Comparing Olympus 4/3lenses to FX "Full Frame" offerings

Started Jan 25, 2014 | Discussions
JiminDenver Veteran Member • Posts: 5,190
Re: Wrong
1

The only reason full frame delivers more light is because it delivers a larger scene. Put a 25mm on 4/3s instead of a 50mm and the 4/3s sensor will see the same light a full frame does at 50mm AND it will require the same f stop to achieve the same shutter speed at a given ISO as the full frame does.

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JimB
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Sergey_Green
Sergey_Green Forum Pro • Posts: 11,536
No, it is not the reason ..
3

JiminDenver wrote:

The only reason full frame delivers more light is because it delivers a larger scene. Put a 25mm on 4/3s instead of a 50mm and the 4/3s sensor will see the same light a full frame does at 50mm AND it will require the same f stop to achieve the same shutter speed at a given ISO as the full frame does.

25 vs. 50 is for DoF and framing, not the light. However, if you set the same numeric aperture on each, each will have different diameter of the entrance pupil (effective aperture), which in turn under the same light intensity will not let the same total amount of light in (it is impossible). So exposure remains the same, total amount of light not.

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- sergey

bobn2
bobn2 Forum Pro • Posts: 57,682
Re: Wrong
2

JiminDenver wrote:

The only reason full frame delivers more light is because it delivers a larger scene.

Well, yes, but it depends what you mean by a 'larger scene'. It delivers a larger image with the same density of light (for the same f-number and shutter speed) so that is more light.

Put a 25mm on 4/3s instead of a 50mm and the 4/3s sensor will see the same light a full frame does at 50mm

No, not at the same f-number. a 25/1.4 on a Four Thirds sensor will deliver one quarter of the light to the sensor that a 50/1.4 does to a full-frame, because the density of the light is controlled by the f-number. Same f-number, same amount of light per unit area of the sensor. Four times larger area is four times more light.

AND it will require the same f stop to achieve the same shutter speed at a given ISO as the full frame does.

That is a circular argument, since ISO is defined by exposure. Same ISO, same exposure but four times as much light on the FF sensor, four times as many photons, twice the SNR. It's that simple.

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Bob

Dimitri_P Regular Member • Posts: 201
Re: totally wrong
2

tko wrote among other wierd stuff :

Do you think a F2.8 cell phone lens, the size of a thimble, is equal to a F2.0 M43rds lens?

Equality is a funny thing. 100 pennies is equal to 4 quarters which is equal to a soiled crumpled up 1 dollar bill.

F-stop is a NORMALIZED number, normalized to the sensor size.

You are funny. .

The Hubble telescope is around F24. How can that be, when the lens is so huge and designed to gather light?

Once you figure this out, you will understand.

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Tiger1 Contributing Member • Posts: 525
Re: Wrong
3

bobn2 wrote:

That is a circular argument, since ISO is defined by exposure. Same ISO, same exposure but four times as much light on the FF sensor, four times as many photons, twice the SNR. It's that simple.

The surface area is four times bigger Bob so what each unit area "sees" is the same amount of photons regardless the format.  Hence the SAME SNR.

The fact that smaller sensors tend to have smaller photosites so they can have the same number as larger sensors is what determines noise characteristics. That is why their SNR is inferior.  Nothing to do with format.  I'll even give you an example to prove this.  Imagine you were able to take a Nikon Df chip and cut it into a 4/3 sensor and then put it into a 4/3 camera.  The resulting image would only be 4MP but its noise and low light performance would be identical to the Df at 100% viewing of their pixels.

Simple.

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bobn2
bobn2 Forum Pro • Posts: 57,682
Re: Wrong
4

Tiger1 wrote:

bobn2 wrote:

That is a circular argument, since ISO is defined by exposure. Same ISO, same exposure but four times as much light on the FF sensor, four times as many photons, twice the SNR. It's that simple.

The surface area is four times bigger Bob so what each unit area "sees" is the same amount of photons regardless the format. Hence the SAME SNR.

You've got yourself confused. Lets go through it one step at a time.

1. For the same f-number the amount of light per unit area projected on the sensor is the same.

2. A FF sensor has four times the area as a FT sensor, so four times the 'unit'.

3. Thus a FF sensor collects four times the light at the same f-number.

4. When you view an image, you will do so at a size independent of the sensor size, the size you want to look at it. Imagine that we view in HD, approximately 2 megapixels. In the viewed image, each output pixel, one 2-millionth of the image represents the photons collected by one 2-millionth of the sensor.

5. Thus each pixel of the output image from FF sensor represents four times the number of photons as that made from a FT sensor, assuming the same exposure (f-number, shutter speed and scene luminance).

6. Hence the FF image will have double the SNR of the FT image.

The fact that smaller sensors tend to have smaller photosites so they can have the same number as larger sensors is what determines noise characteristics.

For what you say to be true, given that most noise in the image is photon shot noise, the smaller pixels would have to have a much lower quantum efficiency (collect a smaller proportion of the photons falling on them) than big pixels. This is not the case, they don't - in fact they typically have higher efficiency.

That is why their SNR is inferior.

Not so, I'm afraid.

Nothing to do with format. I'll even give you an example to prove this. Imagine you were able to take a Nikon Df chip and cut it into a 4/3 sensor and then put it into a 4/3 camera. The resulting image would only be 4MP but its noise and low light performance would be identical to the Df at 100% viewing of their pixels.

No-one in real life is interested in 100% viewing of their pixels. We are interested in what is the effect of a real photograph, and if you thing that a photograph taken from a FT size crop of a Df sensor and the resultant frame viewed same size as one taken from the full frame, would have the same SNR, you are mistaken, in fact it will have half the SNR.

Simple.

But wrong.

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Bob

Tiger1 Contributing Member • Posts: 525
Re: Wrong
3

bobn2 wrote:

Tiger1 wrote:

bobn2 wrote:

That is a circular argument, since ISO is defined by exposure. Same ISO, same exposure but four times as much light on the FF sensor, four times as many photons, twice the SNR. It's that simple.

