That's a valid point.. The inside of the body on my PL-5 is actually a touch lighter than some parts of the body itself. The sensor itself is also quite reflective, think some of the reflected light bounces back on the sensor again?
Your Wikipedia source agrees with me on this. First sentence in the summary at top:
You're absolutely right. When you say the "density of light", allow me to say it otherwise: it's the number of photons that will hit the sensor per square mm.
The flange distance being a lot shorter in m4/3 cameras (compared to reflex cameras) that phenomenon has to be taken into account. That's why Olympus, who has a huge experience in electronic microscopes (there lenses nearly touch the subject), has allowed its camera division to access its technology in order to apply these sophisticated coatings at the back of their lenses in order to minimize that.
No that is not right. The f-number does NOT tell you anything about the density of light or number of photons. That is going to depend as well on the scene luminance and exposure time. The f number only tells you the relationship of the focal length to the aperture.
That's what she says: it's a ratio.
It's more complicated than that. F/2,8 means the focal length of the lens (let's say 50mm) divided by 2,8. It indicates the relative diameter of the opening of the iris. For a given lens, the more that diameter is big, the more the F stop is small. So it's an reverse relation for a given lens.
Please Ken, let's not complicate the discussion. We are talking about the same scene, shot in the same lightning conditions and same ISO setting but taken by two cameras -- m4/3 camera and a Full-frame.
Yes, that's what I said in the first place, but the f in that ratio is for "focal length". All f/6.7 means, is that the aperture is equal to the focal length divided by 6.7.
It's not more complicated than that. All f/6.7 means, is that the aperture is equal to the focal length divided by 6.7.
Exactly what I said in the first place. If you change the focal length, you change the f-number. Unless you fix the f-number instead, which means you are changing the aperture (which also changes the depth of field in the image).
Agree, but you also won't have equivalent images then, other than in angle of view and exposure. Like I said in the first place, if you want to limit it to that, it's easier to simply talk about equivalent angle of view, and leave ISO and f-number out of it.
Bravo. We are all saying the same thing. So sorry if I misunderstood your message.
Say that otherwise. I have no clue about what that means.
Firstly, let me remove all the BS:
I mean that f-number equals f/D. So if you change f, you change the f-number. Unless you choose to change D (the aperture diameter) instead. But as I explained it above in my first post, if you keep D the same and change the f-number, then you have also changed the exposure, and have to increase ISO as well to compensate.
Agree. Though you might maintain an advantage in resolution, if there was any at the sensor level. Since we are comparing different lenses, lens resolution will depend also on the individual lens, but there is no reason in theory why it should be any different at the lens level given equvalent quality glass; that is, the glass does not have to be "twice as sharp" to produce the same resolution. The sensor is where you need more pixel density to match the resolution.
Interesting.
I don't think this is how it works. An MFT f/6.7 will illuminate the sensor with the same amount of light per unit area as an FF f/6.7. The difference is of course that the FF area is four times as large, so an FF sensor should be able to take equally "good" picture with four times less light per unit area. So my idea is that:
That is exactly how it works to achieve equal DOF and Total Light delivered to any two formats.
That sounds like a lot of "ifs and buts", there. Yes, the FF DOF will be shallower (if the MFT F-Number is the same). My point was that if the MFT F-Number equals one-half of the FF F-Number (divided by the ratio of the Crop Factors), then the DOF will be the same (because the diameter of the Entrance Pupil, and thus the Total Light delivered, will be the same).
Of course - assuming the tech is equally old. Where I am the cheapest FF is twice as expensive as the most expensive MFT so not everybody will not be able to upgrade every second year. An OM-D EM-6 or even EM-7 vs Nikon D600 is not necessarily an unfair comparison.
Thanks for your answer!
I meant to use the FF lens on an FF body. Aslo, the body would match the lens in being of lower quality, perhaps due to being older tech. What I want is a point of comaprison between MFT and FF systems. Oly and Panasonic apparently want us to think that a 300mm MFT is basically the same as a 600mm FF. I think there might be some truth to it, but only if both the lens and the body are of a very superior quality
I know. A shallow depth of field is not always desirable, so equivalent DOF is not necessarily relevant.
A 300mm f3.5 would be great, yes
Too bad it's not MFT.draleks wrote:
I meant to use the FF lens on an FF body. Aslo, the body would match the lens in being of lower quality, perhaps due to being older tech. What I want is a point of comaprison between MFT and FF systems. Oly and Panasonic apparently want us to think that a 300mm MFT is basically the same as a 600mm FF. I think there might be some truth to it, but only if both the lens and the body are of a very superior quality
I know. A shallow depth of field is not always desirable, so equivalent DOF is not necessarily relevant.
A 300mm f3.5 would be great, yes
http://www.amazon.com/Olympus-300mm...1&sr=8-1&keywords=Olympus+300mm+f/2.8+ED+Lens
Thank you for a clear response. Of course the FF system will gather four times more light due to a four times larger sensor, so the sensor on the MFT will also be facing a tougher challenge with four times less light per pixel given equal image resolution.
