I want to install a lens built for sensors 1/x" (with 1 <= x <= 3), on a sensor 1/3". I need to have an horizontal field of view of "a cm", when my working distance is "b cm". What is the formula of the focal length (where the lens is initially adapted for sensors 1/x") I need , please ?


https://en.wikipedia.org/wiki/Image_sensor_format#Table_of_sensor_formats_and_sizes offers a chart to determine approximate sensor size from the "1/x inch" descriptions. It may not be precisely exact, since a few sensors might all use the same description. Should be ballpark, close enough.

Crop Factor is (43.266 / that diagonal) The 43.266 mm is the 35 mm film diagonal, used for standard reference.

My site has some crop factor calculators at https://www.scantips.com/lights/cropfactor.html#crop that will do all of this.

For example, a 1/2.3" sensor is about 5.65x crop factor. A smaller 1/3" sensor is about 7.2x crop factor. These crop factors are relative to 35 mm film, as is standard.

Then if you have these two sensors to consider, in that case, the relative Crop Factor of the smaller sensor (larger crop factor) is 7.2/5.65 = 1.27x (relative to this larger sensor instead of 35 mm film). This is also the inverse ratio of the Equivalent focal length (to give the same field of view). This larger sensor needs a 1.27x longer focal length than the smaller one). Or vice versa.

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    Might be worth mentioning that if the lens is for a smaller sensor, it may have too small crop circle, meaning the large sensor will have black corners. The OP specifically excluded this in the question, but someone else considering a different type of conversion could benefit from the information. – juhist Dec 4 '19 at 19:48

For rectilinear lenses, the field of view is a simple similar-triangles geometry problem:

enter image description here

The ratio of the distance to the subject a over the horizontal field of view b at the subject distance, is equal to the ratio of the focal length ƒ to the image sensor width w:

ƒ/w = a/b

Thus, ƒ = (w a) / b

You didn't state any assumption explicitly, but I get the impression that you might have assumed that the denominator of 1/x sensors has a relation to the sensor dimensions. Just to be clear, the denominator in "1/x" sensor size is not dimensional. It is merely nominal, just denoting a sensor type. There is a 16mm:1" rule-of-thumb, but it's not precise or definitive. See, Why is a 1" sensor actually 13.2 × 8.8mm? for more explanation.

Here is a plot of sensor diagonal distances, for several sensors from 1/3"-type to 1" (where the "1/x" is in decimal value):

enter image description here

So if you need a degree of precision, don't rely on a strict equation relating the denominator of 1/x-type sensors to the calculation for needed focal length as presented above.

  • Thanks ! When I put my 1/2" lens on my 1/2" sensor, it's ok, but when I put the same lens on my 1/3" sensor, everything is very blur, even when I focus at maximum, so I need to put an extension tube, but the image seems of lower quality (because of the extension tube ?). But I don't understand why it's blur without extension tube, could you help me please ? – Alfred Dec 7 '19 at 15:45

Math using Excel to find angle of view The image sensor dimensions = D The focal length = F Focal length mm = 50mm Height format mm = 24mm 27.0° vertical angle of view Length format mm = 36mm 39.6° horizontal angle of view diagonal mm = 43.27mm 46.8° diagonal angle of view =((ATAN((F/2/D)))*180/PI()*2 Hope this helps!

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