The surface area is four times bigger Bob so what each unit area "sees" is the same amount of photons regardless the format. Hence the SAME SNR.

You've got yourself confused.

I am not confused.

Lets go through it one step at a time.

1. For the same f-number the amount of light per unit area projected on the sensor is the same.

We agree on something.

2. A FF sensor has four times the area as a FT sensor, so four times the 'unit'.

We agree here too.

3. Thus a FF sensor collects four times the light at the same f-number.

But that light is spread over 4 times the area....

4. When you view an image, you will do so at a size independent of the sensor size, the size you want to look at it. Imagine that we view in HD, approximately 2 megapixels. In the viewed image, each output pixel, one 2-millionth of the image represents the photons collected by one 2-millionth of the sensor.

When you view an image the quality you see is a measure of the number of pixels and the quality per pixel.

5. Thus each pixel of the output image from FF sensor represents four times the number of photons as that made from a FT sensor, assuming the same exposure (f-number, shutter speed and scene luminance).

You are only correct if the number of pixels equate the number of photosites AND that the photosites capture the 4 times the light than the smaller photosites. Photosites capture light, pixels DO NOT.

6. Hence the FF image will have double the SNR of the FT image.

You are being simplistic. SNR is dependent on the size of the photosite, its structure, its circuitry and the circuitry that takes its signal away to be processed.  All of these things coupled with the way the signal is amplified will affect SNR.  It's not a linear relationship to just size of photosite which you are assuming (incorrectly) is four times the size of that found in a 4/3 sensor.

The fact that smaller sensors tend to have smaller photosites so they can have the same number as larger sensors is what determines noise characteristics.

For what you say to be true, given that most noise in the image is photon shot noise, the smaller pixels would have to have a much lower quantum efficiency (collect a smaller proportion of the photons falling on them) than big pixels. This is not the case, they don't - in fact they typically have higher efficiency.

They may be more efficient per unit area but they are less efficient overall. Think about it. A large photosite collects more photons than a small one.  Why else are big photosite sensors less noisy than smaller ones?

That is why their SNR is inferior.

Not so, I'm afraid.

You are wrong here.

Nothing to do with format. I'll even give you an example to prove this. Imagine you were able to take a Nikon Df chip and cut it into a 4/3 sensor and then put it into a 4/3 camera. The resulting image would only be 4MP but its noise and low light performance would be identical to the Df at 100% viewing of their pixels.

No-one in real life is interested in 100% viewing of their pixels. We are interested in what is the effect of a real photograph, and if you thing that a photograph taken from a FT size crop of a Df sensor and the resultant frame viewed same size as one taken from the full frame, would have the same SNR, you are mistaken, in fact it will have half the SNR.

You are changing arguments. I am talking about the SNR per pixel.  Of course if you have the same quality pixel but have more of them you will achieve  better over all image quality.

Simple.

But wrong.

Sorry, but you are wrong Bob.

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Bob

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Ian Stuart Forsyth
Ian Stuart Forsyth Veteran Member • Posts: 3,351
Re: Wrong
1

Tiger1 wrote:

bobn2 wrote:

Tiger1 wrote:

bobn2 wrote:

That is a circular argument, since ISO is defined by exposure. Same ISO, same exposure but four times as much light on the FF sensor, four times as many photons, twice the SNR. It's that simple.

The surface area is four times bigger Bob so what each unit area "sees" is the same amount of photons regardless the format. Hence the SAME SNR.

You've got yourself confused.

I am not confused.

Lets go through it one step at a time.

1. For the same f-number the amount of light per unit area projected on the sensor is the same.

We agree on something.

2. A FF sensor has four times the area as a FT sensor, so four times the 'unit'.

We agree here too.

3. Thus a FF sensor collects four times the light at the same f-number.

But that light is spread over 4 times the area....

Here’s your problem the intensity of the light remainders the same, it’s not spread over 4 times larger surface area but rather that 4 times larger area is being hit with the same light intensity capturing 4 times more light

4. When you view an image, you will do so at a size independent of the sensor size, the size you want to look at it. Imagine that we view in HD, approximately 2 megapixels. In the viewed image, each output pixel, one 2-millionth of the image represents the photons collected by one 2-millionth of the sensor.

When you view an image the quality you see is a measure of the number of pixels and the quality per pixel.

5. Thus each pixel of the output image from FF sensor represents four times the number of photons as that made from a FT sensor, assuming the same exposure (f-number, shutter speed and scene luminance).

You are only correct if the number of pixels equate the number of photosites AND that the photosites capture the 4 times the light than the smaller photosites. Photosites capture light, pixels DO NOT.

6. Hence the FF image will have double the SNR of the FT image.

You are being simplistic. SNR is dependent on the size of the photosite, its structure, its circuitry and the circuitry that takes its signal away to be processed. All of these things coupled with the way the signal is amplified will affect SNR. It's not a linear relationship to just size of photosite which you are assuming (incorrectly) is four times the size of that found in a 4/3 sensor.

The fact that smaller sensors tend to have smaller photosites so they can have the same number as larger sensors is what determines noise characteristics.

For what you say to be true, given that most noise in the image is photon shot noise, the smaller pixels would have to have a much lower quantum efficiency (collect a smaller proportion of the photons falling on them) than big pixels. This is not the case, they don't - in fact they typically have higher efficiency.

They may be more efficient per unit area but they are less efficient overall. Think about it. A large photosite collects more photons than a small one. Why else are big photosite sensors less noisy than smaller ones?

That is why their SNR is inferior.

Not so, I'm afraid.

You are wrong here.

Nothing to do with format. I'll even give you an example to prove this. Imagine you were able to take a Nikon Df chip and cut it into a 4/3 sensor and then put it into a 4/3 camera. The resulting image would only be 4MP but its noise and low light performance would be identical to the Df at 100% viewing of their pixels.

No-one in real life is interested in 100% viewing of their pixels. We are interested in what is the effect of a real photograph, and if you thing that a photograph taken from a FT size crop of a Df sensor and the resultant frame viewed same size as one taken from the full frame, would have the same SNR, you are mistaken, in fact it will have half the SNR.

