# Using FF lens on APS-C changes lens aperture value?

Someone told me that hen using Full frame lens on APS-C camera (1.5x crop factor) in addition to multiplying focal length by crop factor (e.g. 35mm * 1.5 = 52mm actual focal length on APS-C) you also need to multiply maximum aperture by crop factor e.g. if lens is F/1.8 you F/1.8 * F/1.5 = F/2.7 actual aperture of lens when used on APS-C?

Does aperture value also change when FF lens is mounted on APS-C?

Neither the focal length of the lens nor the aperture of the lens change if we change the size of the sensor.

What does change is the angle of view and the depth of field, but not the exposure. (We are assuming neither the focal length, f-number, the camera position, nor the focus distance to our subject changes - only the size of our camera's sensor.)

We often associate a specific angle of view with a particular focal length and film size, and when we say using a crop sensor "changes" the focal length what we are really saying is that using a crop sensor changes the angle of view as compared to using the same focal length with a 36x24 mm piece of film or a "full frame" sensor that is essentially the same size as the 135 format (i.e. "35mm" film).

With regard to "aperture", what doesn't change is exposure. This is because the f-number is a dimensionless ratio. It's used to determine how much light per unit area falls on a sensor or film over a specific time period (the exposure time or "shutter speed"). When we say we used an aperture of f/2.8, for instance, what we're really saying is we used an aperture that was 1/2.8 in diameter compared to the lens' focal length. The lens focal length divided by the diameter of the aperture, more properly called the entrance pupil, is the f-number that we often call the "aperture". But strictly speaking, the aperture is a diameter of an opening, not the ratio of that diameter to the focal length. If our lens is 50mm, f/2.8 has an aperture of 17.9mm. If our lens is 80mm, f/2.8 is an aperture 28.6mm in diameter. If our lens is 80mm and has a diameter of 17.9mm, though, that gives us f/4.5. The number 4.5 is essentially (i.e. "close enough for government - or photographic - work") what we get when we multiply 2.8 by our 1.6X conversion factor (and, of course, 1.6X is the same ratio between 50mm and 80mm).

With regard to "aperture", what does change is depth of field (DoF) or depth of focus (DoF). Remember, our focal length didn't really change, only the angle of view actually changed due to using a smaller sensor. A 50mm lens is still a 50mm lens. So an f/2.8 entrance pupil is still 17.9mm wide for that 50mm lens. The difference is that when we take the image collected using a smaller sensor and enlarge it to be viewed at the same display size as an image taken with the same focal length lens using a larger sensor, we must enlarge the image from the smaller sensor more. When we use a 1.6X cropped sensor, the change in depth of field is very close to the change in depth of field that would be the case if we used a 1.6X longer lens with a 1.6X higher f-number on a FF (i.e. 1.6X wider and 1.6X taller) sensor.

For example, if we use a 50mm lens at f/2.8 on our Canon APS-C camera with a crop factor of 1.6X, then:

• We would get a field of view that would be equivalent to using an 80mm lens on a FF camera.
• We'd also get depth of field that looks about the same¹ as if we had used f/4.5 with the 80mm lens on the FF camera.
• But we'd still get exposure equivalent to f/2.8 with an 80mm lens used on the FF camera.

Remember, in a way DoF is really an illusion that is determined by exactly how large blur can be before our eyes can see that blur as "blurry" versus how small blur must be to still look sharp and in focus to our eyes. The more we must enlarge an image to view it at a specific display size, the larger we make all of the blur in the image, and the easier it is for our eyes to be able to tell that the same amount of blur, as measured on the sensor before we enlarged it, is blurry.

¹ Though the total DoF would be almost the same, the distribution of that DoF in front of and behind the point of focus could be different. Assuming standard display conditions, our 80mm lens used on a FF camera at f/4.5 focused at 50 feet would give us a total DoF of 35.5 feet with 12.1 feet in front of the point of focus and 23.4 ft behind the point of focus. Our 50mm lens used on a 1.6X APS-C camera at f/2.8 focused at 50 feet would give us a total DoF of 36.5 feet with 12.3 feet in front of the point of focus and 24.2 feet behind the point of focus.

No.

Using a crop sensor on a full-frame lens means that you waste a large image circle on a smaller sensor (unless you use a speed booster which changes a whole lot). The f number basically relates the brightness of a scene to the brightness on the sensor, and the ISO number gives the relation of that brightness to full exposure. If you use a crop sensor, all of that stays quite identical. However, the effective focal length becomes larger (since the image is not "as wide") while the aperture opening remains the same. Since the f number is the ratio between focal length and aperture opening, the "effective" longer focal length is not accompanied by a different aperture opening, so the "effective" aperture number corresponding to depth of field when your "effective" focal length would be achieved on a full-frame sensor is larger by the crop factor.

However, if you don't look at depth of field but at the relation between scene brightness and ISO value to be used, the aperture number stays the same regardless of how much or little you crop from the image circle.

Which is not overly surprising: after all, you don't expect brightness to change when cropping the finished photograph, so cropping by using a smaller sensor should also not cause a change in brightness.

Multiplying the focal length and aperture by the crop factor is known as "equivalence." Equivalence is based on recording the exact same image composition on the smaller sensor; say a picture of a lightbulb that fills the frame vertically.

To get the lightbulb to fill the frame with the crop sensor you either have to back up (using the same lens) or use a shorter focal length. That results in an increased focus tolerance and output depth of field.

If you move farther away from the light source (recorded light bulb) the density of light reaching you is reduced (as dictated by the inverse square law). The same occurs if you instead use a shorter FL with the same f# (a smaller entrance pupil letting less light through). I.e. the light bulb is recorded smaller on the smaller sensor and contains/received less total light (but the reduced qtty of light is focused into a smaller area, resulting in the same exposure/area). The aperture value of a FF lens does not change; the smaller sensor just doesn't make use of all the light (crops it away).

And the smaller sensor is not using the lens's full image circle, resulting in a cropped field of view.

All of these effects are roughly equivalent to the crop factor... i.e. a 50% crop (area) equates to 1 stop more DoF, 1 stop less light, and 50% longer FoV/magnification. And a 2x crop (M4/3, 25% area) is 2 stops and 2x FL.

But there are many things that keep the concept of equivalence from being entirely correct... f#'s are calculated and rounded, sensor technology differences, focus tolerance (depth of focus at sensor) vs output DoF, focal length vs hyperfocal distance, etc.

And equivalence is also roughly equivalent to what happens if you crop an image in post, and for the same reasons (discarding image area/light and enlarging more for equivalent output).