# How do fixed lenses "focus" on things at different distances? [duplicate]

I'm trying to understand depth of field and I'm struggling.

If I'm playing with a 50mm fixed lens, and I alter its focus using the little ring around the lens, what is actually changing in terms of depth of field? I'm trying to understand DoF a bit more. From my understanding, it's a calculation of aperature, focal length, and focal distance. By changing the focus manually, what's changing? Is it the focal length?

Sorry if this is vague. I'm just struggling to wrap my mind around the technical details of what happens when you alter a lens's focus. What variable is changing?

• By "fixed", are you referring to what are also called "prime" lenses, which have a fixed focal length, or lenses that don't change focus, like on single-use cameras, or both? Although you mention changing focus, you also put "focus" in quotes. Changing aperture can increase DOF, giving an illusion of "focusing". Aug 20, 2019 at 5:53

Lenses these days are complex contraptions but the principal lens model they try to approximate is a single ideal convex lens of the given focal length. When you are changing focus, you are changing the distance of that lens to your imaging plane.

The focusing model then has rays spreading out from each sensor point to all points on the ideal lens that are reachable through the aperture (which is idealised into the same plane as the ideal lens) and from this sort-of circle shape a cone of viewing rays converges to the corresponding point in the focused plane. "Viewing rays" can be cut short by something before the focusing plane or, if they can continue onward to some background point. So basically you have these cones from the lens/aperture converging to points in the focusing plane and spreading out afterwards again.

The closer your focus plane, the more rapid the cones converge and spread out again. And obviously a larger aperture also makes them wider.

The variable changing as you change focus is the position of the idealised lens before your plate/film/sensor plane. Early cameras had a bellows to accommodate moving the lens. Large frame cameras still have them because there are reasons to not just move the lens but also shift it (for avoiding "tumbling lines" without having to place the horizon in the middle of the image) and tilt it (to get an oblique focus plane rather than one parallel to the image). But the normal focusing change is just varying the distance of the lens to the imaging plane.

Think about a tiny portion of a vista, maybe the gleam in a pretty girl’s eye. If this object is a far distance, the light from this tiny object arrives at the lens as parallel rays. The shape of the glass lenses and their density alters the path of this incoming light. It departs the lens as converting rays. A trace of these rays resembles a cone of light. The apex of this cone will form 50mm downstream from the lens (provided the source was at a far distance away (infinity ∞) and the lens has a 50mm focal lenght. Since this 50mm fixed lens has limited ability to alter the angle the rays exit, should the object be closer than ∞, the apex of the cone will form further downstream. When you focus by turning the focus ring on the lens barrel, you are adjusting the distance lens to image sensor / film in an attempt to kiss off the apex of the cone of image forming light on film or sensor.

Briefly, all objects in a vista can be consider to consists of countless tiny points. Each will be the origin of light rays. If the object is distance, these rays enter the lens parallel rays. If the object is close to the camera, the rays striking the lens are not parallel, they are diverging. The lens redirects all these rays. There new path is a cone of light. The apex of the cone of light from each object has a finite length. The shortest distance lens-to-apex is a trace of a distant object. Objects that are close have longer lens-to-apex distances. When you turn the ring, you are adjusting the distance lens-to-image sensor. An object is in focus only when the apex of the cone of the image forming rays kiss the image sensor. When this happens, the object is imaged as a super tiny circle of light. This tiny circle has indistinct margins; we call them circles of confusion. The entire image of a vista consists of countless such circles. Objects in focus reproduce as super tiny circles. Objects out-of-focus reproduces as larger circles. To appear sharp, the circles of confusion must be super tiny, so tiny they are perceived at points of light and not circles of light.

Depth-of-field is that zone before and behind the distance actually focused upon, in which objects appear to be reasonably in sharp focus. This zone contracts and expands based on the diameter of the camera’s aperture setting. Tiny diameters of aperture expand the zone. Depth-of-field has many elements, aperture, subject distance, acuity of the observer’s eye, lighting level of the image being viewed, contrast of the image being viewed, and the distance of the observer to the image being viewed. The complexity of depth-of-field is too great for these few paragraphs. Why not study up on the subject. In a nutshell, if the circles of confusion are seen as points of light, the image will be declared tack sharp. If the circles are visible to the observer as circles and dimensionless points, the observer will declare the image to be fuzzy.

