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I understand that a T/stop number deals with how much light is let in, instead of the sensor size vs. aperture like the f/stop number. However, what I haven't found is whether or not that T/stop number deals with crop factors. As we all know, on an APS-C or MFT sensor, an f/1.8 50mm lens is actually behaving like an f/2.7 75mm or f/3.6 100mm respectively.

However, what I'm wondering is if my T/2.1 10mm cine lens on my MFT camera (GH5) is behaving like an f/4.2 20mm or an f/2.1 10mm. (Or f/2.1 20mm for that matter.)

Does this question make sense? I'm basically wondering if the T/stop number accounts for the crop factor, since it deals with the amount of light let into the lens.

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Crop factor has nothing to do with T-stop. T-stop is strictly about light transmission which affects exposure. If a lens could be 100% transmissive the T-stop and f-number of the lens would be the same number, but they would still be measures of different things. T-stop measures how much light is passed through the lens, f-stop measures the size of the aperture compared to the focal length of the lens. The f-number of a lens is what determines depth of field. With regards to exposure, crop factor has nothing to do with f-stop either.

The only way a 50mm f/2 lens "acts" like a 100mm f/4 lens on a µ4/3 camera is with regard to angle of view and depth of field when the camera position is the same for the FF camera with 50mm f/2 lens and the µ4/3 camera with the same 50mm f/2 lens. With regard to exposure, the 50mm f/2 lens is still an f/2 lens.

When you use a 10mm T2.1 lens on a µ4/3 camera it will still be a 10mm T2.1 lens. You'll get an angle of view similar to a 20mm lens on a 35mm/FF camera. You'll get DoF similar to an f/4.2 lens on a 35mm/FF camera, but you'll still get the exposure of a 10mm T2.1 lens - because that's the lens you are using.

T-stops and f-stops are measures of how much light per unit area is allowed through the lens. When you use a smaller sensor, the same amount of light per unit area is falling on the sensor. But since the sensor is smaller, it is just collecting less total light.

If you have a one acre field and I have a two acre field and it rains uniformly on both our fields we'll both get 1/2 inch of rain, but I'll have received twice as much total water because I have twice as much surface upon which the same amount of rain per square foot fell.

For more, please see:
How to achieve full-frame look/view on a crop-sensor - without changing the lens?
Why do DPReview's charts show the G7X II as having an aperture around f/8?
Can a smaller sensor's "crop factor" be used to calculate the exact increase in depth of field?

  • The other questions linked by this answer do not answer the question here. – user50888 Nov 30 '17 at 16:35
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    @benrudgers They're not meant to answer this question. If they did then this question would probably be a duplicate. They do relate to how crop factor affects the effect of a lens' T-stop and f-number. – Michael C Nov 30 '17 at 18:42
  • Crop factor does not affect T-stop and f-number. – user50888 Nov 30 '17 at 18:50
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    @benrudgers As the answer unequivocally states. So what's your point? – Michael C Nov 30 '17 at 18:51
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    Crop factor does affect the angle of view and depth of field for a particular focal length and f-number. The effect of the increased magnification needed to display a photo from a smaller sensor at the same display size as a photo from a larger sensor do affect what the photo looks like. – Michael C Nov 30 '17 at 18:54
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The f-stop is a simple ratio. It is the focal length divided by the working diameter. We use this value to compare the image brightness of one lens vs. another. The f-stop is often in error because it does not take into account light loss due to reflections from the polished surfaces and the lack of perfect transparency of the glass. The "T" -top is calculated by actual light measurement. Thus the "T" - stop has higher accuracy. Both the f - stop and the T - stop are a ratio void of dimension thus the format size difference is moot.

  • the most useful answer for me, because it actually explains why a T-stop number can be different from f-stop number – szulat Nov 30 '17 at 10:02
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I'm going to highlight and expand on a bit from my answer to Can a smaller sensor's "crop factor" be used to calculate the exact increase in depth of field?

The key thing is:

In all cases the exposure for a given aperture (in this case, the same t-stop) on any area of a sensor is the same. (I mean, assuming even brightness in the scene.) Consider cutting a photo in half rather than cropping out the middle section: each half, left and right, is the same brightness as the other. But, of course, each really represents half the amount of overall captured light. And this is the same for any portion of the image you want to consider.

Using a smaller sensor is just like that ­— a given aperture gives the same exposure no matter whether you're capturing a full-frame image, or capturing that full-frame image and then cropping later, or "cropping" at capture time with a smaller sensor.

But, of course, a cropped image does have less light. The secret is that we "cheat" when enlarging. We keep the brightness the same, even though the actual number photons recorded per area is "stretched". That is, if on the sensor, 200 million photons collected in a square represents a medium gray, if we print so that square is 10"×10", we don't spread the brightness out making it much dimmer — we instead keep the brightness so it's the same gray.

Make sense?

So, how does this relate to t-stop? Exactly the same as the theoretical value for f-stop. Some amount of light isn't transmitted, but that's the same all across the frame. The sensor doesn't "know" if that's because your lens is near-perfect transmission or if you have a 10-stop neutral density filter. And, cutting the frame in half still means that each half has half as much light as the full frame. So, as in the other answer, you can roughly consider a 1.5× crop factor to be equivalent to 1.5× lower ISO having the equivalent signal-to-noise ratio for the same exposure in a same-sized print.

  • Thanks so much Matt! I appreciate the time you put into this response. – Tyler VanDenBerg Dec 1 '17 at 8:08

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