GH2 Worth buying a Tamron 17-50 2.8?

nothing said:
You misunderstand the basics. The lens to sensor distance does not change, as you had it do in your example. Read up on "flange distance".

The image circle that a lens makes does not change, different cameras just utillize different size chunks of the middle of that image circle.
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I understand it perfectly. Different cameras utilize different chunks of the light circle and therefore capture different amount of light and photons. Hence different brightness, unless adjusted.

The only thing that changes in the image above is the diameter and area of the light circle that the lens outputs. This made the brightness different. If you took a lens for full frame camera, that would be the brightness in the lower image for m4/3 image. Whereas if you took a lens made spesifically for m4/3, that would be the brightness in the top image. Same f-stop.
 
Aki_Hartikainen said:
The only thing that changes in the image above is the diameter and area of the light circle that the lens outputs.
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How did you change the diameter of the circle?

Let me restate it: for a given lens, the distance to the sensor does not change, no matter what camera you mount it on (given that it is mounted correctly). An image from any given lens is the same brightness on a micro 4/3 sensor, as it is on a Super 35 sensor, as it is on a FF sensor. Same brightness.
 
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Same relative brightness. If the smaller film or sensor can capture 10 photons and not 40, the output does not magically turn from 10 photons to 40 photons. You amplify to get the same brightness. Can we agree on this? Because if we can not agree on this, we have a lot to explain to Mr. Einstein.

To crop is to magnify. When you crop and magnify, you will lose light no matter what. But then you compensate. In Photoshop this happens fully automatic, you crop an image for ten pixels and ten photons and enlarge it to full screen, you now use more display power per pixel for same brightness. With film projectors, not so automatic. Generally more candle power is needed for smaller film size or lens must be changed.

True, I made the image above by changing the register distance, but it does not change the principle. It only illustrates what happens when you crop and magnify. Had the lens had a different lens element for the smaller light circle, more brightness would have come from the same distance with the same aperture. The m4/3 lenses have a lot of potential for bright images, but this has not yet fully materialized. It may be difficult to produce very sharp lenses for smaller but brighter image circle.
 
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If you change the magnification of a lens by expanding or contracting the image circle, you effectively change the focal length of the lens. Without a corresponding change to the aperture, the f-stop (focal length divided by aperture) of the lens is altered. Thus you perceive a loss or concentration of light when moving the lens further or closer to the imager plane.

Lenses cannot be mounted at any arbitrary distance from the imager, however, they must be mounted at a specific distance in order to focus correctly. Thus an f2.8 lens is an f2.8 lens on any camera it is correctly mounted on, regardless of imager size.
 
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nothing said:
If you change the magnification of a lens by expanding or contracting the image circle, you effectively change the focal length of the lens. Without a corresponding change to the aperture, the f-stop (focal length divided by aperture) of the lens is altered. Thus you perceive a loss or concentration of light when moving the lens further or closer to the imager plane.

Lenses cannot be mounted at any arbitrary distance from the imager, however, they must be mounted at a specific distance in order to focus correctly. Thus an f2.8 lens is an f2.8 lens on any camera it is correctly mounted on, regardless of imager size.
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fascinating exposé. allow me to help out in your excellent effort?
- "If you change the magnification of a lens by expanding or contracting the image circle"
No such thing is possible. In the paradigm of > Image < Lens < Object , the focal length is defined here...
It's a number in mm that is calculated as the distance from the 'normal' of the lens, where the light rays, converge and then cross over to the image plane. The image circle is the image that is thus produced. The effective 'usable image' is the area that is cropped by a mask that sits behind the camera's lens mount. The Image circle itself cannot be expanded or contracted, just the cropping is. You do effectively change the focal length of a lens by say...adding extention tubes to the rear mount as this increases the distance between the normal of the lens and the image plane.
- "Without a corresponding change to the aperture, the f-stop (focal length divided by aperture) of the lens is altered."
This line contradicts itself. Aperture is a number that is calculated by dividing the focal length of the lens by the diameter of the iris for that given optic.
- "Lenses cannot be mounted at any arbitrary distance from the imager"
sure they can, with extension tubes. This does change the cropping AND magnification of the Image with a given optic.
- "however, they must be mounted at a specific distance in order to focus correctly."
I think we've dealt with that too.
- "Thus an f2.8 lens is an f2.8 lens on any camera it is correctly mounted on, regardless of imager size"
correctly mounted is sort of meaningless and image circle has also been dealt with.
What are you trying to say?
 
