Translucent? No. Semi-transparent or semi-silvered, yes.

Started Aug 24, 2010 | Discussions thread
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Victor Engel
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Re: 30% loss of light = about 2/3 stop!
In reply to hallwal, Aug 27, 2010

No. It works on a logarithmic scale. To determine the number of stops adjustment, take the percentage, expressed as a fraction or decimal, and take the log2 of that number. If you have a calculator that computes log but you don't know what base it is, just divide the result by the log of 2.

log2(0.5) = -1 (1 stop reduced)
log2(0.33) = -1.5995 (over 1 1/2 stops)

... but I guess you meant reduced by 33%, so it's really 67%

log2(0.67) = -0.5778 (reduced just over half a stop)
log2(0.7) = -0.5146

hallwal wrote:
If 50% = 1 stop, then surely 33% = 2/3 stop

Bruce Clarke wrote:
Yes, that's what I would say. 50% loss = 1 stop.

taotoo wrote:
I think dpreview have this wrong - it must be more like 1/2+ of a stop lost.

jonas ar wrote:

Bruce Clarke wrote:

The review doesn't seem to make much of the 30% light lost on its way to the sensor, surely a major penalty?

one third of a stop penalty

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