04-07-2012, 07:21 AM
[quote name='Plochmann' timestamp='1333774268' post='17352']
You do not understand for you do not ask. But why ask something if the answer does not help you. The Buddha once said... <img src='http://forum.photozone.de/public/style_emoticons/<#EMO_DIR#>/tongue.gif' class='bbc_emoticon' alt='
' /> But there is nothing to interpret, the troll only said, the f-stop is so fast because the 4/3 sensor is so puny!! It was many a month ago.
And also you've read to swiftly. I never said anything about 4/3 lenses being smaller because they are more focused. You were right, that is nonsense <img src='http://forum.photozone.de/public/style_emoticons/<#EMO_DIR#>/biggrin.gif' class='bbc_emoticon' alt='
' />
[/quote]
"Can 4/3 lenses be faster because the imaging circle is smaller and therefore more focused?"
[quote name='Plochmann' timestamp='1333774268' post='17352']
The rest is fine, but I don't think technically answers my question. Why is it that a smaller imaging circle allows for a lower f-stop number?
[/quote]
It does not "allow for" a lower f-value. The small sensor crops so much, that the view angle is a LOT smaller. a 17mm lens on MFT has about 54 degrees a view angle. On 135 format, the view angle of a 17mm lens is about 93%. That is a LOT wider. To design a lens with such a wide view angle and then such a big aperture will be so difficult and expensive, that that is the reason why you will never see a 17mm f0.95 FF lens. Likewise, you will never see a 8mm f0.5 lens on MFT (which would be the equivalent of thet 17mm f0.95 lens on FF).
[quote name='Plochmann' timestamp='1333774268' post='17352']
The equation is the same mathematically. Focal length divided by diameter of iris opening. This common and overly simplified equation which I've read doesn't even actually relate to modern optics, is what is confusing. I've read that you have to use a trigonometric equation when dealing with complex elements. I've also read that a lens element's Focal length is something fixed by the shape of a the lens element and determined in factory, and the iris diameter would depend on the dimensions of a lens, neither of these say anything about the image circle.
[/quote]
Aperture is not the size of the opening in the aperture mechanism, but how big the hole looks through the optics. That is meant with that it is complex to calculate. Not the size on the aperture mechanism. So, for the 17mm f0.95 the apparent aperture is 17 / 0.95 = ~17.9mm. Which is NOT the hole size in the aperture mechanism, but how big the hole appears for the light through the aperture.
[quote name='Plochmann' timestamp='1333774268' post='17352']
What I'm trying to say is, if I take that 17mm 0.95 lens and somehow put it on a 135 camera and then go and take a light meter and read a gray card in a dark place, and it says expose to 0.95, so I go and open it all the way up and take a picture, would not it even on 135 film expose the gray card to be 40% when properly printed? I think so. It comes down to that maybe section you mention, the exposure.
[/quote]
A light meter does not say expose to 0.95. You have to tell a light meter which ISO sensitivity of film you are using. A light meter will tell totally different exposure settings for ISO 25 than for ISO 800. That is a key issue if you want to understand exposure.
Take 12mp mft and 12mp FF sensors as example. They both will have a similar number of photo sensitive diodes which collect light. If we put each of these sensors under the same lens, say a 50mm f2 lens with big enough image circle to cover the biggest sensor, for the same duration of time... The photo sensitive diodes of the FF sensor will collect 4 times as much light! Why? Because they are covering 4x the surface of the mft ones. This makes the FF sensor much more sensitive than the MFT sensor. If one would rate the mft sensor ISO 100, the FF sensor would be rated ISO 400. Similar to film really, a bigger grain is more sensitive to a finer grain. Not only the pixels themselves of the FF sensor collect more light per exposure in this example, even the whole image gets formed with 4x more light than the MFT one.
And that is the tricky bit no exposure meter will tell you. Sensors have no fixed sensitivity, though. In fact, their sensitivity is never published. The signal received by the sensor can be amplified to whatever the maker wishes. This happens in two stages. One stage to "equalize" sensors, so they all will produce more or less similar exposure times to make things simple for the user, and one the user has control over, the so called ISO setting (which is not about sensitivity anymore with sensors, as you now should realize).
[quote name='Plochmann' timestamp='1333774268' post='17352']
In fact, whether or not you were trying to say I was wrong, which I can not figure from the semantics of what you wrote, I think it is a matter of focus. Focusing in the most simple terms is the converging of light rays to a point. If we were to take that lens, with a 0.95 aperture, but have to spread it out over a full frame the depth of field would be a lot less and the aperture value seems to also be affected somehow(for whatever mathematical reason or equation that we do not possess). How are size of the aperture opening and image circle related, is that the missing link.
