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Given two lenses: a full-frame lens and a crop lens, both at focal lengths necessary to capture the same frame on their respective systems and at the same f/stop. The physical size of the aperture opening (entrance pupil) has to be larger on the full-frame lens in order to deliver light across the entire larger sensor. When mounting that full-frame lens on a crop body, does that larger aperture opening have any effect on the light captured by the crop sensor?

For example, between a 36mm f/2.8 full-frame lens and a 24mm f/2.8 aps-c lens both mounted on the same aps-c body shooting the same scene at the same f/stop and ISO, does the full-frame lens allow capturing at a faster shutter speed? Or is all the extra light wasted because it falls outside the area covered by the crop sensor and both lenses will perform exactly the same?

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  • \$\begingroup\$ Your question seems to assume the lens on the full frame camera is a longer focal length needed to achieve the same framing from the same distance as a shorter lens on an APS-C camera. Is this, in fact, the case? Or do you intend for both lenses to be the same focal length? (In which case the second half of your first sentence is incorrect) \$\endgroup\$
    – Michael C
    Nov 3, 2015 at 8:24
  • \$\begingroup\$ Updated my question to reflect those assumptions. \$\endgroup\$
    – ystan-
    Nov 4, 2015 at 7:55
  • \$\begingroup\$ You do realize that a 24mm APS-C lens will give the same field of view as a 36mm lens will give on a FF body? But the 36mm lens will give a field of view on an APS-C body that a 54mm lens will give on the FF body? See photo.stackexchange.com/a/38915/15871 \$\endgroup\$
    – Michael C
    Nov 4, 2015 at 8:55
  • \$\begingroup\$ To shoot the same scene on the APS-C body the camera would need to be 1.5X the distance from the subject with the 36mm lens than it would with the 24mm lens. Of course with anything but a flat subject parallel to the image sensor this would result in a change in perspective to preserve the subject size. \$\endgroup\$
    – Michael C
    Nov 4, 2015 at 8:58

3 Answers 3

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The job of the camera lens is to project an image of the outside world onto the surface of the image sensor. All lenses project a fairly large circular image. Only the center portion has good definition. The boundaries of this image are indistinct and dim. The boundaries are blurred because the image forming rays hit obliquely. What should image as points of light appear as ellipses. The edges are dim because the aperture as seen from a view from the edges appears as an oval, not a circle. This oval has less surface area -- thus passes reduced light. This is called a vignette.

The bottom line is: all camera lenses will project an image that is far bigger than the format size. The boundaries are baffled and masked off. This is necessary as this peripheral light would otherwise reverberate about and some will comingle with the desired image forming rays. If not dampened, flare results. Flare is devastating as it robs contrast.

The full frame (Fx) measures 24mm height by 36mm length. The Dx format is taken from a failed film format called Advanced Photo System (APS). This format is 66% of the size of the Fx. The crop factor is 66% = 0.66 (in decimal form) thus 1/0.66 = 1.5, the crop or magnification factor. In other words the imagining chip of the Dx is smaller -- thus 1.5X more magnification is required when making an enlargement the same size as one made from an Fx.

The so-called crop or magnification factor is the cause of lots of confusion. Another way to explain is with this analogy: You are making a presentation in the college auditorium with projector and screen. The projector screen measures 100 inches by 150 inches. The projected images fit the screen exactly. In the middle of your presentation you are interrupted by a rude professor. The screen you are using was reserved and they confiscate it. Happily in the closet is another screen. Sorry to report it is only 66% the size of the one you were using. Undaunted, you set-up the smaller screen. It measures 66 inches by 100 inches. Your audience only sees the central area of your projected images. Skillfully you change the zoom of the projector. This adjusts the magnification to 66% of the original, and you continue your lecture.

If you did not act to change the magnification, your audience would only see the center of your images. Thus there would be image spillover. This spillover light brightens the darkened theater and likely would degrade your images.

The f/number system takes the chaos away, the larger the lens working diameter, the more light the lens can gather. However, a lens with a longer focal length projects a dimmer image. This is because the larger image is spread out over a large surface. This would indeed cause bedlam. The f/number system is a ratio. The focal ratio (f/#) takes into account both focal length and working aperture. Except for trivial stuff, any lens set to the same f/# as another; both deliver the same image brightness.

