$$ D = \frac{f\times H}{h} $$ where

$$ \begin{align} D &= \textrm{distance from lens to object} \\ (d &= \textrm{distance from lens to 35mm film}) \\ f &= \textrm{focal length} \\ H &= \textrm{height of object} \\ h &= \textrm{height of object's image on 35mm film} \end{align} $$

I've seen some form of this equation in several places including these two threads:

The problem I'm having is with \$f\$. From the magnification for a thin lens:

$$ M = \frac{D}{d} = \frac{H}{h} = \frac{f}{f-D} $$

I can't figure out how to arrive at \$D = f\cdot H/h\$.


2 Answers 2


An object 1 meter (1000mm) in height is photographed with a 50mm focal length lens. The object distance is 5meter (5000mm).

H = actual height = 1000

D = actual distance = 5000

We can trace out a triangle with the vertex at the lens. The base of this triangle is the objects height; the height of this triangle is object distance. The ratio of height to distance is 1000 ÷ 5000 = 0.200.

Inside the camera, we can trace a similar triangle. A similar triangle means the angles of both triangles have identical angles and the ratio of corresponding sides will be identical.

The height of this image triangle is the focal length of the lens d = 50

The base of this triangle is unknown. We can calculate; 50 X 0.2 = 10

Thus h = 10mm

Prove formula D = f(H/h)




h = 10

D = 50 X (1000/10)

D =50 X 100

D = 5000 (actual object distance = 5 meters)

  • \$\begingroup\$ Thanks, Alan. I guess maybe I don't understand focal length as well as I thought I did. You appear to be using the distance between lens and film surface as focal length. It was my understanding that the focal length is the distance between lens and focal point. Wouldn't the image projected onto film surface need to be some distance BEYOND the focal point? Thanks, again. \$\endgroup\$
    – Taro
    Commented May 15, 2017 at 16:23
  • 1
    \$\begingroup\$ The focal length is the distance between lens and focused image when the lens is targeting a far away object. As you ocus on nearby objects, the lens to focused image increases. At "unity" (magnification 1), lens to focused image is 2 X the focal length. Technically the measurements you need is distance from rear nodal to focused image and distance to front nodal and the object. I think for your calculations, the differences between object and front of lens and rear of lens to focused image will do just find. OK to use focal length for distance lens to focused image. \$\endgroup\$ Commented May 15, 2017 at 18:21
  • \$\begingroup\$ Thanks again, Alan. This is for an object estimated to be 37-metres from a 55mm lens. So, how would I go about determining the object-to-camera distance where this factor becomes negligible? \$\endgroup\$
    – Taro
    Commented May 15, 2017 at 23:19
  • \$\begingroup\$ You must know focal length of the lens. An object 37 meters distant can be assumed to be at infinity. If we know image height is 18mm than ratio is 18/55=0.3273. Height is 37 X 0.3273 =12.1091meters. Suppose you know the height is 4 meters & distance = 37 meters, object ratio is 4/37=0.1081. Then image size is 55 X 0.1081 = 5.9459mm. \$\endgroup\$ Commented May 16, 2017 at 0:46
  • \$\begingroup\$ Thank you, Alan. Just to be clear, the object is considered at infinity if it is beyond the hyperfocal distance, correct? \$\endgroup\$
    – Taro
    Commented May 16, 2017 at 2:52

What you're trying to arrive at can't be right. Since \$D/d = H/h\$ $$ D = \frac{fH}{h} $$ is equivalent to $$ D = \frac{fD}{d} $$ Dividing both sides by \$D\$: $$ \begin{align} 1 &= \frac{f}{d} \\ d &= f \end{align} $$ Which clearly isn't correct.

  • 1
    \$\begingroup\$ In complement: it's incorrect, except when the object is at infinity (or "far away enough"), in which case d = f (or d ≈ f) holds. \$\endgroup\$ Commented May 15, 2017 at 17:21
  • 3
    \$\begingroup\$ @MatthieuMoy bingo. The OP's "D = f*H/h" only holds for all distances for idealized pinhole cameras (in which case, d = f always). \$\endgroup\$
    – scottbb
    Commented May 15, 2017 at 18:16
  • \$\begingroup\$ One or both of you should add that as an answer and I'll delete mine. \$\endgroup\$
    – MikeW
    Commented May 15, 2017 at 18:32
  • \$\begingroup\$ Thanks, Mike, Matthieu and Scott. All these comments are very helpful. So, what I'm getting is that for a lens, d approaches f as D increases, correct? If that's the case, I'll need to determine the D at which point df. \$\endgroup\$
    – Taro
    Commented May 15, 2017 at 23:19

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