You are changing arguments. I am talking about the SNR per pixel. Of course if you have the same quality pixel but have more of them you will achieve better over all image quality.

Simple.

But wrong.

Sorry, but you are wrong Bob.

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Bob

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The Camera is only a tool, photography is deciding how to use it.
The hardest part about capturing wildlife is not the photographing portion; it’s getting them to sign a model release

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boggis the cat Veteran Member • Posts: 6,329
'Best format' anxiety
1

The 'equivalent photographs' (abbreviated by some as 'equivalence') argument works both ways.

What this boils down to in practical terms is that if you shoot at medium distances at medium to wide angle and want a shallow depth of field (e.g. a typical 'head and shoulders' portrait at 85 mm EFL) then a 135 format system will give you an advantage.  If you want to shoot subjects at longer distances using narrower angle and do not want a very blurred image due to a very narrow DOF (e.g. people 100 ft or more away at 300 mm EFL) then a smaller format system will give you an advantage.

Deciding whether you need to move 'up' or 'down' in format (e.g. between 135 and FourThirds) is easy enough to do.  Review your shots: if you find that you have a lot of shots at wide-open aperture where you would prefer a shallower DOF then you may benefit by moving to a larger format; if you find that you have a lot of shots where you have a small aperture (and high ISO and / or low shutter speed) in order to get a sufficiently deep DOF then you may benefit from moving to a smaller format.

If it appears that you may benefit from changing format, then start looking at the costs and trade-offs involved.

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Tiger1 Contributing Member • Posts: 525
Re: Wrong
2

Ian Stuart Forsyth wrote:

Tiger1 wrote:

bobn2 wrote:

Tiger1 wrote:

bobn2 wrote:

That is a circular argument, since ISO is defined by exposure. Same ISO, same exposure but four times as much light on the FF sensor, four times as many photons, twice the SNR. It's that simple.

The surface area is four times bigger Bob so what each unit area "sees" is the same amount of photons regardless the format. Hence the SAME SNR.

You've got yourself confused.

I am not confused.

Lets go through it one step at a time.

1. For the same f-number the amount of light per unit area projected on the sensor is the same.

We agree on something.

2. A FF sensor has four times the area as a FT sensor, so four times the 'unit'.

We agree here too.

3. Thus a FF sensor collects four times the light at the same f-number.

But that light is spread over 4 times the area....

Here’s your problem the intensity of the light remainders the same, it’s not spread over 4 times larger surface area but rather that 4 times larger area is being hit with the same light intensity capturing 4 times more light

I know.  That is what I am saying!  The intensity of light per unit area is the same...... I thought that I have made that obvious.

4. When you view an image, you will do so at a size independent of the sensor size, the size you want to look at it. Imagine that we view in HD, approximately 2 megapixels. In the viewed image, each output pixel, one 2-millionth of the image represents the photons collected by one 2-millionth of the sensor.

When you view an image the quality you see is a measure of the number of pixels and the quality per pixel.

5. Thus each pixel of the output image from FF sensor represents four times the number of photons as that made from a FT sensor, assuming the same exposure (f-number, shutter speed and scene luminance).

You are only correct if the number of pixels equate the number of photosites AND that the photosites capture the 4 times the light than the smaller photosites. Photosites capture light, pixels DO NOT.

6. Hence the FF image will have double the SNR of the FT image.

You are being simplistic. SNR is dependent on the size of the photosite, its structure, its circuitry and the circuitry that takes its signal away to be processed. All of these things coupled with the way the signal is amplified will affect SNR. It's not a linear relationship to just size of photosite which you are assuming (incorrectly) is four times the size of that found in a 4/3 sensor.

The fact that smaller sensors tend to have smaller photosites so they can have the same number as larger sensors is what determines noise characteristics.

For what you say to be true, given that most noise in the image is photon shot noise, the smaller pixels would have to have a much lower quantum efficiency (collect a smaller proportion of the photons falling on them) than big pixels. This is not the case, they don't - in fact they typically have higher efficiency.

They may be more efficient per unit area but they are less efficient overall. Think about it. A large photosite collects more photons than a small one. Why else are big photosite sensors less noisy than smaller ones?

That is why their SNR is inferior.

Not so, I'm afraid.

You are wrong here.

Nothing to do with format. I'll even give you an example to prove this. Imagine you were able to take a Nikon Df chip and cut it into a 4/3 sensor and then put it into a 4/3 camera. The resulting image would only be 4MP but its noise and low light performance would be identical to the Df at 100% viewing of their pixels.

No-one in real life is interested in 100% viewing of their pixels. We are interested in what is the effect of a real photograph, and if you thing that a photograph taken from a FT size crop of a Df sensor and the resultant frame viewed same size as one taken from the full frame, would have the same SNR, you are mistaken, in fact it will have half the SNR.

You are changing arguments. I am talking about the SNR per pixel. Of course if you have the same quality pixel but have more of them you will achieve better over all image quality.

Simple.

But wrong.

Sorry, but you are wrong Bob.

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Bob

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The Camera is only a tool, photography is deciding how to use it.
The hardest part about capturing wildlife is not the photographing portion; it’s getting them to sign a model release

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Ian Stuart Forsyth
Ian Stuart Forsyth Veteran Member • Posts: 3,351
Re: Wrong
1

Tiger1 wrote:

Ian Stuart Forsyth wrote:

Tiger1 wrote:

bobn2 wrote:

Tiger1 wrote:

bobn2 wrote:

That is a circular argument, since ISO is defined by exposure. Same ISO, same exposure but four times as much light on the FF sensor, four times as many photons, twice the SNR. It's that simple.

The surface area is four times bigger Bob so what each unit area "sees" is the same amount of photons regardless the format. Hence the SAME SNR.

You've got yourself confused.

I am not confused.

Lets go through it one step at a time.

1. For the same f-number the amount of light per unit area projected on the sensor is the same.

We agree on something.

