I have read this information off a popular Photography site ; I am confused because I tought that for a ratio of 3:1 you need to set your main light to say: f/8 and your fill only one full stop down, f 5.6 in this case.
This is what I believe is not true:

"Simple way to set your ratios. Set your camera to the desired F/stop. Let's say F/8 in this example. Set your main light with the meter to that same f/8 setting. Divide your desired ratio (say 3:1) into that F/8 stop. (8 divided by 3 = F/2.6) Set your fill to F/2.6 and take a shot."

But f/2.6 is at least 3) stops difference from f/8, so the ratio would actually be 1 to 9 if I do what the above statement says?


2 Answers 2


You're right to think this is wrong, but wrong to think the ratio is 9:1 as a result. :)

The ratio is amount of light difference and it's worth remembering that change of one stop will halve or double the amount light. So, for example:

  • 2:1 - 1 stop difference between fill and main
  • 3:1 - 1.5 stops (approx.)
  • 4:1 - 2 stops
  • 8:1 - 3 stops

Another way to say that is that is:

  • 2:1 - The fill is half the strength of the main
  • 4:1 - The fill is one quarter the strength
  • and so on...

Basically, 2 to the power of the stop difference is the ratio:


So, to wrap it all up, your fill should meter at f/4.8 for a 3:1 ratio as that is one and half stops from f/8.

  • \$\begingroup\$ Where did I get the notion of to get a 1:3 ratio, only 1 stop difference is needed, cause some parts of the subject, IE, a portrait, the nose or forefront get all the light from the main light plus the light from the fill light. But the areas shaded from the main light get only the fill light. Since the fill light is half of the main light, when working with 1 stop difference, the nose gets 3 times as much light as other areas lighted only with the fill light. May be that this point of view is only used with portrait photography? \$\endgroup\$ Nov 29, 2013 at 3:56
  • \$\begingroup\$ Not sure I follow what you're asking... \$\endgroup\$
    – Joanne C
    Nov 29, 2013 at 4:12
  • \$\begingroup\$ Let's say the main light provides 2 "units" of light, with 1 stop difference, the fill light provides half of that, 1"unit of light. some parts of the subject receive the 2 units from the main and the 1 unit from the fill for a total of 3 units. But some other areas of the subject receive only the 1 unit provided by the fill light. That is a 3 to 1 ratio among the brighter and darker parts of the subject \$\endgroup\$ Nov 29, 2013 at 4:17
  • \$\begingroup\$ Still not following your point. Try reading this: sekonic.com/whatisyourspecialty/photographer/articles/… - it's a good article and really, don't try to overthink it. \$\endgroup\$
    – Joanne C
    Nov 29, 2013 at 4:23
  • \$\begingroup\$ It's not additive, by the way. \$\endgroup\$
    – Joanne C
    Nov 29, 2013 at 4:24

Please do not "score" this - This is really a comment BUT the comment format minges the information.

FYI only:

Stops = log_base_2(Light1/Light2 )
= log_base_10(light1/light2) / log_base_10(2)
= log_base_10(l1/L2)/0.3010
= log_base_10(L1/L2) x 3.32

His 3:1 comments on fill versus main can be explained as follows.
Note that this is just clarification of what he says - not input per se by me.

Main = 2
Fill = 1
Illuminated by Main + Fill = 2 + 1 = 3
Illuminated by fill only = 1
So for (Main + fill): fill of 3:1 you want Main:fill of 2:1 or 1 stop.


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