Optical formulasWhile photo lenses contain multiple lens elemens they can be viewed as a single optical element. The laws of geometric optics can be applied to a lens like they can be applied to a single element.
The following diagram shows how a point is projected by a converging lens onto its image. The lens lies on the optic center H perpendicular to the optic axis O. The two focal points F and F' are located left and right to the lens element at the distance f which is the focal length. The distance of the object from the optic center is the object distance g, and the distance of the image from the optic center is the image distance b. The photographic distance d is the distance between object and image, i. e. g+b.
- Angle of view
- Field of view
- Depth of field
- Hyperfocal distance
- Optical power
- Focal length of lens with close-up lens
- Magnification of lens with close-up lens
Angle of viewThe angle of view of a lens depends on its focal length and the size of the image. In lens makers' data sheets you most often find the diagonal angle of view. The formula to calculate the angle of view is
α = 2 × arctan(B / (2 × f))using:
- Angle of view
- Image size
- Focal length
f = B / (2 × tan(α / 2))Sometimes the horizontal angle of view is given, or sometimes you like to calculate one angle from the other. The following formulas will help you with that:
Diagonal image size: B = √( Bx² + By² ) Horizontal image size: B = Bx Vertical image size: B = Byusing:
- Horizontal image size
- Vertical image size
|35 mm||43 mm||36 mm||24 mm|
|6×4.5||69 mm||56 mm||41 mm|
|6×6||79 mm||56 mm||56 mm|
|6×7||86 mm||66 mm||56 mm|
With these formulas it becomes clear why different focal lengths are considered “standard” for different image formats. If you calculate the angle of view from these focal lengths, you will get a value of about 50°. So what these different “standard” focal lengths have in common is a similar angle of view
Field of viewThe field of view is the size (e. g. the width) of an object, filling a frame when photographed from a certain distance.
G = 2 × d × tan(α / 2)using:
- Field of view
- Angle of view
α = αhor
MagnificationThe magnification is the relationship of the size of an object and the size of its image on film:
A = B / GWith 35 mm film, for example, a frame-filling image (36 mm wide) of a 36 mm wide object is said to be of the magnification 1. A smaller image has a magnification of less than 1. The value is usually given as a fraction, e. g. 1:1 instead of 1, or 1:2 instead of 0.5.
You can insert the formula for G into the above formula and thus can calculate magnification using focal length and distance. However, for close distances you will get incorrect values. This has the following reasons:
- The formula for G is only an approximation for the real relationship and is only accurate enough for large distances (i. e. d >> f).
- Many modern photo lenses use internal focusing, and with that the focal length changes with the focus distance. This effect is stronger for shorter distances. So f is actually smaller than its nominal value when you focus to shorter distances.
The calculator below makes accurate calculations.
EnlargementThe enlargement value is mostly given for binoculars and telescopes. It's simply the relationship between the field of view using binoculars and the field of view using the bare eye. The formula is:
V = Gf1 / Gf2using:
- Field of view using a “standard” focal length
- Field of view using a “longer” focal length
V = f2 / f1using:
- “standard” focal length
- “longer” focal length
So for example, using a 400 mm lens on a 35 mm camera (with its “standard” focal length of 50 mm) will result in an 8 times enlargement, comparable to binoculars with 8× enlargement.
Depth of fieldWhen you set up the camera to use a certain aperture and you focus to a given distance, there will be a range around the given distance that can be considered “in focus”. This distance range is the “depth of field” (or sometimes also called “depth of focus”). Depth of field depends on magnification and aperture. Magnification, as we've seen, depends on focal length and distance. The formulas to calculate the depth of field are as follows:
front DOF = c × F × d² / (f² + c × F × d) rear DOF = c × F × d² / (f² − c × F × d)using:
- Diameter of circle of confusion
- Aperture number
The front DOF is the range in front of the object that is in focus, the rear DOF is the range behind the object that is in focus.