[/quote]
You mix things up and get a totally wrong idea of what happens. Lets take a 50mm f2 with big enough image circle to cover the biggest sensor, as example.
The lens just does what the lens does. It captures light from as wide an angle as its image circle allows. How wide that is is only limited by its image circle. When we lay the FF sensor under it, NOTHING what so ever changes in how light converges (how can it, the sensor does not pull light towards it like a magnet). The only thing that happens is that the FF sensor ONLY captures the part of the projected image that gets projected on its surface. Whatever falls outside its surface gets "cropped".
When we now lay an MFT sensor in its place, the MFT sensor ONLY captures the part of the projected image that gets projected on its surface. The rest of the image gets "cropped".
Nothing got spread out more on the bigger sensor, the aperture did not get affected (still 25mm for both sensors). The focal length of the lens did not change either (remained a steady 50mm). Focus did not change either. The image circle of the lens did not change either. No matter how small a sensor we lay under that 50mm f2 lens, with image circle big enough to cover the biggest sensor, nothing changes in focus, in focal length, in aperture, in image circle of the lens.
The only thing that changes is how much of the projected image gets captured by the sensor, how much of it gets cropped.
What DOES change is the angle of view, then. For the MFT sensor, the angle of view with this 50mm lens will be about 20 degrees. The rest of the lens' angle of view remains UNUSED. For the FF sensor, the angle of view used is about 40 degrees. It collects light from a MUCH wider angle.
So, we are back from where we started. The reason you see a lens with a focal length of 17mm with an f-ratio of 0.95 on MFT lays in the very small image circle. A bigger image circle would mean MUCH larger elements having to refract light from a MUCH wider angle, making it a super complex and super heavy and expensive lens.
What you have to remember then is that focal length on its own does not mean anything. What actually matters is the angle of view captured, NOT the focal length number.
The 17mm f0.95 has to be compared to a lens on FF which provides the same angle of view for FF, and with a similar aperture. This then means it should be compared to a 35mm f1.8-f2 lens. As you might be well aware of, there is no shortage of 35mm f2 lenses for 135 format FF cameras.
You do not understand for you do not ask. But why ask something if the answer does not help you. The Buddha once said... <img src='http://forum.photozone.de/public/style_emoticons/<#EMO_DIR#>/tongue.gif' class='bbc_emoticon' alt='
' /> But there is nothing to interpret, the troll only said, the f-stop is so fast because the 4/3 sensor is so puny!! It was many a month ago.And also you've read to swiftly. I never said anything about 4/3 lenses being smaller because they are more focused. You were right, that is nonsense <img src='http://forum.photozone.de/public/style_emoticons/<#EMO_DIR#>/biggrin.gif' class='bbc_emoticon' alt='
' /> [/quote]
"Can 4/3 lenses be faster because the imaging circle is smaller and therefore more focused?"
[quote name='Plochmann' timestamp='1333774268' post='17352']
The rest is fine, but I don't think technically answers my question. Why is it that a smaller imaging circle allows for a lower f-stop number?
[/quote]
It does not "allow for" a lower f-value. The small sensor crops so much, that the view angle is a LOT smaller. a 17mm lens on MFT has about 54 degrees a view angle. On 135 format, the view angle of a 17mm lens is about 93%. That is a LOT wider. To design a lens with such a wide view angle and then such a big aperture will be so difficult and expensive, that that is the reason why you will never see a 17mm f0.95 FF lens. Likewise, you will never see a 8mm f0.5 lens on MFT (which would be the equivalent of thet 17mm f0.95 lens on FF).
[quote name='Plochmann' timestamp='1333774268' post='17352']
The equation is the same mathematically. Focal length divided by diameter of iris opening. This common and overly simplified equation which I've read doesn't even actually relate to modern optics, is what is confusing. I've read that you have to use a trigonometric equation when dealing with complex elements. I've also read that a lens element's Focal length is something fixed by the shape of a the lens element and determined in factory, and the iris diameter would depend on the dimensions of a lens, neither of these say anything about the image circle.
[/quote]
Aperture is not the size of the opening in the aperture mechanism, but how big the hole looks through the optics. That is meant with that it is complex to calculate. Not the size on the aperture mechanism. So, for the 17mm f0.95 the apparent aperture is 17 / 0.95 = ~17.9mm. Which is NOT the hole size in the aperture mechanism, but how big the hole appears for the light through the aperture.
