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With a Canon 550D camera and a Tamron 18-270 mm zoom lens, I tried to fix shutter time, aperture (f/8.0 to cover all zoom range), and ISO, so the exposure meter read 0.0 with 18 mm focal length and while focusing on a uniform white wall. Then I changed to 270 mm focal length, but the exposure reading did not change significantly (only to -0.7).

However, I had expected exposure to change, since only (18 mm / 270 mm) ** 2 = 1/225 of the wall is now covered in compare the coverage when focal length is 18 mm, thus only 1/225 of the light gets to the sensor after the change of focal length.

Why does exposure not change when narrowing the field of view with change of focal length?

EquipDev
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1 Answers1

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It is because the aperture value of f/8.0 actually gives an aperture hole of f/8.0 over the entire focal length range, thereby opening up when the focal length gets larger, whereby the change in the aperture hole exactly balances out the change in wall area that is covered.

So at 18 mm focal length the aperture hole diameter is 18 mm / 8.0 = 2.3 mm, and 270 mm focal length the aperture hole diameter is 270 mm / 8.0 = 33.8 mm, so the aperture hole diameter grows a factor 15 in diameter, or factor of 225 in area, when the focal length narrows down the area on the wall with a factor 15 in both x and y direction, or factor 225 in area, whereby the total amount of light is the same over the focal length range.

The -0.7 change in exposure I saw is probably a result of poor lens design, so some light is lost at larger focal length.

EquipDev
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    which is why apertures are usually written as f/8.0! – ths Jan 27 '15 at 13:54
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    Actually, the metering change probably has a lot more to do with the fact that the content of the scene changes significantly when you zoom from 18mm to 270mm. Even if you're using spot metering, your 18mm spot is 15X larger at 270mm, so you're only metering a part of it (less than 0.5% of the original spot). – user35658 Jan 27 '15 at 16:04
  • Aperture should be in mm^2 – Alec Teal Jan 27 '15 at 16:50
  • @AlecTeal In some sense it technically is. However, the f/scale is used more frequently because exposure is a more common concern, and here, the exact property in question is incredibly useful. – mattdm Jan 27 '15 at 18:25
  • @mattdm that's not the point - you've given an area in a linear unit – Alec Teal Jan 27 '15 at 18:52
  • @AlecTeal/@mattdm: Thanks for the comments; I have improved the description to be accurate with respect to units and hopefully apply to common terminology. – EquipDev Jan 27 '15 at 19:41
  • @EquipDev you haven't fixed the units, there are 3 fundamental units, Length, Mass and Time, speed is L/T (length per time), an area is L^2 (length by a length). If I have units C for "cartons of orange juice" and B for bits (as in the bits in orange juice) the unit of B/C is bits per carton. If I multiply that by Cartons I'm back with bits, this is dimensional analysis. You're talking about an area using a length. "How big is that hole?" "13mm" makes no sense. That's what I mean by units. – Alec Teal Jan 27 '15 at 21:45
  • @EquipDev because the amount of light coming through is proportional to the aperture size SQUARED not the aperture size - the result follows. (that is if you double the diameter of the aperture, 4x as much light comes in) – Alec Teal Jan 27 '15 at 21:48
  • @AlecTeal: I have reread my updated description before your comment, but I believe it is accurate with respect to units. Where exactly do you think I get the unit wrong? – EquipDev Jan 28 '15 at 09:53