The parameter c is the diameter of the disc that is produced when a sharp dot of the object is out of focus and rendered as unsharp disc in the image. Discs smaller than c are considered “sharp enough” and hence “in focus”, discs larger than c are considered “not sharp enough” and hence “out of focus”. The circle of confusion is basically a threshold value for acceptable unsharpness. Technically, only the plane of focus is really sharp in an image, and everything before and behind the plane of focus is more or less unsharp. So the value of c is more or less arbitrary. If you are more critical with sharpness, c should be smaller, otherwise it can be larger. There is also considerable discussion if c should be chosen differently for different film formats. The proponents say that you will likely make smaller enlargements of larger film formats, and hence c should be bigger. The opponents say that DOF is only influenced by the lens, aperture and distance and not by the film format (see formula), and hence c should be constant.
Hyperfocal distanceWhen you set the focus distance to the hyperfocal distance with a given aperture, the depth of field will range from about half the hyperfocal distance to infinity. This is also the maximum depth of field for the given aperture. The formula for the hyperfocal distance is:
dh + rear DOF = ∞If you then insert the formula for rear DOF and isolate dh you will get:
dh = f² / (c × F)
The optical power (unit: diopter) is reciprocal to the focal length (in meters), i. e.
D = 1 / f
This means a close-up lens with +2 dpt has a focal length of 500 mm.
Focal length of lens with close-up lens
A close-up lens (maybe in the form of reverse-mounted lens) changes the focal length of the combined lens system. The optical power of the close-up lens is added to the optical power of the base lens, i. e.
Dtotal = Dbase + Dclose-upConverted to focal length, the formula is
1 / ftotal = 1 / fbase + 1 / fclose-up ↔ ftotal = 1 / (1 / fbase + Dclose-up)(with focal length in meters!)
If you prefer to use mm, add a factor of 1000:
ftotal = 1000 / (1000 / fbase + Dclose-up)
For example, a 100 mm lens with a +2 dpt close-up lens has a total focal length of 83 mm, i. e. it is reduced by the close-up lens.
But how does a smaller focal length result in a larger magnification? Well, the maximum extension of the lens is still unchanged. For example, if the above 100 mm lens has 1:1 magnification, it has 100 mm of extension. With the focal length reduced to 83 mm, maximum extension is now larger than the focal length, allowing it to focus closer and resulting in magnification larger than 1:1. So adding the close-up lens alone is not enough to increase magnification, you also have to get closer to the subject, and the close-up lens is just making this possible.
Magnification of lens with close-up lens
The magnification of a lens with a close-up lens is always the same when set to infinity. The formula is:
A = fbase / fclose-up ↔ A = fbase × Dclose-up
50 mm macro lens, focused by extensionMany lens makes offer such a lens, and many of them can focus up to 1:1 magnification. Focusing is purely by extension, i. e. the entire lens element package is moved forward. The spacing of the individual elements does not change.
With the calculator we can check the data in the lens' data sheet. We choose any film format and enter the focal length “50” and the distance “0.2” (the shortest possible distance of the lens). This results in a magnification of 1:1 as expected. We also see that the object distance and image distance are the same, and both are exactly half of the focus distance. The focal points of this system are 50 mm away from the optic center (hence “focal length”). Between the rear focal point and the image there's another 50 mm distance. This distance is the lens' extension. The focal length and the extension add up to the image distance. This confirms the rule of thumb that “with an extension as long as the focal length you will get a magnification of 1:1”.
100 mm macro lens with internal focusingMany lens makers offer macro lenses with a longer focal length. These are, however, usually not focused by extension only, but internally (possibly by a combination of both). With this the lens elements move not only away from the film plane but move also relative to each other. The advantage is that you don't need as much of extension. The disadvantage is that the focal length is reduced when focusing at shorter distances.
Again we check the data in the data sheet of a lens. The focal length is 100 mm, and the shortest focus distance is 0.35 m. This is supposed to result in a magnification of 1:1. However, the above calculator displays a much larger value. This shows that at this distance the focal length is actually shorter than 100 mm. We can now reduce the focal length step by step until we get a magnification of 1:1. We will end up at a focal length of 87.5 mm. So this lens, focused at its shortest distance, has an actual focal length of 87.5 mm instead of 100 mm.
Please note that the resulting value is not exact using the above calculator. The value, however, indicates that indeed the true focal length is reduced, which is correct.
© 2011 Michael Hohner; This page was last changed on 2014-08-25