# How does sensor size affect minimum front element size for a given f-stop and focal length?

Does the sensor size affect the needed size of the front element given a focal length and f-stop? My intuition tells me that a bigger image circle would require a bigger front element to maintain the same ”light density” but I can not figure out what the exact relation is.

• This is another one of those questions easily answered by the thin lens equation but complicated in the light of an optical assembly. Ultimately the FOV will be determined by the focal length and the sensor size (simple trig) but location of the entrance pupil determines where in the view cone each element is located. Perhaps with further consideration a monolithic equation is possible but right now I can only think to describe the dependent relationship. Sensor irradiance and stop size are not directly germane to element size, though. May 30, 2018 at 16:51

How does sensor size affect minimum front element size for a given f-stop and focal length?

It affects the width of the angle from which the entrance pupil should be visible when looking through the front of the lens.

Does the sensor size affect the needed size of the front element given a focal length and f-stop?

It affects the width of the angle from which the entrance pupil should be visible when looking through the front of the lens.

For most practical lens designs this means a larger element when using similar materials with similar refractive indices and similar shapes to construct the front elements of the lenses.

What, you may well ask, is the entrance pupil? It is basically the aperture opening as viewed through the front of the lens. The size of the physical aperture is usually magnified (either negatively or positively - i.e. reduced in size or enlarged in size) by the lens elements between the front of the lens and the physical location of the aperture. The actual physical size of the aperture diaphragm is not used to calculate f-number. The size of the entrance pupil, sometimes called the effective aperture, is what is actually used.

My intuition tells me that a bigger image circle would require a bigger front element to maintain the same ”light density” but I can not figure out what the exact relation is.

The bigger front element is not required to maintain light density in the areas of the scene that are captured by both lens and sensor combinations.¹ Light density is maintained by an entrance pupil that is the same size for both lenses.¹ The bigger front element is required to collect light, at the same magnification, from a wider enough angle of view to provide an image circle larger enough to cover the larger sensor.

When both are the same focal length and f-number, lenses for larger sensors don't collect more light from the same field of view as lenses for smaller sensors. They collect "more" light only in the same proportion, in terms of area, to the respective sizes of the fields of view needed by the two respective lenses.²

Given the same exact scene, the lens for the larger sensor will collect the same amount of light from the center of its field of view that is equal to the total field of view of the lens for the smaller sensor.¹ Not only will the amount of light "collected" from that part of the lens' field of view be the same, but so will the size of the virtual image projected onto the sensor of objects in the scene that are within the field of view of the lens needed for the smaller sensor.¹

But the lens for the larger sensor will also "collect" light from the parts of its required field of view that are not within the required field of view of the lens for the smaller sensor.¹ That's where the "more light it collects" comes from - the wider parts of the scene that will be needed to provide an image circle wide enough to cover the larger sensor.²

Where some people get confused is when we view two images - taken from the same shooting positions with cameras of two different sensor sizes using lenses of the same focal length - at the same display size. The reason the same objects look larger in the image from the smaller sensor is not because the same focal length lens magnifies those objects more. It doesn't. Both lenses project the same objects onto their respective sensors at the same size.¹ It is because we have enlarged the size of the image collected by the smaller sensor more to display it at the same size as the image collected by a larger sensor.

If we enlarge the images from both sensors by the same enlargement factor, the image from the larger sensor will be larger than the image from the smaller sensor. For instance, if we enlarge an image from a full frame (FF) sensor to view it at 18x12 inches and use the same amount of enlargement (12.7X) for an image from a 1.5X APS-C camera, we will be viewing the second image at a size of 12x8 inches. The objects in both images will be the same size when the FF image is viewed at 18x12 and the APS-C image is viewed at 12x8.¹ If we then place the 12x8 image from the APS-C camera on the middle of the 18x12 image from the FF camera they will match up and it will look the same as the image from the FF sensor did before we superimposed the image from the smaller sensor onto its center (if both lenses use the same type of projections and are free from geometric distortion).¹ But when we view the image from the APS-C camera at 18x12 inches, we've used an enlargement ratio of 19.05X for the APS-C image compared to an enlargement ratio of 12.7X to view the FF image at 18x12 inches. The larger size of the same objects in the APS-C image viewed at 18x12 compared to the image from the FF camera viewed at 18x12 doesn't come from any difference in the lenses or the sensors, the larger size of the same objects is the result of our higher enlargement ratio.¹

The key to understanding the difference between the light needed to be "collected" for a larger and smaller sensor with lenses of the same focal length and apertures is to understand that the light density will be the same for both lenses.¹ It is the angle of view that must be larger for the larger sensor. The parts of the scene that are needed for both sensors will be projected by both lenses (of the same focal length) at the same size.¹ But the objects that are on the periphery just outside the smaller sensor's required angle of view will need to be projected onto the larger sensor by the lens for the larger sensor.¹ Thus the front of a lens for a larger sensor must be able to collect light from a wider angle of view than the front of a same focal length lens would need to collect for a smaller sensor.