2. A FF sensor has four times the area as a FT sensor, so four times the 'unit'.

We agree here too.

3. Thus a FF sensor collects four times the light at the same f-number.

But that light is spread over 4 times the area....

Here’s your problem the intensity of the light remainders the same, it’s not spread over 4 times larger surface area but rather that 4 times larger area is being hit with the same light intensity capturing 4 times more light

I know. That is what I am saying! The intensity of light per unit area is the same...... I thought that I have made that obvious.

Then why would you say just above “But that light is spread over 4 times the area “ when responding to uncle Bobs “ thus a ff collect’s 4 times the light at the same F number”

Don’t you see the fallacy of say” but that light is spread over 4 time the area” when the intensity remains the same ?

4. When you view an image, you will do so at a size independent of the sensor size, the size you want to look at it. Imagine that we view in HD, approximately 2 megapixels. In the viewed image, each output pixel, one 2-millionth of the image represents the photons collected by one 2-millionth of the sensor.

When you view an image the quality you see is a measure of the number of pixels and the quality per pixel.

5. Thus each pixel of the output image from FF sensor represents four times the number of photons as that made from a FT sensor, assuming the same exposure (f-number, shutter speed and scene luminance).

You are only correct if the number of pixels equate the number of photosites AND that the photosites capture the 4 times the light than the smaller photosites. Photosites capture light, pixels DO NOT.

6. Hence the FF image will have double the SNR of the FT image.

You are being simplistic. SNR is dependent on the size of the photosite, its structure, its circuitry and the circuitry that takes its signal away to be processed. All of these things coupled with the way the signal is amplified will affect SNR. It's not a linear relationship to just size of photosite which you are assuming (incorrectly) is four times the size of that found in a 4/3 sensor.

The fact that smaller sensors tend to have smaller photosites so they can have the same number as larger sensors is what determines noise characteristics.

For what you say to be true, given that most noise in the image is photon shot noise, the smaller pixels would have to have a much lower quantum efficiency (collect a smaller proportion of the photons falling on them) than big pixels. This is not the case, they don't - in fact they typically have higher efficiency.

They may be more efficient per unit area but they are less efficient overall. Think about it. A large photosite collects more photons than a small one. Why else are big photosite sensors less noisy than smaller ones?

That is why their SNR is inferior.

Not so, I'm afraid.

You are wrong here.

Nothing to do with format. I'll even give you an example to prove this. Imagine you were able to take a Nikon Df chip and cut it into a 4/3 sensor and then put it into a 4/3 camera. The resulting image would only be 4MP but its noise and low light performance would be identical to the Df at 100% viewing of their pixels.

No-one in real life is interested in 100% viewing of their pixels. We are interested in what is the effect of a real photograph, and if you thing that a photograph taken from a FT size crop of a Df sensor and the resultant frame viewed same size as one taken from the full frame, would have the same SNR, you are mistaken, in fact it will have half the SNR.

You are changing arguments. I am talking about the SNR per pixel. Of course if you have the same quality pixel but have more of them you will achieve better over all image quality.

Simple.

But wrong.

Sorry, but you are wrong Bob.

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Bob

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The Camera is only a tool, photography is deciding how to use it.
The hardest part about capturing wildlife is not the photographing portion; it’s getting them to sign a model release

-- hide signature --

The Camera is only a tool, photography is deciding how to use it.
The hardest part about capturing wildlife is not the photographing portion; it’s getting them to sign a model release

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dave gaines
OP dave gaines Veteran Member • Posts: 9,183
It's all about the lenses
2

Messier Object wrote:

Hi Dave,

I bet you didn't expect the Equvalence evangelists and denyers to have a battle in your thread.
Once again, neither side made any ground on the other, and hopefully they'll withdraw before the MODs shut down the thread

Thanks Peter,

Yes, what a bunch of hooey. There's no winning this argument. I don't know why anyone responds to GB and others when they raise this subject. It's sounds like a broken record (for you strictly digital age kids that's an LP vinyl disc used for music).

This thread was really about lens IQ, fast glass, focal length options and the advantages of 4/3 on the long end. Olympus shooters should know they have lots of options with just 20 good lenses. Some of these are redundant between SG, HG and SHG. So they really only need 10 or 12 to handle every contingency available.

...
Since getting my 5D3 just over a year ago I've slowly built up my Canon FX system without any thought or reference to my 4/3 kit . I just know what I want and need to get the best out of the 5D.

This is a good approach. I haven't tried to replicate the focal lengths so much as the IQ and the FOV options. I've covered 14 mm to 200 mm with 3 good zooms that have slightly different ranges than the Olympus. Now I need a lens for UWA that accepts filters. That's a tough nut to crack with different solutions in Nikon FX.

As for Oly, I've not sold any of it, but I know for sure that I'll never buy another Olympus lens, however I just might buy the next E-Mx if it handles better and does a better job with PDAF.  Having the ZD150 and ZD300 means I still have a hand in the Olympus 4/3 game and still want to get a better sensor behind my 2 SHG primes.

I could not afford to keep both systems with the caliber of lenses I had and wanted from Nikon. If I had the long tele's you have I might have kept those and the E-5 too. I've given up on having a fast tele with FX. It's just too expensive, more than an Olympus 300 mm f/2.8.

I hope you find satisfaction with your D800.

Very happy so far, thanks.

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Dave

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Tiger1 Contributing Member • Posts: 525
Re: Wrong
3

Ian Stuart Forsyth wrote:

Tiger1 wrote:

Ian Stuart Forsyth wrote:

Tiger1 wrote:

bobn2 wrote:

Tiger1 wrote:

bobn2 wrote:

That is a circular argument, since ISO is defined by exposure. Same ISO, same exposure but four times as much light on the FF sensor, four times as many photons, twice the SNR. It's that simple.

The surface area is four times bigger Bob so what each unit area "sees" is the same amount of photons regardless the format. Hence the SAME SNR.

You've got yourself confused.

I am not confused.

Lets go through it one step at a time.

1. For the same f-number the amount of light per unit area projected on the sensor is the same.

We agree on something.

