When using a full frame lens on a crop body, does the lens require more light?

If I use an f/2.8 full frame lens on a crop body, does it become an f/4.2 (*1.5)?

  • @PhilipKendall // How is this a possible duplicate? I don't see mention of light gathering on your link. – Moon Jan 11 '16 at 9:54
  • Hint: crop factor affects field of view (apparent focal length). "f" in f/X is physical focal length. – user Jan 11 '16 at 10:03
  • @MichaelKjörling // That is not related to this question. – Moon Jan 11 '16 at 10:25
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    Actually, I believe it's very related. If you know that it isn't, why don't you tell us why not? – user Jan 11 '16 at 10:28
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The amount of light passed through the lens stays the same, the lens will still be a F/2.8 lens.

Since the smaller sensor only crops out a different area from the illuminated circle, the exposure related properties of the image taking process will stay the same, regardless of the crop-factor.


The same light will pass through the lens regardless of the type of camera to which it is attached. Less of that total amount of light will land on the smaller sensor. But exposure, when discussed in terms of varying sensor/film sizes, is not about the total amount of light falling on the sensor. It is about field density, or the amount of light falling on the sensor per unit of area. Since the smaller sensor has less total area, it takes less light falling on it to give the same exposure value. Thus the lens has the same f-number, regardless of the size of the sensor.

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    The f-number is the leveling factor. I find your explanation confusing as we are mainly concerned with the amount of light energy that plays on any given unit area. We are not that concerned with the amount total that plays on the expanse. If I am wrong – clarify for me! – Alan Marcus Jan 11 '16 at 17:22
  • Does your browser allow you to see the formatting in the answer that renders one sensor in bold text? – Michael C Jan 11 '16 at 17:26
  • Understanding that it takes less total light falling on the smaller sensor to maintain the same field density on a smaller area is crucial to the concept. So is understanding that the lens acts the same regardless of the sensor size. It collects the same amount of light and projects it to the same sized image circle. The smaller the sensor, the less of that light falls on the sensor. But the amount of light needed to fall on the sensor is directly proportional to the size of the sensor. – Michael C Jan 11 '16 at 17:37
  • Explain to me how the photographer measures and adjusts to controls the total light falling on the sensor. – Alan Marcus Jan 11 '16 at 18:25
  • This question doesn't ask that. It doesn't ask that at all. It asks if the crop factor needs to be applied to an f-number to determine exposure when using a lens that casts a larger image circle than the size of the sensor on the cameras to which the lens is attached. The answer to that question is stated above. It explains why the f-number does not need to be converted by a crop factor. Although less of the light collected by the lens is projected on the sensor, since exposure is a function of field density, or light per unit area, the exposure doesn't change. – Michael C Jan 11 '16 at 18:33

The result of using the lens on a crop body is the same as that of using it on a full-frame body and then cropping.

To possibly clarify: A picture taken with 1/100s at 50mm/2.8 and ISO 100 on a crop body will have

  • the field of view of a 75mm lens on an FF body,
  • the exposure of an image taken at f/2.8 and ISO 100 on an FF body,
  • the depth of field of an image taken at 75mm/4.2 on an FF body.

So indeed it will be quite like an image taken with 1/100s at 75mm/4.2 and ISO 225 on an FF body.

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    This is the only answer so far that touches on DOF. In that respect, and assuming shooting the same scene with both a crop body and a full frame body with the same perspective (camera position doesn't change), then as Carsten indicated, the DOF will be as if the lens had a smaller aperture by a factor of the crop sensor. – scottbb Jan 12 '16 at 3:28

Getting the exposure spot-on is the key to the kingdom in photography. To accomplish this we regulate the amount of light energy projected by the lens and adjust the length of time this projected image is allowed to play on the surface of the digital sensor or film. Too much and the image will be washed out, too little and the image will be gloomy.

Now the camera lens acts just like a funnel it that it gathers light from the outside world and projects this as a tiny image. We regulate the exposure by adjusting the time the shutter stays open and regulating the working diameter of the lens (aperture).

The timing of the exposure is easy; we generally we set the shutter so it opens only for a fraction of a second. Setting the lens diameter to yield the correct image intensity is not so simple. That’s because two factors intertwine to regulate how much light transverses the lens. 1. The working diameter of the lens. To accomplish this adjustment we mimic the human eye endowing the camera with a regulating iris. The size of this moveable aperture regulates how much light the lens can gather. 2. The focal length of the lens. The longer the focal length, the more the projected image is magnified. A highly magnified image is very dim vs. a tiny image which is bright. Consider a slide projector moved further and further away from a white wall. The projected image gets bigger (more magnification) and the light is spread over more and more wall, it dims.

The two interwoven factors that control the projected image brightness are conquered by using a mathematical ratio. The ratio concept tames the chaos. Allow me to explain: We measure the focal length and the working iris diameter and divide. As an example, a lens 100mm in focal length with a working diameter of 25mm, performs at 100 ÷ 25 = 4. We label this value as f-4 . The f is an abbreviation for focal ratio. Now a giant astronomical telescope camera might have a focal length of 10,000mm with a giant working diameter of 2,500mm. Since 10000 ÷ 2500 = 4, this setup also operates at f-4. Both pass the same amount of light. It is the focal ratio that we use to set our camera lenses. The f/# system eliminates chaos. Stated differently, any lens set to the same f-number projects the same image brightness.

The f=number setting takes into account both working diameter and focal length. It makes no difference if the camera is a super miniature or a giant. The f-number is a ratio and that is the leveling factor.


A full-frame lens at f/2.8 will deliver light corresponding to an aperture number of 2.8. It will just deliver it to a wastefully large area.

Let's go through the motions for a full-frame 50mm/2.8 lens on an APS-C body with crop factor 1.5: the delivered light will correspond to f/2.8. Now the effective focal length corresponding to the framing of the image is 75mm. If we had a full-frame camera with an actual f=75mm lens, we'd get the same perspective/bokeh at f/4.2 but would need 2.25 times the ISO value to maintain the same shutter speed (and if we have comparable pixel counts, the larger pixel area of a full-frame camera should actually deliver that kind of ISO).

So the 50mm/2.8 full-frame lens on the 1.5× crop camera gives us the framing of a 50mm/2.8 crop-camera lens which corresponds to a 75mm focal length on full-frame, with the light density corresponding to f/2.8 on either camera but the depth of field corresponding to 75mm/4.2 on a full-frame camera.

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