-size, magnification one, written as 1:1 or “unity, the lens extension is one focal length and the object-to-sensor distance will be 4X the focal length. The magnification will be 1X. If we are working with a 50mm, at “unity”, it will be positioned 100mm from the sensor, the sensor-to-object-distance will be 200mm.

As you focus closer and closer, the magnification increases. The image forming rays dim because they are tasked to cover more surface area. At “unity”, the loss of brilliance at the image plane is 4X (2 f/stops). An under-exposure is likely unless compensation is deployed. For this reason, lens makers usually stop the forward movement (close focus), when the exposure error is about 1/3 f/stop. Thus the typical lens stops its forward movement so the closest you can get is about 20 to 36 inches (500 thru 1000mm. This minimum focus distance is likely to remain the same throughout the zoom.

The macro lens design is optimized for close focusing. The front lens group magnifies the lens aperture so that its apparent size appears to expand as you close focus. This cancels image dimming that normally accompanies close focusing. Some designs close focusing to “unity” and beyond.
The mechanism of the zoom is quite complex. Good ones are expensive. A macro zoom will be your best bet.

• Thank you for sharing your knowledge. The Image dimming cancellation of macro lenses is interesting and I didn't know about that. That may explains why they are expensive. Oh, and in the 1x example, won't the sensor-to-object-distance be 200mm? Commented Sep 12, 2016 at 0:05
• You are correct -- 50mm at unity - lens 100mm from sensor - object 100mm from lens - sensor to object 200mm - will stand in the corner for 1 hour. Commented Sep 12, 2016 at 3:48

Here is a extension tube calculator to compute all these for you. The underlying formula is explained below.

An e mm extension tube added to a f mm lens with m magnification results in a magnification of m+(e/f). So a 20mm extension tube on 50mm lens will add a 0.4x magnification.

Worked out example. Canon 80-200mm f/2.8 has a native magnification of 1:7.7 = 1/7.7 = 0.13, adding 32mm of extension tube gives it additional 32/200 = 1/6 = 0.16 magnification, so resulting magnification is 0.13+0.16 = 0.29. Notice that I am using the maximum focal length in the calculation, it works on other lengths too.

Now for minimum focussing distance. The new MFD = (f + (f / (m + (e/f))))(1 + m + (e/f)).

Worked out example. Canon 80-200mm f/2.8 with 32mm of extension tube has MFD = (200 + 200 / (0.13 + 32/200)) (1 + 0.13 + 32/200) = 1147mm = 1.15m.

All of this is based on the thin lens equation.