What size lens will give a 2ft by 2 ft view at a range of 2 ft?

I was considering getting a 35 or 50mm lens or similar and am not sure how to choose the right one. Cannon 60d.

TL;DR: you need a 14.9 mm lens to cover your requirements

The TL;DR is the "fish". Below is the "teaching one to fish" explanation...

In general, other than fisheye lenses, cameras basically follow the pinhole projection model, meaning that the angle of rays entering the lens's aperture (which, for basic purposes such as this question, we can model as a tiny pinhole) equals the angle of rays leaving the lens into the camera. That means that elementary geometry can be used to answer your question.

A picture is worth a thousand words. In the image below, F is the measure of the field of view (FoV), and L is the distance from the camera to the object. In your case, you stated F = 2 ft, and L = 2 ft. But by similar triangles in geometry, the ratio of the black lines, L / F, is the same as the ratio of the red lines, f / S. Well, f is just the focal length of the camera's lens that you're asking about. S is the dimension of the camera's sensor corresponding to the measurement of F. That is, if you care about the height of the FoV, then S is the height of the camera's sensor. Similarly, if you cared about the width of the FoV, then S is the width of the camera's sensor.

Thus, the relation f / S = L / F holds. Solving for f,

f = SL / F

Plugging in your specifics, the sensor on a Canon 60D has a height (smallest dimension) of 14.9 mm; your subject is 2 ft from the camera, and you desire a FoV of 2 ft high. Therefore, your desired focal length f = 14.9 mm ∙ 2 ft / 2 ft = 14.9 mm.

Focal lengths aren't very precise, so you won't find a 14.9 mm lens. You can find 14 mm and 15 mm lenses, and of course zoom lenses that cover that range. If your requirements are very tight, I'd recommend going with a 14 mm lens (and cropping slightly, if you need to limit the FoV to exactly 2 feet).

Note that the angle of view (AoV), 𝜃 (greek 'theta'), labeled in the image does not come into play in the similar-triangles calculation. If you need to convert angle of view to focal length, then trigonometry will be needed, resulting in:

𝜃 = 2∙atan(S / 2 f) = 2∙atan(L / 2 F)

However, your requirements don't specify angle of view; the angle is implied by the requirement of field subtended at a particular subject distance. Therefore, you can ignore the AoV calculations.

Because the Canon 60D has the ability to set an image's aspect ratio to 1:1, you can set the camera for 1:1, and ignore cropping issues, etc. See the 60D Instruction Manual, pages 157–158, for the menu settings to set the aspect ratio to 1:1.

A focal length equal to the short dimension of the sensor or film frame will produce an image that can (nominally) capture a square that is the same height and width as the distance from the camera.

For example, 35mm film (or a "full frame" sensor) is 24mm high and a 24mm lens will fit a 1mx1m square at a distance of 1m. However because 35mm film frame is (typically) 36mm wide, it will capture 1.5m in the other direction at 1m distance.

The only commonly available native square format is 6x6 medium format film. This is nominally 56mm x 56mm. So a 56mm lens would capture a 1m x 1m square at 1m. Such lenses are not common. 55mm lenses have been manufactured but tend to command a price premium in the market.

The Canon EOS 60D has a sensor size of 22.3 x 14.9mm. A 14.9mm lens will capture a 1m x 1m square at 1m. However, such lenses are not widely available. 15mm will be slightly long. 14mm will be slightly wide. These focal lengths tend to command a price premium.

There is a Field of View calculator here.

A 1:1 relationship between focal length and sensor or film frame dimension has historically been exploited in street photography. On 35mm film, 35mm and 24mm lenses make zone focusing easier because the relationship of size of subjects to the frame lines in the viewfinder facilitates estimation of distance...when the subjects are of known size of course which people and urban furniture usually are.

The easiest way is put a zoom lens on your camera and figure out what focal length you need empirically.

Since you don't say what your sensor size or camera model is, you'll need to do the maths yourself.

Angle of view for your situation is about 2 * arctan(1/2) = 53°.

From a Wikipedia article: angle of view = 2 * arctan(d/(2*f)
Where: d = sensor size; f = focal length
d, for a full frame is 24mm.