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  • \$\begingroup\$ So, I think the question is: when you zoom the projector in to concentrate the image in that 66% area, doesn't it naturally become brighter, because the same amount of light is now in a smaller area? \$\endgroup\$
    – mattdm
    Nov 3, 2015 at 16:39
  • \$\begingroup\$ Because the projector has no aperture adjust. for the camera we adjust aperture. Say it was f/8. The f/# is the focal length divided by working diameter. A 100mm with working aperture of 12 ½ mm = 100 ÷12/5 = f/8. If we substitute a 50mm lens and retain the 12 ½ mm aperture the f/# is 50 ÷ 12.5 = f/4. Now f/4 is two stops (4X) brighter.More likely we re-set the working diameter to 6 ¼ mm. Thus 50 ÷ 6.25 = f/8. Double the focal length we get a decrease in light.The reduction to 25%. Half the focal length and the increase in light energy is + 4X. the f/# system takes this all into account. \$\endgroup\$ Nov 3, 2015 at 17:28
  • \$\begingroup\$ The f/# system is based on the geometry of circles. Multiply the diameter of any circle by 1.4 (square root of 2) and you have calculated a revised circle with twice the surface area (capture are for a lens). Thus the number set: 1 – 1.4 – 2 – 2.8 – 4 – 5.6 – 8 – 11 – 16 -22 each is 1.4 times or divided by its neighbor This number set delivers a 2X incremental change to the capture of light. \$\endgroup\$ Nov 3, 2015 at 17:31
  • \$\begingroup\$ @AlanMarcus: very detailed post, though a bit overwhelming in technicalities. two questions: [1] does your answer agree with Ross Millikan's summary in the previous answer? and [2] which lens/sensor behaviour does your projector analogy accurately describe, if any, since you've mentioned it doesn't have the concept of aperture? \$\endgroup\$
    – ystan-
    Nov 4, 2015 at 7:03
  • \$\begingroup\$ The ability to set a high shutter speed depends on the image brightness.The f/# equalizers, any lens set to the same f/# projects the same brightness (trivial difference exist). Thus the choice, one lens over another set to same f/# is moot. Note: a shorter lens yields a smaller object size, thus blur is less noticeable, we can use a slower shutter. The projector analogy helps you understand “crop/magnification” factor. I think its real value is for old gray-hairs like me. We are familiar with the angles-of-view the full frame delivers. People who never used the full frame are bored. \$\endgroup\$ Nov 4, 2015 at 16:39
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Your last sentence gets it right. The extra light is wasted because it falls outside the sensor area. In the approximation that your lens is a single thin lens (it is not, but it is a useful way to think about it) the rays that pass through the center of the lens are not changed in direction. In full frame, you need rays that pass at a wider angle so you can cover the larger sensor.

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FF sensors behave better in low light situations because of the pixel pitch, and inherent physical size of the photo sites. It does not have to do with the size of the image circle being created by the lens.

If an aps-c sensor and a full frame sensor are the same MP, then the area of the image circle we can capture increases, causing the density of the photosites to decrease. If each individual photo site is physically larger, then more photons can fall on that site, and each one can gather more light - therefore the sensor as a whole is gathering more light.

The aperture is not bigger on a FF - the max aperture is the max aperture, and is calculated by the diameter of the aperture/the focal length. That ratio does not change when going from an aps-c to a ff. The effective focal length is what changes, simply cropping in by 1.5/1.6x (depending on manufacturer) what the full frame equivalent of that focal length would capture.

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  • \$\begingroup\$ Of course it has to do with the size of the image circle! A FF lens is designed to create an image circle large enough to cover the surface area of a FF sensor and likewise for a APS-C lens. The FF sensor cannot possibly gather more light if a APS-C lens is used, creating an image circle not large enough to cover the entire surface area of the FF sensor. \$\endgroup\$
    – ystan-
    Nov 4, 2015 at 6:55
  • \$\begingroup\$ the title of this post is "Light Capture of Full-Frame Lens on Crop Body" - my post simply explains how given the same lens, focal length, aperture, and ISO, FFs and APS-C sensors will yield varying exposures. \$\endgroup\$ Nov 4, 2015 at 14:41

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