It's pretty easy to observe. Just compare a fairly wide angle lens for a FF sensor next to a lens of the same focal length for, say, a Micro Four-Thirds sensor that is half as large linearly and one-fourth as large in area. For our example we'll use 17mm f/4.

A 17mm lens with an image circle large enough to cover an µ4/3 sensor must provide a 52° angle of view.
A 17mm lens with an image circle large enough to cover a FF sensor must provide a 104° angle of view.

From any angle within each lens' FoV (for their respective sensor sizes), the entrance pupil is the same size for the same f-number.

A 17mm f/4 FF lens has an entrance pupil 4.25mm wide.
A 17mm µ4/3 lens has an entrance pupil 4.25mm wide.

There's no difference in the size of each entrance pupil. The difference is that the µ4/3 lens only needs a front element wide enough so that the entrance pupil may be seen anywhere from within a cone 26° from the center of the lens' optical axis. This gives the lens a 52° angle of view. The FF lens, on the other hand, must provide a cone with twice the angle. The front element for that lens must be larger enough, or have a higher enough refractive power, that the entrance pupil is visible at an angle 52° from the center of the lens' optical axis. This gives the lens a 104° angle of view.

¹ Given the same scene under the same illumination, the same camera position, the same focal length, and the same f-number.

² Given the same scene under the same illumination that is constant within the entire field of view, the same camera position, the same focal length, and the same f-number.

The camera lens projects a circular image of the outside world on the surface of film or digital sensor. The resulting image will be brightest and sharpest at the center. This image falloff is typically called vignette. The film camera, loaded with negative film has the advantage. The resulting negative exhibits a vignette. Now when we print this negative by projection, using an enlarger, its lens also vignettes. I am describing the negative/positive system. It has a subtle advantage in that the vignette of the camera and the vignette of the enlarger often counter. True when printing a negative but the vignette is amplified when printing a positive (slide) on reversal paper. A digital system displays is amplification unless adjusted by software.

The optical vignette has many ingredients. The chief responsibility is due to the geometry of how the image is projected by camera lens. The edges of the projection are further away from the lens then the center. The law of the inverse square takes its toll, the edges receive less light. The light that plays on film or sensor at the edges arrives at an angle. Looking back at the lens from the boundaries, we see an oval aperture, not a circle. The oval aperture has less surface area, thus less light plays at the boundaries. Think of a flashlight beam, dead on, it’s a circle. Hold the flashlight at an angle and the light plays as an oval and its less bright. This has a name “cosign error”.

The wide-angle lens suffers the most from a lack of uniformity of field. The uniformity of the wide-angle lens can be improved if somehow we could mount the lens further away from film or digital sensor. Increased back focus will temper the angle of the arriving light at the boundaries. Wide-angles are thus constructed with increased back focus. An inverted telephoto design does this trick. The measuring point for focal length is the rear nodal. If a telephoto lens is reversed, we get a wide-angle plus the rear nodal is shifted forward. The result is a longer back focus and greater uniformity of image.

If the wide-angle is not designed to give higher uniformity, a 70° angle of view will result in a loss, the difference between center and margin will be more than 45%. We are also taking about loss of acuity at the boundaries.

To get more uniform coverage and acuity the back focus is lengthened. The lens barrel is constructed so that it gives a clear view, no shadowing of the full diameters of the lenses. Symmetrical lenses of medium aperture deliver larger circles of good definition. To gain a larger circle of good definition, the lens designer will sacrifice corrections for the intermediate zones (slightly off axis) to gain improved marginal definition. In other words, a good lens is a series of compromises.

The simple answer this respondent wants was developed by the Zeiss engineers at Jena before WWII. To get M=1, the focal length of the lens should be equal to the diagonal of the sensor size. So, in film terms, a 35 mm film has an image (sensor) size of 36 mm x 24 mm. The diagonal = sq rt (36^2 + 24^2) = 43 mm, exactly the focal length of the first 35 mm camera lens. The world's reliance on calling the 50 mm lens the 'standard' actually gave M = 1.16.

• @ Brian --- Please site your source -- Zeiss M=1 Jun 1, 2018 at 15:07