If I have a +1, +2, +4, or +10 close-up lens which attaches to the front of my regular lens, how can I calculate its effect on close focusing distance, figure out how much magnification I can get from the combination, and tell what size of object will fill the frame at the closest distance?
(We have similar questions for reversed lenses and extension tubes but not for secondary lenses.)
It is really rather simple. The diopter value will tell you how close you must be to the subject of your photograph, the farthest distance you can get a sharp image with the close-up lens on. you divide one metre by the diopter value to get the distance, which is the focal distance of the close-up lens.
Hence a +1 lens will give a far distance of 1 m and the +2 lens will give a far distance of ½ m the +4 lens will provide a far distance of 25 cm (¼ m) and finally the +10 lens will have a far distance of 10 cm
All those distances are measured from the attached lens.
The closest distance is a trickier subject. It depends on how close your original lens will focus and how far it extends from the sensor. If you measure the distance from its front end at its closest focusing distance and divide one metre by this distance, you get a diopter value that corresponds to the focusing range of the lens.
If you have one of the more common kit zoom lenses, it might be 18-55 mm and focus down to about 30 cm from the sensor. The front end of the lens is about 20 cm from the subject when it is at its closest, which corresponds to about five diopters or perhaps somewhat less. Hence adding one or two diopters won't make much difference on its closest distance, mostly it will just degrade image quality.
Only the ten diopters lens will provide a substantial advantage regarding image size and close distance, but at the price of severely degraded performance, as simple lenses like this are rife with lens aberrations, as colour aberration, coma and distortion. The result is that you cannot take a sharp picture at all with the ten diopters lens attached.
Looking at it from a brighter side, you might have a tele lens or a tele zoom lens, which focuses only down to about one metre. Adding a one diopter lens will take you to distances between one metre and a half metre and two diopters will give you even closer range, without sacrificing so much image quality as with the ten diopters lens. The one diopter and two diopters lenses can be very useful with a tele zoom. The one diopter will double the reproduction scale if your lens otherwise focuses only to one metre. And the two diopters lens will more than triple it.
It should be remembered that you can never focus to a larger distance than the focal length of the attached close-up lens, so with one diopter it is one metre from the lens and with two diopters it is 50 cm from the lens. Farther distances are not possible.
For higher diopter values than 2 you would need an achromatic close-up lens to get sharp images. Those are more expensive than simple lenses, but perform much better. With a tele zoom and an achromatic close-up lens, you may achieve performance much like a real macro lens when you take pictures of natural objects, where straight lines and curvature of field are not important. However, it is not suitable for reproduction of flat or rectilinear objects, as it will still give distortion and curvature of field, and it will have some coma error as well. The only correction of an achromatic close-up lens is colour correction, which makes images sharper with less fringing toward the edges of the image.
Yes you can mount supplemental close-up lenses over your camera lens. These lenses are akin to ordinary reading eyeglass, they work but they are not photo-grade. That said, you can go to an eyeglass counter at the drugstore and place one of the reading glass lenses over your camera lens and test and see how much closer you can achieve suitable focus. The power of the reading glass lens is marked on the earpiece or on the tag. This value is a measure of the focal length using a unit of measure called the diopter. Commonly, reading glasses are labeled 2+ or 2.5+ or maybe up to 5+.
1 = 1000mm = 39.37 inches -- +2 = 500mm = 19.7 inches -- +3 = 333mm = 13 inches -- +4 = 250mm = 9.8 inches -- +5 = 200mm = 7.9 inches -- +6 = 167mm = 6.6 inches -- +7 = 143mm = 5.6 inches -- +8 =125mm = 5 inches -- +9 = 111mm = 4.4 inches -- +10 = 100mm = 4 inches.
While the lenses in store bought reading eyeglasses will work ok, best you buy photo grade supplemental close-up lenses from a photo supply store. All close-up lenses induce lens degrading
The worst is chromatic. This aberration shows itself by colored hallows surrounding objects. Photo grade close-up lenses reduce the effect of aberrations. The better ones are two lenses mounted together. This is called an achromatic lens (Latin for colorless).
A lens's diopter is just the reciprocal of the focal length of the lens, with units of 1/[meters]; equivalently, the focal length of a lens is just the reciprocal of the diopter value, in meters. So a +1 diopter corresponds to a focal length of 1 meter; +2 diopter corresponds to 0.5 m (500 mm); +10 diopter corresponds to 0.1 m (100 mm); etc.
Diopter units are particularly convenient to use when combining lenses together: their equivalent diopter value is just the sum of the individual diopters. For example, suppose a 50 mm lens has a +2 diopter close-up lens mounted on the front. The diopter value of the lens is 1/50mm = 1/0.05m = 20. So the total equivalent diopter value is 20 + 2 = 22, which corresponds to a focal length of about 0.045 m, or 45 mm.
But wait, if the diopter of the combined lens plus close-up lens is just the sum of the diopters, does that work for more than two lenses combined? Absolutely. Adding another +3 diopter to the previous combination results in a total diopter of 25, which corresponds to a focal length of 0.04 m, i.e., 40 mm.
How can I calculate its effect on close focusing distance?
Roughly speaking (see note below), the close focusing distance will be reduced according to:
$$ x_\mathrm{new} = \frac{x_\mathrm{old}}{D\cdot x_\mathrm{old} + 1} $$
where
For example, suppose a lens can focus as closely as 500 mm. With a +4 diopter close-up lens attached, the combination can now focus as close as 500 mm / (4 m-1 ∙ 0.5 m + 1) = (500/3) mm ≈ 167 mm.
Note: The reason I say "roughly speaking" is that these distances are based on distances from the center of an idealized single thin lens with negligible thickness, such that the thin lens formula can be used. Real-world optical lens elements have thickness, and of course real-world photographic lens systems consist of several individual lens elements and lots of air space in between many of them. Additionally, when talking about a camera lens's Minimum Focus Distance (MFD), the MFD is measured from the camera's film/sensor plane. Thus, the MFD includes the image-side focus object-side focus distances in the thin lens formula, as well as the substantial length of most of the lens assembly (which is usually not constant as a lens focuses). But the "close focus distance" discussed above is merely the mathematically-described object-distance side of the then lens. Thus, this formula is a guide to judge what diopter you might need.
Infinity focus
With a close-up lens, you will no longer be able to focus far away. The furthest you will be able to focus is simply the focal length of the close-up lens – i.e., the reciprocal of the diopter, in meters. So with a +4 diopter you cannot focus any further away than 0.25 meters (250 mm).
How much magnification can I get from the combination?
Assuming you're working at the closest focus distance possible, both before using the close-up lens, and with the new closer focus distance with the close-up lens, then the image will be magnified by a factor of \$(D \cdot x_\mathrm{old} + 1)\$ times larger than it was without the close-up lens.
For example, if a lens's closest focus distance is 500 mm, then with a +4 diopter close-up lens and focus at the new close focus distance of \$x_\mathrm{new}\$, the image will be (4 m-1 ∙ 0.5 m + 1) = 3 times larger than without the close-up lens.
Note that the imaging system's magnification is not 3:1. Rather, the ratio of magnification after vs. before, \$M_\mathrm{new} / M_\mathrm{old}\$, is 3.
