I was looking at an old question about bigger pixels and whether they make for a better picture here

Why and how do "bigger pixels" make a better picture?

and one user, Philip Kendall, says that he calculated the new sensor size from two parameters: the sensor-area-increase percentage and the fact that a pixel is 1.5 microns.

How does the pixel size here figure? How do you use it to get the dimensions of the new sensor?

Wouldn’t you just calculate the new sensor area using the increase percentage and then use the fact that its aspect ratio is 4 : 3?

I don’t get it how he got the sensors dimensions first and how he used the pixel size to determine them.


2 Answers 2


They are two independent sums which give (roughly) the same number:

  • 15% bigger sensor: 15.5 mm2 * 1.15 = 17.8 mm2.
  • 1.5 micron pixels: knowing that the iPhone 5S has an 8 MP camera and a 4:3 aspect ratio means you know that it's a 3.3 x 103 by 2.4 x 103 array. Multiplying those numbers by 1.5 microns gives you 18 mm2.

None of that is rocket science, but remember that comment was posted on the iPhone 5S launch day when you couldn't just go and look this stuff up on the web. The fact that both numbers are in the same ballpark means that it was likely that the calculation was approximately right.


Calculating sensor size from pixel size is not precise, because there are empty gaps between pixels, unaccounted for.

If you know an accurate crop factor for the lens (from specifications), it is the ratio of the sensor size to 35mm film size, and will give good results. Diagonals must be computed, but if the crop factor is say 7x, then the sensor is 1/7 the dimensions of 35 mm film, which we know. Option 3 at http://www.scantips.com/lights/fieldofview.html

If you can measure the field of view (out at several feet distance), you can compute the sensor size. Option 8 at http://www.scantips.com/lights/fieldofview.html

  • Most "pixel size" specifications are actually "pixel pitch" specification or areas derived from using pixel pitch as the linear dimensions.
    – Michael C
    Dec 21, 2019 at 0:17

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