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No, the size of the sensor has nothing to do with the light exposure. The amount of light captured is determined by the lens aperture and the photosensitive ele

No, the size of the sensor has nothing to do with the light exposure. The amount of light captured is determined by the lens aperture and the photosensitive element size. The problem with small sensors is that their pixel density is much higher and this increases the noise. Also to have the same resolution a bigger sensor can use bigger pixel and that increases the amount of light captured by the single element. For example a full frame sensor with 12MP has a better signal to noise ratio than a 48MP one. If you use a sensor that is 4 times smaller then if you want to maintain the same resolution the signal to noise ratio is much smaller, if you want to maintain the same low light performance then your resolution will be 4 times smaller. Just think about a full frame sensor, if you crop the image in the center the final result will be exactly the same as an image captured with a smaller sensor with the same pixel density.

TomTomXC 9y ago
I don't think thats true. You need to calculate equivalent f-stop for smaller sized sensors to compare the captured light. The f stop mentioned on the smartphone relates to the physical focal length while the 35mm-equivalent focal length is adjusted by the crop factor. As a result the sensor actually covers a small part of the area that could theoretically be lit with such a short focal length (light falling in from the side at large angles).
- tigershark 9y ago
  There is no equivalent f/stop. There is an equivalent focal length but as I explained it has nothing to do with the amount of light captured by the photovoltaic elements. The equivalent focal length is used only to infer the field of view.
  - TomTomXC 9y ago
    Imagine a 14mm 2.8 on full frame - that's a very wide angle. Of course a 14mm 2.8 on a mobile phone size sensor will not be used a wide angle, as the senor will only cover a tiny amount of the arc behind the lens. While the two lenses theoretically "transport" the same amount of light, the sensor of the smart phone only sees a tiny amount of it in the center of the theoretical image circle. In practical terms of course the smaller-sensor lens will already be constructed to only have a very small image circle and therefore let through nowhere near as much light as the lens constructed for a larger sensor. That's why it makes sense to consider equivalent f-stops for calculations of light amount passing to the sensor, it also has its place in calculation of DOF.

TomTomXC 9y ago

I don't think thats true. You need to calculate equivalent f-stop for smaller sized sensors to compare the captured light. The f stop mentioned on the smartphone relates to the physical focal length while the 35mm-equivalent focal length is adjusted by the crop factor. As a result the sensor actually covers a small part of the area that could theoretically be lit with such a short focal length (light falling in from the side at large angles).

tigershark 9y ago
There is no equivalent f/stop. There is an equivalent focal length but as I explained it has nothing to do with the amount of light captured by the photovoltaic elements. The equivalent focal length is used only to infer the field of view.
- TomTomXC 9y ago
  Imagine a 14mm 2.8 on full frame - that's a very wide angle. Of course a 14mm 2.8 on a mobile phone size sensor will not be used a wide angle, as the senor will only cover a tiny amount of the arc behind the lens. While the two lenses theoretically "transport" the same amount of light, the sensor of the smart phone only sees a tiny amount of it in the center of the theoretical image circle. In practical terms of course the smaller-sensor lens will already be constructed to only have a very small image circle and therefore let through nowhere near as much light as the lens constructed for a larger sensor. That's why it makes sense to consider equivalent f-stops for calculations of light amount passing to the sensor, it also has its place in calculation of DOF.