Zephyrnoid said:
fascinating exposé. allow me to help out in your excellent effort?
- "If you change the magnification of a lens by expanding or contracting the image circle"
No such thing is possible.
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Ever heard of a zoom lens?

Zephyrnoid said:
You do effectively change the focal length of a lens by say...adding extention tubes to the rear mount as this increases the distance between the normal of the lens and the image plane.
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I thought you just said that this wasn't possible? And have you ever tried to focus to infinity when mounted on an extension tube? That is what I mean by not corrected mounted - not able to be used normally.

Zephyrnoid said:
- "Without a corresponding change to the aperture, the f-stop (focal length divided by aperture) of the lens is altered."
This line contradicts itself. Aperture is a number that is calculated by dividing the focal length of the lens by the diameter of the iris for that given optic.
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No, aperture merely means a hole of a certain size. The f-stop number is what is calculated by dividing focal length and iris diameter (ie the APERTURE).

Zephyrnoid said:
- "Lenses cannot be mounted at any arbitrary distance from the imager"
sure they can, with extension tubes. This does change the cropping AND magnification of the Image with a given optic.
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Again, a lens mounted as such cannot be used normally. What is the point of bringing it up? Besides, extension tubes alter the focal length of a lens, and consequently the f-stop number.

Zephyrnoid said:
- "however, they must be mounted at a specific distance in order to focus correctly."
I think we've dealt with that too.
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Yes, now I have dealt with it here - they cannot be used normally.

Zephyrnoid said:
- "Thus an f2.8 lens is an f2.8 lens on any camera it is correctly mounted on, regardless of imager size"
correctly mounted is sort of meaningless
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Correctly mounted means mounted at the proper flange distance and hence able to be used normally - focusing to infinity and close focus as designed.

Zephyrnoid said:
and image circle has also been dealt with.
What are you trying to say?
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I don't understand your point at all. Are you trying to agree with Aki? Although we've already gone way off-topic as is, your post doesn't seem to have much relevance at all and it just comes across as disagreeing with everything I say just to disagree.
 
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The lens is f2.8 on any camera it is installed on, but any camera is not the same. That's why it is called a system camera. It is a system that consists of the lens and the camera. The area of the imaging device plays a role in the brightness of the image. Even if relative brightness is the same on the surface of the imaging device, amount of total brightness will not be. 10 captured photons are not as bright as 40 captured photons. That would be against laws of physics. We need to amplify and adjust, which can be well done as was stated many times already.

You are being ignorant of the camera part of the system and insist on focusing on the lens only, which is not enough to determine the brightness of the final image.
 
Aki_Hartikainen said:
You are being ignorant of the camera part of the system and insist on focusing on the lens only, which is not enough to determine the brightness of the final image.
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So explain to us what you think the camera is doing please.

I can tell you that a 16mm camera loaded with any particular film and a 35mm camera loaded with the exact same film, with the exact same lens set to the same f-stop, shooting the exact same test chart (or even a plain wall) under the exact same lighting, will both produce the exact same image density on the negative. That is a fact. How do you explain that?
 
Surely any film size will expose and develop to the same density behind a certain lens.

But when we had to produce similar sized images for viewing, and only had certain amount of wattage or developer activity to spare, final brightness for images will be different from different film sizes. And this situation is the same when comparing full frame and cropped frame digital cameras. Now the development will happen electronically and digitally inside the camera, but the principle remains the same. The smaller image has to be magnified more, thus needing either more wattage or more developing for the same brightness.
 
They do not contradict at all. You crop the sensor, you will take a smaller image. But we are going to view the images at the same size. The smaller image has to be magnified more, and that's what happens inside the camera. You crop the image, you will have to magnify more and spare more wattage. More magnification means more wattage or development required for similar viewing brightness. We will view the images from a certain sized display, and not the negatives.

You crop the sensor or film, you reduce light gathering power for the system. Brightness of the final image will suffer, unless adjusted and amplified.
 
If you put a smaller window in your home, is the brightness inside going to remain the same regardless of what's in front of the window? The window is the camera sensor or film size and the indoor is your final image that you view. Sensor or film is camera's window to the world. :smile:

I am sure we are beginning to get on the same page now.
 
