Instead of looking up a chart, how can I calculate Lux from EV?

  • \$\begingroup\$ Whilst the chart mentioned above is no longer on Sekonic's website, it is still available by requesting it from [email protected]. \$\endgroup\$
    – mooie
    Commented Jan 31, 2023 at 21:50

1 Answer 1


The short answer is that you can't, really, because they don't measure the same thing Want details? Read on!

From the chart you link to, that looks really easy: 0 EV is 2.5 Lux, and each +1 EV doubles Lux. So, 2.5 × 2^EV.

The wikipedia article on EV and on the article on light meters gives some helpful background information. The chart and that equation are only valid for incident-light meters, and the value of 2.5 is correct for flat sensors (it's a common choice out of a valid ranges of about 2.4 to 4, as recommended by the ISO standard). This constant is called "C" in the standard, and obtained through empirical testing. For a hemispherical sensor, values of 3.2-3.4 are apparently used. (As below, the actual standard uses 100× that value, but that's just a matter of decimal places.)

When measuring reflected light, as with a camera's built-in light meter, a different equation is required, and a different constant. This constant, "K", is expressed as numbers in the range 10.6 to 13.4 in the standard, but it hurts my brain less to think of them as fractions 0.106 to 0.134 when working out the EV-at-ISO-100 math. Canon, Nikon, and Sekonic use the value of 12.5, which is conveniently ⅛, which, since we're working in powers of two, is -3. So, that formula is L = 2^(EV-3) — which you can get from the Wikipedia article, but it took me a bit to figure out where the -3 came from.

Kenko and Pentax apparently use 14 instead of 12.5, which is a difference of almost exactly ⅙ of a stop, and giving an approximate equation of L= 2^(EV-2.84).

There's a very important thing to note, though — the units aren't the same, because something different is being measured. Lux measures illuminance, the perceived brightness of the light on a surface (like, your incident-light sensor). But since we're working with reflected light, L is luminance, which is the light leaving the measured surface in a given direction (i.e., towards your meter). This is in candela per square metre (cd/m2), or "nits". For more on the difference, see What is the difference between luminance and illuminance?

So, the point is: if you're using the camera's built-in meter, you're measuring something different from what Lux measures, so even though you can easily figure out an EV number from aperture and shutter speed, that number won't necessarily translate into Lux in a meaningful way.

  • \$\begingroup\$ Since the chart is provided by Sekonic it is a fair assumption that it is right. \$\endgroup\$
    – labnut
    Commented Mar 24, 2011 at 8:35
  • 1
    \$\begingroup\$ @labnut Since the chart is supplied by appeal-to-authority there is a good chance that it is useful and accurate enough in many situations. cf Box's "All models are wrong. Some models are useful". \$\endgroup\$ Commented May 21, 2016 at 1:00
  • \$\begingroup\$ What if you're taking a picture of the sky? Would the camera-metered light be considered incident? (Technically sky light is reflected/scattered, but let's pretend it's a giant area light.) \$\endgroup\$ Commented Jun 16, 2022 at 16:19

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