2. A FF sensor has four times the area as a FT sensor, so four times the 'unit'.

We agree here too.

3. Thus a FF sensor collects four times the light at the same f-number.

But that light is spread over 4 times the area....

Here’s your problem the intensity of the light remainders the same, it’s not spread over 4 times larger surface area but rather that 4 times larger area is being hit with the same light intensity capturing 4 times more light

I know. That is what I am saying! The intensity of light per unit area is the same...... I thought that I have made that obvious.

Then why would you say just above “But that light is spread over 4 times the area “ when responding to uncle Bobs “ thus a ff collect’s 4 times the light at the same F number”

Don’t you see the fallacy of say” but that light is spread over 4 time the area” when the intensity remains the same ?

Uncle Bob is just plain wrong.  He claims that each pixel receives 4 times as much light. Pixels don't receive light, photosites do.  Even given this error and assuming he means photosites he makes the assumption that each photosite is four times larger in a FF sensor than in a 4/3 sensor.  This ignores gaps that may exist between photosites and how these gaps may vary from one sensor to another.

I also am the one that pointed out that the intensity of light per unit area is the same.

4. When you view an image, you will do so at a size independent of the sensor size, the size you want to look at it. Imagine that we view in HD, approximately 2 megapixels. In the viewed image, each output pixel, one 2-millionth of the image represents the photons collected by one 2-millionth of the sensor.

When you view an image the quality you see is a measure of the number of pixels and the quality per pixel.

5. Thus each pixel of the output image from FF sensor represents four times the number of photons as that made from a FT sensor, assuming the same exposure (f-number, shutter speed and scene luminance).

You are only correct if the number of pixels equate the number of photosites AND that the photosites capture the 4 times the light than the smaller photosites. Photosites capture light, pixels DO NOT.

6. Hence the FF image will have double the SNR of the FT image.

You are being simplistic. SNR is dependent on the size of the photosite, its structure, its circuitry and the circuitry that takes its signal away to be processed. All of these things coupled with the way the signal is amplified will affect SNR. It's not a linear relationship to just size of photosite which you are assuming (incorrectly) is four times the size of that found in a 4/3 sensor.

The fact that smaller sensors tend to have smaller photosites so they can have the same number as larger sensors is what determines noise characteristics.

For what you say to be true, given that most noise in the image is photon shot noise, the smaller pixels would have to have a much lower quantum efficiency (collect a smaller proportion of the photons falling on them) than big pixels. This is not the case, they don't - in fact they typically have higher efficiency.

They may be more efficient per unit area but they are less efficient overall. Think about it. A large photosite collects more photons than a small one. Why else are big photosite sensors less noisy than smaller ones?

That is why their SNR is inferior.

Not so, I'm afraid.

You are wrong here.

Nothing to do with format. I'll even give you an example to prove this. Imagine you were able to take a Nikon Df chip and cut it into a 4/3 sensor and then put it into a 4/3 camera. The resulting image would only be 4MP but its noise and low light performance would be identical to the Df at 100% viewing of their pixels.

No-one in real life is interested in 100% viewing of their pixels. We are interested in what is the effect of a real photograph, and if you thing that a photograph taken from a FT size crop of a Df sensor and the resultant frame viewed same size as one taken from the full frame, would have the same SNR, you are mistaken, in fact it will have half the SNR.

You are changing arguments. I am talking about the SNR per pixel. Of course if you have the same quality pixel but have more of them you will achieve better over all image quality.

Simple.

But wrong.

Sorry, but you are wrong Bob.

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Bob

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The Camera is only a tool, photography is deciding how to use it.
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Ian Stuart Forsyth
Ian Stuart Forsyth Veteran Member • Posts: 3,351
Re: Wrong

Tiger1 wrote:

Ian Stuart Forsyth wrote:

Tiger1 wrote:

Ian Stuart Forsyth wrote:

Tiger1 wrote:

bobn2 wrote:

Tiger1 wrote:

bobn2 wrote:

That is a circular argument, since ISO is defined by exposure. Same ISO, same exposure but four times as much light on the FF sensor, four times as many photons, twice the SNR. It's that simple.

The surface area is four times bigger Bob so what each unit area "sees" is the same amount of photons regardless the format. Hence the SAME SNR.

You've got yourself confused.

I am not confused.

Lets go through it one step at a time.

1. For the same f-number the amount of light per unit area projected on the sensor is the same.

We agree on something.

2. A FF sensor has four times the area as a FT sensor, so four times the 'unit'.

We agree here too.

3. Thus a FF sensor collects four times the light at the same f-number.

But that light is spread over 4 times the area....

Here’s your problem the intensity of the light remainders the same, it’s not spread over 4 times larger surface area but rather that 4 times larger area is being hit with the same light intensity capturing 4 times more light

I know. That is what I am saying! The intensity of light per unit area is the same...... I thought that I have made that obvious.

Then why would you say just above “But that light is spread over 4 times the area “ when responding to uncle Bobs “ thus a ff collect’s 4 times the light at the same F number”

Don’t you see the fallacy of say” but that light is spread over 4 time the area” when the intensity remains the same ?

Uncle Bob is just plain wrong. He claims that each pixel receives 4 times as much light. Pixels don't receive light, photosites do. Even given this error and assuming he means photosites he makes the assumption that each photosite is four times larger in a FF sensor than in a 4/3 sensor. This ignores gaps that may exist between photosites and how these gaps may vary from one sensor to another.

I also am the one that pointed out that the intensity of light per unit area is the same.

What he states if “Thus each pixel of the output image from FF sensor represents four times the number of photons as that made from a FT sensor, assuming the same exposure (f-number, shutter speed and scene luminance)”

Stupefied down, If we view a FF image at 16mp output and compare it to FT image of equal output each pixel will contain 4 time more light information than the FT image

4. When you view an image, you will do so at a size independent of the sensor size, the size you want to look at it. Imagine that we view in HD, approximately 2 megapixels. In the viewed image, each output pixel, one 2-millionth of the image represents the photons collected by one 2-millionth of the sensor.