If you put a 50 mm f2 lens in front of the window in your home, and all of the light comes through the lens and the window, you will get certain image brightness in the house. Right? So you reduce the size of the window to one quarter and the lens remains the same. What kind of brightness can you expect inside your home?

That would be the equivalent of 50 mm f4. But now from the smaller window you also have reduced field of view. What does this mean? Your 1/4 window made your image the equivalent of 100 mm f4. It was the window, not the lens that caused this.

At least your depth of field remains the same regardless of the size of the window.
 
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Nnot a relevant example - you aren't lighting a whole interior of a house with a lens, you are just illuminating a certain area of the sensor - which will still be as bright whether you crop the rest of the illumination away or not. This is assuming you are talking about direct light through the window - if you are talking about indirect light, then the window size is more analogous to the aperture than the sensor size.
 
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We are lighting the whole house through a window, that has a lens in front of it. If you resize the window, all of the aforementioned will start to have an impact on the indoor brightness and field of view. Now we are considering the whole system, and not just one part of it.
 
How about this for an example.

You have 3 things when you are taking a picture: the light source, the lens (in this case we specifically talk about the aperture), and finally the sensor.

As an analogy, we have a house, with a sunny scene outside through the window, and people inside looking outside. The scene is our light source. The window is our lens, and in this case we have a fixed window size, relating to the fixed aperture when talking about the tamron on two different cameras. Finally we describe our different sized sensors by changing the number of people who are looking outside.

Let's say every eye is a pixel. Well the cannon body with its APS-C sensor will have more photosensitive pixels than the gh2 and it's micro-four thirds sensor. Even for the sake of this argument, we can say that two cameras with the same sensor size but different pixel counts can be represented by this analogy.

I think that we can all agree that as long as the scene and the window stay the same, the light coming into the room stays the same aswell. Now it is all about the perception of the light by the people in the room. We can conclude that no matter how many eyes are looking outside, they all see the same image at the same brightness, and that the number of people looking outside does not effect the brightness of the scene. That is to say that if 1 person looked outside vs 100 people looked outside, the 100 people would NOT have seen the scene 100x less brightly. If this was the case, I would argue we all should have a very hard time seeing anything at concerts, plays, or any group gatherings.

Since light essentially travels in rays, any light that goes to one person's eye gets trapped and lost to our eye, but no other eye will use these particular rays to form their image. Similarly, having a camera with a bigger or smaller sensor, or a higher or lower pixel count will NOT affect the brightness of the captured image. Assuming the pixels are just as light sensitive from one sensor to the other, there will be NO difference in the brightness.

Finally we must address this very subtle assumption. Not all pixels are created equal. For example we know for fact that the GH2 is in general darker than the GH1, but this is not because the light hitting the sensor changed at all in brightness, it is because the actual pixels themselves are not as sensitive, or are calibrated differently. What does this mean about the argument? Well, the tamron at 2.8 on a cannon at iso 100 and shutter speed 60 COULD be brighter than the GH2 at all the same settings, but this would not be a result of the size of the sensor, but rather the sensitivity of the sensor. So it would be just as fair to think that it is possible the GH2 will actually produce a brighter image.

And to the person who argued about the light having to travel a longer distance, yes this will in fact lower the brightness of the light, but when we are talking about an inch at most of space between the back element and the sensor, the difference in light intensity would be incredibly negligible. If you had one camera mount the lens a quarter mile away from the sensor, and one camera mount the lens an inch from the sensor, you would certainly tell a difference then, but from a cannon or nikon distance away to a microfour thirds, the difference is more than negligible.

If this was at all unclear, ask away. I will be happy to elaborate further.

By the way i'm a physics student at a top tier univeristy.
 
nothing said:
Absolutely irrelevant example then. No relation to a camera, which isn't concerned with the illumination on side walls, only on the imager plane.
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Take a solar panel then. Is a quarter sized solar panel going to produce the same brightness as full size panel, when you hook a lamp on it?
 
A solar panel will produce no brightness because that is not what it does. It absorbs energy. And in this case the size makes a huge difference in the amount of energy gathered. However the energy captured per unit area will not change, and this is proportional to the brightness of the image if we are making comparisons to the camera sensor.
 
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