When does the distance formula D = fH/h apply in photography?
Asked 5/15/2017
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2 answers
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I’m trying to understand the common approximation
D = fH/h
where D is the distance from the lens to the subject, f is focal length, H is the subject’s real height, and h is the height of the subject’s image on the sensor/film.
Starting from similar triangles or lens magnification, I’m confused because magnification gives D/d = H/h, which seems to imply D = fH/h only if the image distance d equals f. Is this formula exact, or only an approximation? Under what conditions is it valid?
Originally by Photography Stack Exchange contributor. Source · Licensed CC BY-SA 4.0
Photography Stack Exchange contributor
9y ago
2 Answers
1
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)
Solve:
f=50
H=1000
h = 10
D = 50 X (1000/10)
D =50 X 100
D = 5000 (actual object distance = 5 meters)
Originally by user44949. Source · Licensed CC BY-SA 4.0
user44949
9y ago
0
Generated from our catalog & community — verify before relying on it.
The key point is that D = fH/h is not exact in general. It comes from similar triangles only when you treat the image distance d as approximately equal to the focal length f.
From magnification:
M = h/H = d/D
So the exact relationship is:
D = dH/h
not D = fH/h.
To replace d with f, you must assume the subject is far enough away that the lens is focused near infinity. In that case the sensor/film sits very close to one focal length behind the lens, so d ≈ f, giving the useful approximation:
D ≈ fH/h
That’s why it often works for distant subjects and basic field-of-view calculations.
But for closer subjects, macro work, or any case with significant lens extension, d is not equal to f, so the formula becomes inaccurate. One of the community answers correctly points out that if you assume D = fH/h exactly while also using D/d = H/h, you are forced into d = f, which shows the limitation.
So:
- exact: D = dH/h
- approximate for distant subjects: D ≈ fH/h
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