When you view an image the quality you see is a measure of the number of pixels and the quality per pixel.

5. Thus each pixel of the output image from FF sensor represents four times the number of photons as that made from a FT sensor, assuming the same exposure (f-number, shutter speed and scene luminance).

You are only correct if the number of pixels equate the number of photosites AND that the photosites capture the 4 times the light than the smaller photosites. Photosites capture light, pixels DO NOT.

6. Hence the FF image will have double the SNR of the FT image.

You are being simplistic. SNR is dependent on the size of the photosite, its structure, its circuitry and the circuitry that takes its signal away to be processed. All of these things coupled with the way the signal is amplified will affect SNR. It's not a linear relationship to just size of photosite which you are assuming (incorrectly) is four times the size of that found in a 4/3 sensor.

The fact that smaller sensors tend to have smaller photosites so they can have the same number as larger sensors is what determines noise characteristics.

For what you say to be true, given that most noise in the image is photon shot noise, the smaller pixels would have to have a much lower quantum efficiency (collect a smaller proportion of the photons falling on them) than big pixels. This is not the case, they don't - in fact they typically have higher efficiency.

They may be more efficient per unit area but they are less efficient overall. Think about it. A large photosite collects more photons than a small one. Why else are big photosite sensors less noisy than smaller ones?

That is why their SNR is inferior.

Not so, I'm afraid.

You are wrong here.

Nothing to do with format. I'll even give you an example to prove this. Imagine you were able to take a Nikon Df chip and cut it into a 4/3 sensor and then put it into a 4/3 camera. The resulting image would only be 4MP but its noise and low light performance would be identical to the Df at 100% viewing of their pixels.

No-one in real life is interested in 100% viewing of their pixels. We are interested in what is the effect of a real photograph, and if you thing that a photograph taken from a FT size crop of a Df sensor and the resultant frame viewed same size as one taken from the full frame, would have the same SNR, you are mistaken, in fact it will have half the SNR.

You are changing arguments. I am talking about the SNR per pixel. Of course if you have the same quality pixel but have more of them you will achieve better over all image quality.

Simple.

But wrong.

Sorry, but you are wrong Bob.

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Bob

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The Camera is only a tool, photography is deciding how to use it.
The hardest part about capturing wildlife is not the photographing portion; it’s getting them to sign a model release

-- hide signature --

The Camera is only a tool, photography is deciding how to use it.
The hardest part about capturing wildlife is not the photographing portion; it’s getting them to sign a model release

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The Camera is only a tool, photography is deciding how to use it.
The hardest part about capturing wildlife is not the photographing portion; it’s getting them to sign a model release

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bobn2
bobn2 Forum Pro • Posts: 57,682
Re: Wrong
2

Tiger1 wrote:

Ian Stuart Forsyth wrote:

Tiger1 wrote:

Ian Stuart Forsyth wrote:

Tiger1 wrote:

bobn2 wrote:

Tiger1 wrote:

bobn2 wrote:

That is a circular argument, since ISO is defined by exposure. Same ISO, same exposure but four times as much light on the FF sensor, four times as many photons, twice the SNR. It's that simple.

The surface area is four times bigger Bob so what each unit area "sees" is the same amount of photons regardless the format. Hence the SAME SNR.

You've got yourself confused.

I am not confused.

Lets go through it one step at a time.

1. For the same f-number the amount of light per unit area projected on the sensor is the same.

We agree on something.

2. A FF sensor has four times the area as a FT sensor, so four times the 'unit'.

We agree here too.

3. Thus a FF sensor collects four times the light at the same f-number.

But that light is spread over 4 times the area....

Here’s your problem the intensity of the light remainders the same, it’s not spread over 4 times larger surface area but rather that 4 times larger area is being hit with the same light intensity capturing 4 times more light

I know. That is what I am saying! The intensity of light per unit area is the same...... I thought that I have made that obvious.

Then why would you say just above “But that light is spread over 4 times the area “ when responding to uncle Bobs “ thus a ff collect’s 4 times the light at the same F number”

Don’t you see the fallacy of say” but that light is spread over 4 time the area” when the intensity remains the same ?

Uncle Bob is just plain wrong.

No, Uncle Bob is right. Think what 'per unit area' means. For each similar unit of area, the two sensors receive the same light. Let's choose the unit to be a square millimetre. The Four Thirds sensor is 225 square mm, the FF one is 864, nearly four times (actually 3.84). So if they both receive the same light per square mm, then the FF sensor receives 3.84 times the light.

He claims that each pixel receives 4 times as much light. Pixels don't receive light, photosites do.

Only if the sensors have the same number of pixels, but the number of pixels is irrelevant to the discussion, and it doesn't matter whether you call them 'pixels' or 'photosites' - it doesn't make any difference.

Even given this error and assuming he means photosites he makes the assumption that each photosite is four times larger in a FF sensor than in a 4/3 sensor. This ignores gaps that may exist between photosites and how these gaps may vary from one sensor to another.

That is a whole new and similarly fallacious argument. There isn't any systematic difference in the photon collecting efficiency of FF and FT pixels.

I also am the one that pointed out that the intensity of light per unit area is the same.

No, you aren't - it's been a part of the discussion from the start.

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Bob

Tiger1 Contributing Member • Posts: 525
Re: Wrong
1

Ian Stuart Forsyth wrote:

Tiger1 wrote:

Ian Stuart Forsyth wrote:

Tiger1 wrote:

Ian Stuart Forsyth wrote:

Tiger1 wrote:

bobn2 wrote:

Tiger1 wrote:

bobn2 wrote:

That is a circular argument, since ISO is defined by exposure. Same ISO, same exposure but four times as much light on the FF sensor, four times as many photons, twice the SNR. It's that simple.

The surface area is four times bigger Bob so what each unit area "sees" is the same amount of photons regardless the format. Hence the SAME SNR.

You've got yourself confused.

I am not confused.

Lets go through it one step at a time.

1. For the same f-number the amount of light per unit area projected on the sensor is the same.

We agree on something.

