However, I talked to one of my best friends today...He gave me his professional opinion that with current technology (coatings, etc.) the minimum effect on image quality per air to glass interface is 3%.
For example, after 10 filters there would be 73.7% transmission with an optically perfect set of filters (.97)^10 = 73.7%.
Depending on the angle of the light ray passing through the filter, the abberations / MTF deficiency will vary, but once again, 3% is basically a minimum level of effect.
Sorry, going to disagree again. First off, I'm not sure you're distinguishing between transmission loss and IQ decrement. Transmission loss with modern lens coatings is much less than 1% per interface.
Now, take your friend's value of 3% IQ loss per interface leading to a 6% loss of IQ from a filter. Roger Cicala's (lensrentals.com) tests of large filter stacks refutes that. Moreover, optically there's no difference between a curved glass-air interface and a flat one - the magnitude of refraction and reflection is the same, only the vector direction differs. So, let's look at the 70-200mm f/2.8L IS II - arguably the best zoom lens in existence. It has 19 groups, meaning 38 glass-air interfaces. At 3% loss per interface, that means absolutely no light hitting the sensor if you mean transmission, or over 100% IQ loss if you don't mean transmission. I don't think either is even close to true.
You are confusing transmittance vs. absorbance.
Let's say I take a one-inch pane of glass. I cut it in half so that it is 1/2 inch thick. I have doubled something as a result. Perhaps the amount of light that gets through??? Then I cut it in half again, so now I have doubled something again. 2*2 = 4. Does that mean that four times as much light gets through? So I could clearly get a large amount of light through if I kept on doubling the transmission, doubling the transmission, and on and on., and I guess it would be infinitely bright light when I ran out of glass to cut in half... Wrong.
But that's the mistake you are making by saying that 100% IQ loss would occur based on the 23 elements / 19 groups in the 70-200mm II lens.
You are adding the 5% 23 times and getting more than 100%, right?
The correct way to compute it is not by subtraction but my multiplication. As I did above, if there are 10 surfaces, then there is about 73% transmittance.
That is only for flat air to glass transitions.
With binoculars they have nitrogen filled optical designs, and there are similar tricks for lenses.
And another thing is that optical groups (lenses directly combined with other lenses) count as a single element.
The answer to your question is to look at the Tstop value for the 70-200mm lens.
This Tstop value is 3.4.
Calculate the number of stops lost: log_2((3.4)^2/(2.8 )^2) = .5602 stops
This translates into a 47% loss of light vs. a hollow tube of the same diameter as the lens.
We can find the amount of light lost at each of the 19 groups as follows:
1 - .47 = (1 - x) ^ 19
.53 = (1-x)^19
.53^(1/19) = 1-x
x = 1 - .53^(1/19)
x = 3.286%.
That's almost exactly what my expert friend quoted.
You can't beat someone who works for NASA, and designed lenses for the space program at age 23 back in the 1970s.
Have a great day!