Assume that I'm taking the photo of a lamp in an outdoor cloudy day. The lamp itself has the luminance corresponding to 17EV. The environment has the illuminance corresponding to 14EV.

What will be the exposure value of the lamp?


1 Answer 1


The math on two fstops is the square root of the sum of the two squares, but I don't know about EV.

If the lights were equal, and if they lighted the same overlapping area, twice the light is one stop additional. But you say the ambient is 3 stops down from the lamp, so it won't have much effect. This ambient adds less than 0.2 stop increase, not over 17.2EV.

  • \$\begingroup\$ Thank you for your answer. From your anwer I figure out my own answer :D The additional amount of light from the environment is 1/8 of the lamp. So the additional stops by the environment light is log2(1+1/8)=0.1178. Therefore the EV of the lamp is 17.1178 :D \$\endgroup\$
    – opmfan
    Commented Aug 16, 2015 at 14:21
  • \$\begingroup\$ I cannot fault that at all, but my own number came out trivially different. 3 stops is 1/8, and (1+8)/8 = 112.5% (also another way to compute percent is 1 - 2^3 = 12.5%). I am Not confident of my math, but I see that as log2(9/8) = 0.17. Either way, between 0.1 and 0.2 EV. \$\endgroup\$
    – WayneF
    Commented Aug 16, 2015 at 17:46
  • \$\begingroup\$ You're right. log2(9/8)=0.17 :p \$\endgroup\$
    – opmfan
    Commented Aug 16, 2015 at 17:54

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