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Setting White Level with an SLR Exposure Meter (1 Viewer)

Brent Hutto

Supporting Actor
Aug 30, 2001
I had asked the following question in another thread and now I think I've found the answer. I'd love feedback from anyone who understands the situation better than I do (probably most people who read this fall in that category).

I found two definitions of Exposure Value (EV), one based on luminance and film speed and the other on f-stop and shutter speed. Equating them led me to the following conclusion:

ISO=Film Speed (ISO Units)
L=Luminance (foot-lamberts)
S=Shutter Speed (seconds)


Or in other words, divide shutter speed times film speed times 0.32 into the square of the f-stop and you'll get luminance in foot-lamberts. Here's some examples:

F=f/4, S=1/100 and ISO=200 yields a luminance of 25.1 ft-l
F=f/4, S=1/60 and ISO=200 yields a luminance of 14.9 ft-l
F=f/8, S=1/100 and ISO=400 yields a luminance of 50.0 ft-l

Here's my question. How much do I have to allow for light losses in the lens and camera? These calculations are all designed around the light that gets delivered to my SLR's meter and, presumably, the film. I think using this approach will make my TV output more light that the formula would indicate.

For instance, I could set my camera to ISO=200 and f/4 and focus a telephoto lens on an 100 IRE window. If I adjust the "Contrast" setting until the meter says use a shutter speed of 1/100 the formula would say that the TV is producing 25ft-l, right? In reality, maybe 10-20% of the light is lost before it gets to the film/meter so I'd really be driving the TV to something like 28ft-l, wouldn't I? What magnitude should I assume for this effect, assuming no filters on the lens?

I'm inclined to calibrate my 36XBR800 to 30ft-l using the formula and figure that ought to be safe for my tube even if it's really putting out closer to 35ft-l or so.

Scott Kimball

May 8, 2000
The f-stop is the ratio of the focal length to the lens aperture... it really says nothing directly about the amount of light transmission (that would be a t-stop... and unfortunately there are few lenses manufactured with t-stop scales). Using the "stops" on the ratio scale, we know that:

Each increment of the f-stop allows half the light to pass as the preceding f-stop.

Each decrement of the f-stop allows twice the light to pass as the succeeding f-stop.

If we were to assume that f1 is a zero-light loss lens (which I don't think is true, but close enough for argument), you can estimate the light loss for each stop on the scale. Problem is, most lenses start at f1.4, since there is no perfectly transmissive lens. You'd have to do some fancy math to convert actual light to the transmission characteristics of your particular f-scale.

Even so, given that each stop is 50% down or 100% up from the next, that's a significant change in the amount of light. Film emulsions and processing procedures have an exposure lattitude to compensate.

Using a light meter calibrated to f-stops isn't going to give you a specific enough measurement to calibrate a display device.


Jorge M

Stunt Coordinator
Feb 27, 2001
Well said, Scott. I have no idea what you're talking about, and it could be total bull for all I know, but it was very well said! :D

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