Digital BW, The Print

Bill,

In going from f/4 to f/2.8 (one stop) the area, hence, the energy or power
received at the detector is increased by a factor of 2. Wouldn't the
increase then be 10log2 = 3db?  While, in your example, going from f/4 to
f/2 (two stops) increases the area by a factor of 4, so that the received
power also increases by a factor of 4, hence 10log4 = 6db.

Or did I misinterpret your statement?

Harry
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Harry F. Lockwood