Most depth-of-field tables are based on a circle size of 1/1000 of the focal length. Thus for a 50mm lens, that's 0.05mm. Such a value assumes an 8x10 inch print or image will be made. For 35mm film that requires about 8X magnification. For critical work Kodak used 1/1750 thus 0.028mm.

• DOF tables are based upon sensor/format size (the CoC limit), not FL. I.e. .03mm for 35mm/FF, and .02mm for APS-C. This is because the smaller format has to be enlarged more to make that 8x10 print. It also assumes "standard viewing", which is viewing distance equal to the image diagonal, and not an 8x10 print. But it could be an 8x10 print viewed from 12". Aug 20, 2019 at 17:39
• @ Steven Kersting -- Based on 1/1000 of the focal length does take into account the degree of magnification needed to make an enlargement. A 50mm is considered "normal" for a FX 24mm height by 36mm length. To make an 8X10 requires a minimum of 8.5X enlargement. Thus 50/1000X8.5=0.42mm on final image. For the DX 16mm height by 24mm length. Norma is a 30mm lens.To make 8X10 requires 12.7X enlargement. The math is 30/1000X12.7=0.038mm circle of confusion. Accepted value is 0.5mm on the finished image viewed from standard reading distance. The 1/1000 of focal length works! Let's keep it simple! Aug 20, 2019 at 20:55
• @ Steven Kersting -- The industry standard for viewing is 3.4 minutes of arc which is a circle whose diameter is 1/1000 of the viewing distance. That's equivalent to 1/100 of an inch viewed from 10 inches and this value has been widely adopted. The whole thing is a variable based on the acuity of the observer's vision, the contrast of the image being viewed, and the amount of light in the viewing area. Aug 20, 2019 at 21:06
• @ Alan Marcus -- Do you have any references? Because in 40+ yrs of photography I have never heard/read that. And your math doesn't seem to work either... i.e 10" = 254mm, ÷ 36mm = 7.055x. If I multiply the standard 35mm COC of .03 x 7.055 I get .21mm in the final image. Aug 20, 2019 at 21:51
• @ Steven Kersting -- An 8 x 10 inch print measures 203mm by 254mm. The full frame 35mm measures 24mm by 36mm. 230 ÷ 24 = 8.46 and 36mm ÷ 254 = 7.06. These two values are the magnification needed, you must use the larger to fill the 8x10. Thus we use 8.46X or the height will not be achieved. Data to use 1/1000 of focal length was from several textbooks Photographic lenses by C.B. Neblette. Introduction to Photographic Principles by Lewis Larmore. Photographic Optics by Arthur Cox – to name a few. By the way, hyperfocal distance, a subset of DofF is also based on F ÷ 1000. Aug 20, 2019 at 23:34

With a simple lens, when you turn the focus ring you are moving the whole lens closer or further away from the sensor. Much like you would focus the sun to a spot of light using a magnifying glass.

The sun is effectively infinitely far away. A simple 50mm lens would be 50mm from the sensor when focussed on the sun (not normally true for camera lenses due to complex lens design).

As the subject to wish you focus on gets closer to the camera the lens needs to be moved further away from the sensor.

Whenever you focus on an object some of the rest of the scene is also in focus. The amount of the scene that is in focus is referred to as the depth of field.

Depth of field depends on one thing only - the size of the aperture as seen by the subject. As this gets bigger the depth of field gets shallower.