2. A FF sensor has four times the area as a FT sensor, so four times the 'unit'.

We agree here too.

3. Thus a FF sensor collects four times the light at the same f-number.

But that light is spread over 4 times the area....

Here’s your problem the intensity of the light remainders the same, it’s not spread over 4 times larger surface area but rather that 4 times larger area is being hit with the same light intensity capturing 4 times more light

I know. That is what I am saying! The intensity of light per unit area is the same...... I thought that I have made that obvious.

Then why would you say just above “But that light is spread over 4 times the area “ when responding to uncle Bobs “ thus a ff collect’s 4 times the light at the same F number”

Don’t you see the fallacy of say” but that light is spread over 4 time the area” when the intensity remains the same ?

Uncle Bob is just plain wrong. He claims that each pixel receives 4 times as much light. Pixels don't receive light, photosites do. Even given this error and assuming he means photosites he makes the assumption that each photosite is four times larger in a FF sensor than in a 4/3 sensor. This ignores gaps that may exist between photosites and how these gaps may vary from one sensor to another.

I also am the one that pointed out that the intensity of light per unit area is the same.

What he states if “Thus each pixel of the output image from FF sensor represents four times the number of photons as that made from a FT sensor, assuming the same exposure (f-number, shutter speed and scene luminance)”

Stupefied down, If we view a FF image at 16mp output and compare it to FT image of equal output each pixel will contain 4 time more light information than the FT image

So he knows the photon capture efficiency of each photosite in aFF vs a FT sensor?  Wow!!!!!  Were did he get that information from?  To know the output representation from the pixel you must know the capture efficiency of each photosite. Agree?

4. When you view an image, you will do so at a size independent of the sensor size, the size you want to look at it. Imagine that we view in HD, approximately 2 megapixels. In the viewed image, each output pixel, one 2-millionth of the image represents the photons collected by one 2-millionth of the sensor.

When you view an image the quality you see is a measure of the number of pixels and the quality per pixel.

5. Thus each pixel of the output image from FF sensor represents four times the number of photons as that made from a FT sensor, assuming the same exposure (f-number, shutter speed and scene luminance).

You are only correct if the number of pixels equate the number of photosites AND that the photosites capture the 4 times the light than the smaller photosites. Photosites capture light, pixels DO NOT.

6. Hence the FF image will have double the SNR of the FT image.

You are being simplistic. SNR is dependent on the size of the photosite, its structure, its circuitry and the circuitry that takes its signal away to be processed. All of these things coupled with the way the signal is amplified will affect SNR. It's not a linear relationship to just size of photosite which you are assuming (incorrectly) is four times the size of that found in a 4/3 sensor.

The fact that smaller sensors tend to have smaller photosites so they can have the same number as larger sensors is what determines noise characteristics.

For what you say to be true, given that most noise in the image is photon shot noise, the smaller pixels would have to have a much lower quantum efficiency (collect a smaller proportion of the photons falling on them) than big pixels. This is not the case, they don't - in fact they typically have higher efficiency.

They may be more efficient per unit area but they are less efficient overall. Think about it. A large photosite collects more photons than a small one. Why else are big photosite sensors less noisy than smaller ones?

That is why their SNR is inferior.

Not so, I'm afraid.

You are wrong here.

Nothing to do with format. I'll even give you an example to prove this. Imagine you were able to take a Nikon Df chip and cut it into a 4/3 sensor and then put it into a 4/3 camera. The resulting image would only be 4MP but its noise and low light performance would be identical to the Df at 100% viewing of their pixels.

No-one in real life is interested in 100% viewing of their pixels. We are interested in what is the effect of a real photograph, and if you thing that a photograph taken from a FT size crop of a Df sensor and the resultant frame viewed same size as one taken from the full frame, would have the same SNR, you are mistaken, in fact it will have half the SNR.

You are changing arguments. I am talking about the SNR per pixel. Of course if you have the same quality pixel but have more of them you will achieve better over all image quality.

Simple.

But wrong.

Sorry, but you are wrong Bob.

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Bob

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The Camera is only a tool, photography is deciding how to use it.
The hardest part about capturing wildlife is not the photographing portion; it’s getting them to sign a model release

-- hide signature --

The Camera is only a tool, photography is deciding how to use it.
The hardest part about capturing wildlife is not the photographing portion; it’s getting them to sign a model release

-- hide signature --

The Camera is only a tool, photography is deciding how to use it.
The hardest part about capturing wildlife is not the photographing portion; it’s getting them to sign a model release

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Ian Stuart Forsyth
Ian Stuart Forsyth Veteran Member • Posts: 3,351
Re: Wrong

Tiger1 wrote:

Ian Stuart Forsyth wrote:

Tiger1 wrote:

Ian Stuart Forsyth wrote:

Tiger1 wrote:

Ian Stuart Forsyth wrote:

Tiger1 wrote:

bobn2 wrote:

Tiger1 wrote:

bobn2 wrote:

That is a circular argument, since ISO is defined by exposure. Same ISO, same exposure but four times as much light on the FF sensor, four times as many photons, twice the SNR. It's that simple.

The surface area is four times bigger Bob so what each unit area "sees" is the same amount of photons regardless the format. Hence the SAME SNR.

You've got yourself confused.

I am not confused.

Lets go through it one step at a time.

1. For the same f-number the amount of light per unit area projected on the sensor is the same.

We agree on something.

2. A FF sensor has four times the area as a FT sensor, so four times the 'unit'.

We agree here too.

3. Thus a FF sensor collects four times the light at the same f-number.

But that light is spread over 4 times the area....

Here’s your problem the intensity of the light remainders the same, it’s not spread over 4 times larger surface area but rather that 4 times larger area is being hit with the same light intensity capturing 4 times more light

I know. That is what I am saying! The intensity of light per unit area is the same...... I thought that I have made that obvious.

Then why would you say just above “But that light is spread over 4 times the area “ when responding to uncle Bobs “ thus a ff collect’s 4 times the light at the same F number”

Don’t you see the fallacy of say” but that light is spread over 4 time the area” when the intensity remains the same ?

