What focal length and aperture do I need to photograph an eye from 80–120 cm away on a 1/2" sensor?

Question

I’m using a Thorlabs DCC1545M camera with a 1/2" sensor (6.66 × 5.32 mm, 1280 × 1024). I want to capture a close-up of a human eye from about 80–120 cm away. My goal is for a 3 mm pupil to appear at least 25 pixels across in the image.

Using a simple imaging calculation, I estimated that at 120 cm I would need about a 50–52 mm lens. Then I checked a depth-of-field calculator and found that to keep roughly the whole 80–120 cm range acceptably sharp, I would need something like f/32.

Most lenses I see around 50 mm are described by their maximum aperture and have much wider settings available, which made me wonder if I’m misunderstanding something. If I want a fixed-focus, fixed-aperture setup, am I on the right track looking for a ~50 mm lens stopped down heavily, or is there something important I’m missing about focal length, aperture, or depth of field?

user44949 · Accepted Answer

Depth-Of-Field calculations are daunting. We are talking about the distance before and behind the point focused upon that will have acceptable sharpness. Math models are based on the tolerable size of the circle of confusion. This is a tiny diffused point of light that comprises the optical image. I calculated this span for you using a circle diameter of 1/1000 of the focal length lens chosen. In this case, it’s 52mm. Thus the circle size I used is 0.052mm. Since the accepted circle size is 0.5mm when the image is viewed from standard reading distance, 0.052 is about 10X smaller. This will allow 10X enlargement to be deemed acceptably sharp.

Depth of field table 52mm lens focus point 120mm camera to subject:

f/32 ----- 690mm thru 4557mm

f/22 ---- 795 mm thru 2429mm

f/16 ---- 875mm thru 1897mm

f/11 ---- 955mm thru 1604mm

Let me add, depth-of-field is subjective, based on the acuity of the eyesight of the observer, the contrast of the image being examined, and the brightness of the image being viewed. The circle of confusion diameter is 3.4 minutes of arc or stated differently, a circle whose diameter is 1/1000 of the viewing distance. Using 1/1000 is deemed OK for most situations, as it roughly takes into account the degree of magnification applied to view the image. Leica, for critical purposes uses 1/1500 and Kodak uses 1/1750. Most depth of field tables are based on 1/1000 of the focal length.

Formula for DOF

P = point focused upon

Pd = distant point sharply defined

Pn = near point sharply defined

D = diameter of circle of confusion (usual 1/1000 of F)

f = f-number

F = focal length

Pn = P/ 1 + PDf/F²

Pd = P/ 1 – PDf/F²

The image sensor measures 5.32mm by 6.66mm. The pixel count is 1024 by 1280. Simple division shows that the span of a single pixel 0.0052mm. This is true for both the horizontal and vertical. If the goal is an image size spanning 25 pixels, then 0.0052 X 25 = 0.13mm. This is the minimum permissible diameter of the image of a pupil. Using the pupil size of 3mm, the magnification for this image is 3/0.13 = 23 written as -23. The negative sign indicates that this will be a reduction.

So the question is, what lens (focal length to use to obtain a -23X. Further you state the subject to camera distance is to be between 800mm thru 1200mm. The center of this image to camera span is 1000mm.

We position the camera 1000mm from the eye to be photographed. Using the 3mm pupil diameter, we trace a triangle, base 3mm, height 1000mm. The ratio of height to base is 1000/3 = 333.3

The camera lens projects a similar triangle. By similar, all angles are the same. The base of this triangle is the image of the pupil. Per above calculation, this will be 0.13mm. Now the height of this triangle is approximately the focal length of the lens. Using this ratio, the focal length will be 333.3 X 0.13 = 43.3mm.

If a 50mm lens is mounted you can obtain a similar image size by simply changing the camera to subject distance. Assuming a 50mm is used, the image triangle base is 0.13mm and the height is 50mm. The ratio is 50/0.13 = 385. To calculate lens to subject distance, the object triangle has a base of 3mm the camera to subject distance is 3 X 385 = 1155mm.

If a 45mm lens mounted, then the image triangle base is 45/0.13 = 346. The object triangle is 3 X 346 = 1038mm. This will be approximately the camera to eye distance.

Assuming you mount a 45mm lens: Set to f/5.6 using a circle of confusion of 1/000 of 45 = 0.045mm.

Point focused upon distance = 1038mm

f/5.6 aperture setting

Distant point sharply defined = 1192mm

Near point sharply defined = 919mm

Span of depth of field = 273mm

What focal length and aperture do I need to photograph an eye from 80–120 cm away on a 1/2" sensor?

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