I believe a smaller sensor gives a 'cropped' effect so that a scene viewed from a full frame camera will have only a 'cropped' portion photographed by a smaller sensor camera located at the same distance from the subject.

Is that right?

It seems simple. The large sensor at any range subtends an angle at the subject and we can draw a cone. A smaller sensor fitting in that cone can only touch the edges if it is moved in to the subject.

So at any distance you 'see' more with the larger sensor than the smaller.

They why is the situation reversed with the two cameras I have here: A canon powershot A520 with a 5.8 - 23.2mm lens and a Canon EOS 350D with an EF 35-80mm lens ?

The Powershot roughly seeming to have a FOV about 4 x that of the EOS?

Is it something to do with the EF lens or have I just everything wrong, seriously wrong?

  • 3
    \$\begingroup\$ Does this answer your question? What is crop factor and how does it relate to focal length? \$\endgroup\$
    – xiota
    Mar 1, 2020 at 22:55
  • 1
    \$\begingroup\$ You are correct in that it has something to do with the lens, field of view is a combination of the focal length of the lens and sensor size. \$\endgroup\$
    – lijat
    Mar 2, 2020 at 0:22
  • \$\begingroup\$ @scottbb, I figure that this question probably is a duplicate, and I don't really have the expertise to give a proper answer. I'll just remove my comment. \$\endgroup\$ Mar 2, 2020 at 13:51
  • \$\begingroup\$ This is only true if the lenses are identical. \$\endgroup\$ Mar 26, 2020 at 18:21

2 Answers 2


You would be right, if both sensors were exposed through the same lens or at least lenses with the same focal length. But they aren't.

As you write yourself, the A520 has a lens with a focal length of 5.8mm at the wide end, while the 350D has 35mm. Scaling those with the sensor sizes (A520 crop factor 6=34.8mm, 350D crop factor 1.6=56mm) shows that the A520 has a shorter focal length and thus a wider FoV.

That said, it's not a factor of 4 difference. If you compare fully zoomed-in pictures, the A520 should be actually tighter than the 350D.

  • \$\begingroup\$ I think I get it. I got it wrong. Despite what you say. I said 'if it is moved in to the subject' and this is not really right. sure it is moved in to the subject but mainly it is given a shorter focal length. Right? I need to draw a diagram. I'm talking about 'cone' diagrams that perhaps others haven't seen and about 'moving' etc. - it needs a picture so's you know what I am thinking, how I"m seeing it. We don't get pictures in stack exchange do we, but we can post links, I'll do that. \$\endgroup\$
    – user577111
    Mar 4, 2020 at 9:52
  • \$\begingroup\$ if you use the same lens, you have the same cone. the smaller sensor would need to be farther back, of course, to produce the same picture, or alternatively, at the same position, it will only record a smaller part of the image circle. "you 'see' more with the larger sensor than the smaller." is correct, in that regard. \$\endgroup\$
    – ths
    Mar 4, 2020 at 10:00
  • \$\begingroup\$ Here's my picture. It makes sense to me. Almost. Just a niggling uncertainty. imgur.com/a/FxwlSZH \$\endgroup\$
    – user577111
    Mar 4, 2020 at 10:24
  • \$\begingroup\$ if you move the sensor relative to the focal point of the lens, you'll get an unsharp image. (track different rays from the same point of the subject to see this). the movement i was talking about concerns the whole system (i'll admit i didn't think of this explicitly before). \$\endgroup\$
    – ths
    Mar 4, 2020 at 10:40
  • \$\begingroup\$ Yep. Well that's it. 'you see more with the larger sensor'. They generally display a diagram that is simply that. It shows the larger sensor and rays to the lens and then they draw a smaller sensor on the larger and there you are - a demonstration of how the smaller sees less. Making me think the smaller sensors automatically show a cropped image: 'see less'. But such a setup requires the smaller sensor to be using that focal length which is unreal. It uses its own, of course. Showing more fov from that location. \$\endgroup\$
    – user577111
    Mar 4, 2020 at 19:24

You just need to brush up on what is "normal" -- wide-angle - telephoto.

When examining a camera to find out what’s wide-angle, what’s normal, and what is telephoto, we need to know the diagonal measure of the film or digital sensor. The convention is to use the diagonal measure to gauge a “normal” angle of view which is about 45⁰, camera held horizontal. A wide-angle will be a shorter lens, about 70% of this value or shorter. A telephoto is about twice this value or longer.

The PowerShot A520 sports a sensor that measures 4.39mm height by 5.76mm length. The diagonal measure of this rectangle is 7.24mm. Thus for this model camera, a lens setting of about 7 ¼ mm is considered to have a “normal” angle of view. Wide-angle is 5 ¼ mm or shorter, telephoto is 14.5mm or longer = telephoto. The crop factor for this camera is 6. Thus these values stated in 35mm equivalent is wide-angle = 31.5mm or shorter, “normal” = 43.5mm and telephoto is 87mm or longer.

Applying this same math to the Cannon EOS350D: This camera sports an sensor that measures 14.8mm by 22.2 (APS size). The diagonal measure and thus the “normal” focal length is 26.7mm. The crop factor is 1.6 so the 35mm equal of “normal” is 42.7mm (traditionally rounded up to 50mm) . Wide angle is 18.6mm or shorter, equivalent is 30mm. Telephoto is 55mm or longer, in 35mm equivalent it’s 110mm or longer.

  • \$\begingroup\$ I've often seen 2/3 (67% or .67X) and 3/2 (150% or 1.5X) of normal used as the "boundaries" of wide angle and telephoto. Thus for the 135 format it would be 30mm and between 65-70mm. \$\endgroup\$
    – Michael C
    Mar 8, 2020 at 21:17

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