Solving for f: focal length = d/(2 * tan({angle of view}/2))
For a full frame sensor and using the 24mm dimension, you'll need a focal length of 24mm.
For an APS-C sensor, 16mm.

Some caveats:

• distance is measured from the film (sensor) plane to object. Decent cameras have a marking on the body where the film plane is located.
• The equation for finding focal length is for a simple lens. Camera lenses aren't simple, especially when shooting close up.
• When focusing at closer ranges, the magnification of your lens will change. Unless you calibrate your lens, you'll be slightly off from the equation.
• Edit: Corrected definition of d.
– qrk
Jul 26 '21 at 20:18

The focal length needs to be the sensor height (so 24mm full-frame equivalent) in order to have something of height x in distance x cover the sensor.

• the 60D doesn't have a full frame sensor. So instead of equating to full frame, what is the focal length needed for a Canon 1.6x crop sensor?
– scottbb
Jul 28 '21 at 19:26
• @scottbb The 24mm full-frame equivalent on an 1.6x crop sensor is 24mm/1.6, namely 15mm. I was assuming that relation to be known. Jul 28 '21 at 21:00
• I understand the assumption, but I'm not sure the assumption is known by the OP, or at least the general reader.
– scottbb
Jul 28 '21 at 23:36

The format size of the 60d = 14.9mm by 22.33mm. You are asking for the angle of view needed to exactly image a 2ft. by 2ft. vista camera 2ft. distant. This works out to a diagonal angle of view of 35.26 degrees. Mount a lens with a focal length of 33.2mm and the angle of view realized will be 35.2 degrees.

• The diagonal of a square is the square root of two longer than it’s sides. The diagonal is about 2.828 feet. Jul 27 '21 at 16:07
• @Bob -- If you do the math - a 33.2mm lens, mounted on this camera, 2 feet from subject, just envelopes a square 2.828 feet height and width. This square image will project 14.l9mm square on the image sensor. Jul 27 '21 at 18:35
• The diagonal of 33.2mm focal length on the crop sensor is 44 degrees. Because it easy to make errors, I let someone else do the math. scantips.com/lights/fieldofview.html#top Jul 27 '21 at 22:08
• 33.2mm is approximately the normal lens for the sensor. Jul 27 '21 at 22:09
• @Bob - I worked this problem - vertical and horizontal angle of view not diagonal angle of view. Jul 27 '21 at 22:33

A tip of the hat to Bob McStevens who caught my math error. My apologies to all, especially Dale, the original poster – I can do better – will put on a dunce cap and stand in the corner.

My error: I wanted to find the horizontal angle of view, object 2 ft. by 2 ft. as viewed from 2 ft. I created an imaginary triangle, apex origin the lens. The base is 2 ft. The height is 2 ft. I bisected the base to create a right triangle. The base of this right triangle is 1 ft. the height is 2 ft. Angle of view is arctan of opposite side over adjacent multiplied by 2. I forgot to multiply by 2.

However there is a simpler way to solve: in optics there is an subject triangle and an image triangle. These are similar in that all angles are the same thus the sides are ratios of each other.

The base of the subject triangle is 2 ft. = 609.6mm. The height of the subject triangle is 1 ft. = 304.8mm. The ratio of these two sides is 609.6 ÷ 304.8 = 2.

Inside the camera, the base of the image triangle is the height of the imaging chip = 14.9mm.

If the camera is exactly imaging a 2 ft. by 2 ft. subject and projecting its image as a 14.9mm square image, the height of the image triangle is 14.9 X 2 = 29.8mm. This distance is the back-focus distance.

In other words, mount a lens, focal length 29.8mm and this will do the deed.

• You still have some math errors. See the other answers that state 14.9 mm (or roughly 15 mm), including mine.
– scottbb
Jul 28 '21 at 19:24
• Hey Alan, I think it's much better to edit an incorrect answer, than simply add another. The rationale for a second answer would be if there are two correct answers, and you want to offer both, separately. That's not the case here. Jul 30 '21 at 11:52