When you bring the lens closer to the subject the physical aperture appears bigger, reducing the DOF. If you use a longer focal length lens at the same distance the physical aperture is bigger for a given f stop, so the DOF is again reduced.

physical aperture = focal length / f-stop

e.g. for a 50mm lens setting the aperture to f/4 means the hole through which light enters is 12.5mm

• Except it is the entrance pupil, not the physical aperture that determines f-number. The proper equation is "f-number = focal length/entrance pupil diameter. Aug 22, 2019 at 5:24
• Depth of field depends on two things: total magnification and f-number. Aug 22, 2019 at 5:25
• I was using appropriate language. But how the aperture appears to the subject is the entrance pupil. Magnification depends on subject distance, which affects the appearance of the aperture. Aug 23, 2019 at 7:27
• When one uses "physical aperture" one is referring to the actual physical size of the aperture opening, as measured using the physical aperture mechanism itself, not the virtual aperture or entrance pupil as measured from in front of the lens. The two are often very different, depending on the amount of magnification between the physical aperture and the front of the lens. Entrance pupils are typically much larger than physical apertures at longer focal lengths, while they are usually much smaller with retrofocus wide angle lenses. Aug 23, 2019 at 21:47
• "Magnification depends on subject distance, which affects the appearance of the aperture." Magnification also depend upon the enlargement ratio between the negative/sensor and the display size, as well as the viewing distance. The same image with the same subject distance, aperture, and focal length will have different DoF when displayed at different sizes and/or viewed from different distances. Aug 23, 2019 at 21:54

When you change lens focus it does not affect the DOField. That is simply a factor of aperture and magnification.

Lens magnification (focal length) determines how large a point will be when it reaches the sensor. Subject distance determines how large that point is initially (i.e. larger if closer), and therefore affects the magnified result. It is not focus distance per-se, but if the subject is in-focus they are effectively the same thing. And display magnification (size/viewing distance) affects the relative size/magnification by which you judge sharpness/acceptable sharpness.

Aperture determines the DOFocus at the image plane. A smaller aperture has a narrower DOFocus, which means more of the scene is in-focus at the image plane rather than more of the scene being in-focus behind, or in-focus in front of the image plane (recording as out of focus).

DOField is a perceptual quality... basically, the larger a point appears to you the less sharp/acceptably sharp it will seem to be.

A simple experiment could make this easier to understand: Take an image of marginal sharpness and view it small on your computer... it will seem sharper. Then increase its' size until the lack of sharpness becomes apparent... it is now "less acceptably sharp." Now move some distance away from your computer and the image will again appear sharper. That is DOField in action... and the same applies to the image as recorded on the sensor (relative size/magnification and DOFocus) because that affects the initial source you're viewing.

When you focus a lens you are moving the DOFocus in relation to the image plane. This can be done externally with a bellows by moving the lens forward/backwards, to move the DOFocus forward/backwards on the image plane; or by moving the image plane/body forwards/backwards w/in the DOFocus. With an internal focus lens, focusing elements move in order to move the DOFocus in relation to the image plane. You can also move the camera and lens together w/o focusing in order to move the DOFocus w/in the scene (i.e. a macro focusing rail).

• Magnification in the context of DoF includes everything that affects the size of the actual object compared to the size of the final image. Focal length affects magnification. So does subject distance/focus distance. That's why a macro lens with 1:1 maximum magnification can make the same object look larger than a lens with the same focal length but only 1:4 MM - because it can focus the object at a closer distance. Ultimately magnification includes subject distance, focal length, and magnification ratio between the image projected onto the image plane and the display size. Aug 21, 2019 at 5:00
• I don't see where I said anything different. Aug 21, 2019 at 14:35
• "When you change lens focus it does not affect the DOField. That is simply a factor of aperture and magnification." Changing lens focus affects magnification and DoF, because it affects the distance upon which magnification is based. If a lens is focused twice as close as it was before, the magnification of a subject twice as close as before will be twice as much as before. Aug 21, 2019 at 21:54
• I included that as subject distance... and if the subject is in focus that is also focus distance. But it is the subject distance that matters in terms of DOF, not whether it is actually in focus (or how much of it is in focus). Magnification is constant... it will be twice as large because it is twice as close. Aug 21, 2019 at 22:16
• **Other than lenses that exhibit focus breathing Aug 21, 2019 at 22:45