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A photo shot with an 1:1.8 lens contains the following exif data:

ApertureValue: 1695994/1000000  
MaxApertureValue: 16/10

How is this possible? Both values are <1.8.

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1 Answer 1

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You are confused because ApertureValue and MaxApertureValue are APEX values, not f-stops.

An f-number of 1.8 is equivalent to an APEX aperture value of about 1.695994. (The formula is \$\mathrm{A_v} = \log_2{\mathrm{A}^2},\$ where \$\mathrm{A_v}\$ is ApertureValue and \$\mathrm{A}\$ is f-number. See the Wikipedia APEX system page for a full explanation.)

Some Exif readers (like ExifTool) convert these to f-stops for the user, but others do not.

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    \$\begingroup\$ Woah! I can't believe I didn't realize that. That's crazy but indeed seems to be correct. See EXIF standard here. \$\endgroup\$
    – mattdm
    Mar 12, 2015 at 18:52
  • \$\begingroup\$ According to the ExifTool docs, FNumber and ExposureTime correspond to the values we're used to working with, and ApertureValue and ShutterSpeedValue are the APEX values. \$\endgroup\$
    – mattdm
    Mar 12, 2015 at 19:00
  • \$\begingroup\$ FWIW, my Pentax camera doesn't seem to set the APEX tags at all. My Nexus phone does, however. Both ExifTool and Exiv2 convert ApertueValue to f/number for display (in this case, f/2.4), but if I look at the raw values, Fnumber is set to 240/100 and ApertureValue to 252/100 — the corresponding APEX value. Hmmm. I'm not sure if these tools are making it more or less confusing by converting. \$\endgroup\$
    – mattdm
    Mar 12, 2015 at 19:10
  • \$\begingroup\$ How the heck did this system survive long enough to appear in digital cameras? Do we blame the Germans since DIN is theirs? \$\endgroup\$
    – davolfman
    Apr 1 at 17:34

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