[quote name='Plochmann' timestamp='1333774268' post='17352']
What I'm trying to say is, if I take that 17mm 0.95 lens and somehow put it on a 135 camera and then go and take a light meter and read a gray card in a dark place, and it says expose to 0.95, so I go and open it all the way up and take a picture, would not it even on 135 film expose the gray card to be 40% when properly printed? I think so. It comes down to that maybe section you mention, the exposure.
[/quote]
A light meter does not say expose to 0.95. You have to tell a light meter which ISO sensitivity of film you are using. A light meter will tell totally different exposure settings for ISO 25 than for ISO 800. That is a key issue if you want to understand exposure.
Take 12mp mft and 12mp FF sensors as example. They both will have a similar number of photo sensitive diodes which collect light. If we put each of these sensors under the same lens, say a 50mm f2 lens with big enough image circle to cover the biggest sensor, for the same duration of time... The photo sensitive diodes of the FF sensor will collect 4 times as much light! Why? Because they are covering 4x the surface of the mft ones. This makes the FF sensor much more sensitive than the MFT sensor. If one would rate the mft sensor ISO 100, the FF sensor would be rated ISO 400. Similar to film really, a bigger grain is more sensitive to a finer grain. Not only the pixels themselves of the FF sensor collect more light per exposure in this example, even the whole image gets formed with 4x more light than the MFT one.
And that is the tricky bit no exposure meter will tell you. Sensors have no fixed sensitivity, though. In fact, their sensitivity is never published. The signal received by the sensor can be amplified to whatever the maker wishes. This happens in two stages. One stage to "equalize" sensors, so they all will produce more or less similar exposure times to make things simple for the user, and one the user has control over, the so called ISO setting (which is not about sensitivity anymore with sensors, as you now should realize).
[quote name='Plochmann' timestamp='1333774268' post='17352']
In fact, whether or not you were trying to say I was wrong, which I can not figure from the semantics of what you wrote, I think it is a matter of focus. Focusing in the most simple terms is the converging of light rays to a point. If we were to take that lens, with a 0.95 aperture, but have to spread it out over a full frame the depth of field would be a lot less and the aperture value seems to also be affected somehow(for whatever mathematical reason or equation that we do not possess). How are size of the aperture opening and image circle related, is that the missing link.
[/quote]
You mix things up and get a totally wrong idea of what happens. Lets take a 50mm f2 with big enough image circle to cover the biggest sensor, as example.
The lens just does what the lens does. It captures light from as wide an angle as its image circle allows. How wide that is is only limited by its image circle. When we lay the FF sensor under it, NOTHING what so ever changes in how light converges (how can it, the sensor does not pull light towards it like a magnet). The only thing that happens is that the FF sensor ONLY captures the part of the projected image that gets projected on its surface. Whatever falls outside its surface gets "cropped".
When we now lay an MFT sensor in its place, the MFT sensor ONLY captures the part of the projected image that gets projected on its surface. The rest of the image gets "cropped".
Nothing got spread out more on the bigger sensor, the aperture did not get affected (still 25mm for both sensors). The focal length of the lens did not change either (remained a steady 50mm). Focus did not change either. The image circle of the lens did not change either. No matter how small a sensor we lay under that 50mm f2 lens, with image circle big enough to cover the biggest sensor, nothing changes in focus, in focal length, in aperture, in image circle of the lens.
The only thing that changes is how much of the projected image gets captured by the sensor, how much of it gets cropped.
What DOES change is the angle of view, then. For the MFT sensor, the angle of view with this 50mm lens will be about 20 degrees. The rest of the lens' angle of view remains UNUSED. For the FF sensor, the angle of view used is about 40 degrees. It collects light from a MUCH wider angle.
So, we are back from where we started. The reason you see a lens with a focal length of 17mm with an f-ratio of 0.95 on MFT lays in the very small image circle. A bigger image circle would mean MUCH larger elements having to refract light from a MUCH wider angle, making it a super complex and super heavy and expensive lens.
What you have to remember then is that focal length on its own does not mean anything. What actually matters is the angle of view captured, NOT the focal length number.
The 17mm f0.95 has to be compared to a lens on FF which provides the same angle of view for FF, and with a similar aperture. This then means it should be compared to a 35mm f1.8-f2 lens. As you might be well aware of, there is no shortage of 35mm f2 lenses for 135 format FF cameras.