Uncle Bob is just plain wrong. He claims that each pixel receives 4 times as much light. Pixels don't receive light, photosites do. Even given this error and assuming he means photosites he makes the assumption that each photosite is four times larger in a FF sensor than in a 4/3 sensor. This ignores gaps that may exist between photosites and how these gaps may vary from one sensor to another.

I also am the one that pointed out that the intensity of light per unit area is the same.

What he states if “Thus each pixel of the output image from FF sensor represents four times the number of photons as that made from a FT sensor, assuming the same exposure (f-number, shutter speed and scene luminance)”

Stupefied down, If we view a FF image at 16mp output and compare it to FT image of equal output each pixel will contain 4 time more light information than the FT image

So he knows the photon capture efficiency of each photosite in aFF vs a FT sensor? Wow!!!!! Were did he get that information from? To know the output representation from the pixel you must know the capture efficiency of each photosite. Agree?

With some math and information that is readily available from companies testing the performance of sensors

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bobn2
bobn2 Forum Pro • Posts: 57,682
Re: Wrong
4

Tiger1 wrote:

What he states if “Thus each pixel of the output image from FF sensor represents four times the number of photons as that made from a FT sensor, assuming the same exposure (f-number, shutter speed and scene luminance)”

Stupefied down, If we view a FF image at 16mp output and compare it to FT image of equal output each pixel will contain 4 time more light information than the FT image

So he knows the photon capture efficiency of each photosite in aFF vs a FT sensor?

Funny you should say that:

http://www.sensorgen.info

Wow!!!!!

Glad you're impressed.

Were did he get that information from?

The raw data from DxOmark, it needs some processiing to calculate the efficiency.

To know the output representation from the pixel you must know the capture efficiency of each photosite. Agree?

I think most manufacturers will try to get the efficiency of each 'photosite' in each of the Bayer submatrices the same, otherwise there will be a lot of pattern noise on the sensor.

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Bob

Great Bustard Forum Pro • Posts: 41,751
Noise as a function of DOF.
1

Doctor Lecter wrote:

"So larger formats gain their noise advantage over smaller formats by using a more shallow DOF for a given shutter speed."

???

Did you just make that one up? You did, didn't you.

In fact, not only did I not make it up, but it should be intuitively obvious. That is, the wider the aperture (entrance pupil) diameter, the more shallow the DOF and the more light that will pass through the lens.

Larger formats gain a noise advantage because they are larger. Period. End of story. There is a larger surface area that produces more information, be it film or digital. Sort of like a computer hard drive. A 1TB drive can hold more data than an 80GB drive.

The larger size of the sensor comes into play, directly, only for base ISO shooting. For example, a person shooing a scene at 12mm 1/200 f/5.6 ISO 100 on 4/3 would shoot the scene at 24mm 1/50 f/11 ISO 100 on FF for the same DOF (assuming 1/50 is fast enough to mitigate motion blur).

In this case, the larger sensor can use one-fourth the shutter speed resulting in four times as much light falling on the sensor, and it is because the sensor is 4x larger that it can absorb 4x as much light without oversaturating.

However, in lower light, where the photographer would use a higher ISO to increase the LCD playback / OOC jpg brightness of the photo, to get less noise, the FF photographer would often choose a wider aperture (e.g. 50mm / 1.4 = 36mm on FF vs 25mm / 1.4 = 18mm on 4/3), but this would also result in a more shallow DOF.

It's pretty simple.

It is, isn't it?

You'll get there... you're almost grasping these basic concepts.

In your particular case, I'm not so convinced. However, I'll make every effort not to give up hope.

Great Bustard Forum Pro • Posts: 41,751
Apreture, HST, and context.
2

Dimitri_P wrote:

tko wrote among other wierd stuff :

Do you think a F2.8 cell phone lens, the size of a thimble, is equal to a F2.0 M43rds lens?

Equality is a funny thing. 100 pennies is equal to 4 quarters which is equal to a soiled crumpled up 1 dollar bill.

Exactly. So, for example, it wouldn't make sense to say "2 coins = 2 coins", would it? That is, it would only be true if each coin had the same value.

F-stop is a NORMALIZED number, normalized to the sensor size.

You are funny.

Ah, here we are:

http://www.josephjamesphotography.com/equivalence/#aperture

Understanding the fundamental concepts of Equivalence requires making important distinctions between various terms which people often take to mean the same thing. It is very much akin to making the distinction between "mass" and "weight", two terms which most people take to mean the same thing, when, in fact, they measure two different (but related) quantities. While there are circumstances where making the distinction is unnecessary, there are other times when it is critical.

The first of these distinctions that needs to be made is between aperture and f-ratio. The term "aperture", by itself, is vague -- we need a qualifying adjective to be clear. There are three different terms using "aperture":

  1. The physical aperture (iris) is the smallest opening within a lens.
  2. The virtual aperture (entrance pupil) is the image of the physical aperture when looking through the FE (front element).
  3. The relative aperture (f-ratio) is the quotient of the focal length and the virtual aperture.

Thus, the "f" in an f-ratio stands for focal length. For example f/2 on a 50mm lens means the diameter of the virtual aperture (entrance pupil) is 50mm / 2 = 25mm. Likewise, a 50mm lens with a 25mm virtual aperture has an f-ratio of 50mm / 25mm = 2.

So, the f-ratio is the aperture normalized to focal length, which will result in the same light per area on the sensor for all formats and focal lengths, but not the same total amount of light on the sensor for different formats.

The Hubble telescope is around F24. How can that be, when the lens is so huge and designed to gather light?

Once you figure this out, you will understand.

If I had to guess, I'd say he already had it figured out. The focal length of the HST is 76.6m and the aperture diameter is 2.4. This results in a relative aperture (f-ratio) of 57.6 / 2.4 = 2.4.

I'm sure you'll agree that the huge light gathering ability of the HST is a function of both the incredibly wide aperture (2400mm) and the incredibly long exposure times it can use.

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