Suppose we have a convex lens for a middle school physics experiment, a screen, and an object being imaged, under what state will we find the hyperfocal state?

The screen is the imaged screen. It can be a movie screen or a white piece of paper.

The lens can be imaged on the screen. In hyperfocal distance, what is the distance between the object, lens, and screen?

Is it calculated using the following formula?


Are clear images from half of H to infinity?

  • I'm assuming you know the definition of hyperfocal distance, that the formula depends on only focal length, f-number, and circle of confusion size. Given that, it's not clear what you are asking. – scottbb May 20 '20 at 7:11
  • @scottbb What is the relationship between the screen and the lens and the imaged object is the hyperfocal distance? – enbin May 20 '20 at 7:17
  • What screen? Please describe the geometry. Have you read the Wikipedia article I linked to? If so, what part of it do you not understand? – scottbb May 20 '20 at 7:42
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    I’m voting to close this question because it is not about producing photographs. – Michael C May 20 '20 at 8:28
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    NO. If ƒ changes relative to v, then using the lensmaker's equation, please show me, which parameter changes: the refractive index of the glass, the radius of the front of the lens, or the radius of the back of the lens. The answer is that none of them change, because the physical characteristics of the lens (refractive index, and shape) are not variable. – scottbb May 21 '20 at 15:43

"Hyperfocal" refers to the condition where depth of field allows the lens to be "in focus" from some minimum distance to infinity.

This depends on a core assumption: the size of the acceptable "circle of confusion," which is determined by the actual aperture diameter and lens focal length, but also by the amount of enlargement the image will receive before final viewing. That is, a 250 mm lens at f/11 will have greater depth of field on an 8x10 negative than it will on an APS-C or Micro 4/3 digital sensor, because the 8x10 is likely to be viewed as a contact print, while the crope sensor image will be enlarged at least to screen viewing size (around 20:1, give or take).

Once this acceptable circle of confusion is determined, it's a fairly simple calculation to determine how far in front of or behind the plane of critical focus objects can be and still produce images with this size or smaller circles of confusion. This takes the form of a table or graph that is condensed into a depth of field marking on a lens or focusing rack.

Finally, one can then set a focus that, at the specified aperture setting, will produced "in focus" depth of field that just extends to infinity, and generally to half the set focus distance (many old box cameras were actually focused at about 10 feet, but had aperture that made them hyperfocal, so they'd say "five feet to infinity" or "place your subject at least five feet from the camera").

It's important to remember that "depth of field" isn't a physical condition or quality -- it's a measure of how much defocus the user is willing to accept. If you scan an 8x10 negative that looks razor sharp all over to the eye at high enough resolution and examine it at 1:1 on a good monitor, you'll find that even at f/32 there's still a plane of critical focus, and everything not in that plane is at least a little bit fuzzy -- but if the fuzziness isn't visible in normal viewing, we consider that "within depth of field," and if that depth of field extends just to infinity, the setup was hyperfocal.

  • Can we use the imaging of the convex lens on the screen to explain hyperfocal distance? – enbin May 20 '20 at 15:56
  • In a demonstration, you could image point light sources (small LEDs, perhaps) at various distances and measure the size of the circular spot they produce, against an arbitrary "in focus" size, and demonstrate how "in focus" and "out of focus" are distributed with distance to the light source. – Zeiss Ikon May 20 '20 at 16:08
  • I use the formula H = f^2 / Nc + f to calculate H, then calculate v according to the formula 1 / H + 1 / v = 1 / f, and then image according to f, H, v. Is this okay? – enbin May 20 '20 at 21:11
  • If we assume that the distance H from the object to the screen is constant, then we can go to the following formula: 1 / (H-v) + 1 / v = 1 / f. From this formula, we can see that change v and change f are the same thing. This is the so-called change v which is the relative change f. – enbin May 21 '20 at 2:55

Your question seems to be based upon an assumption that a human viewer can see the difference between "blurry" and "in focus" at the system limits of a lens system. This is usually far from the case without magnifying the results by a large factor.

Can we use the imaging of the convex lens on the screen to explain hyperfocal distance?

Not very well. Why?

Because the image on your viewing screen will not be very large. The human eyes observing it will not be able to discriminate between slight blur and the most sharply focused parts of the image. Thus pretty much everything will look in focus at that very small display size, even parts of the image that, according to your depth of field and hyperfocal distance formulae should be outside the depth of field. Keep in mind that hyperfocal distance is a corollary of depth of field, which is an illusion in the context of humans viewing an image.

In the context of optical physics, depth of field and the corollary hyperfocal distance are determined by the limit of the system to discriminate blur that is less than a certain size. But the limits of such optical systems are usually far finer that the ability of human eyes to see the difference unless we highly magnify the image so that our eyes can perceive the system limits of the lenses and imaging medium (film, digital sensor, etc.).

When we talk about a human observer viewing an image, the system limit is almost always the viewer's vision. The results of photography are meant to be observed by human eyes. The ability of the viewer's eyes to discriminate fine details is usually the weakest link in the full system that produces a perception in the mind of the viewer of a subject imaged by a lens. Photographs are thus judged in terms of depth of field and hyperfocal distance by the limits of the viewer's ability to discriminate fine details. If you take the exact same captured image and the same viewer looks at it at different display sizes from the same distance the depth of field and hyperfocal distance will be different for each display size! This is because as the image in enlarged by an increasing factor, blur that was too small to be seen as blur by the viewer is eventually large enough that the viewer can tell it is blurry.

For your construction to work as a way to demonstrate, based on human perception, depth of field and hyperfocal distance you need to be able to magnify the resulting image sufficiently that human observers can perceive blur at the system limits of the lens.

  • When humans use the lens to image on the screen, they can also see whether the image is blurred. So is the position of a certain lens a hyperfocal position? – enbin May 21 '20 at 2:28
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    They can't see it very well though. For most lenses pretty much everything will appear to be in focus at the hyperfocal distance calculated by your formulas, which are concerned with the limits of your lens, not with the limits of human perception. – Michael C May 21 '20 at 2:51
  • Lenses of different focal lengths have different hyperfocal distances. – enbin May 21 '20 at 15:35
  • The same lens has different hyperfocal distances depending upon the amount of final enlargement of the resulting image, the distance from which it is being viewed, and the visual acuity of the viewer. – Michael C May 21 '20 at 19:47

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