Correction: Canon is bringing us an RF 24-105mm f/4-7.1 IS STM Macro

Shellbo6901

EOS M50
Jan 16, 2015
26
5
Price is king. It's initial price is 400, which is a lot lower than the RF 24-105L. Just wait a couple years after the lens price falls. Something like the RP and this 24-105 lens for less than $1000. That is the target market.

Canon was criticized for leading with L-glass because there weren't enough consumer-grade options, and now that consumer grade options are coming it, it's not good enough...
Its just the one zoom options that already been done, and then even added the 24 to 240, and when they took the 28-70, the next version was 24-70. but I guess taking it from non-macro to macro was the way they made it a tad different(so I kind of answered my question, but originally I didn't see macro... i will point out so far both the $400-500 RF lenses are macro, so I wonder if most will be, or what their resoning behind it is). The original 24-105 is normally $1099, but with the kit and when they put it on sale is around 899(I believe). I'm sure when the next version around the RP is out it will be a good fit.
 

goldenhusky

EOS RP
Dec 2, 2016
278
72
For $400 this sounds like a great starter lens. Canon can give it away pretty much for free along with cheaper RF mount cameras.
 

Sdiver2489

EOS M50
Apr 17, 2013
27
5
Difficult it is... (I had to look it up myself to understand it better)

Let's say you shoot with an MFT (crop factor 2) with a 50mm f1.4. To get the same angle of view (the same frame) with a FF you need a 100mm lens. The aperture size (f-number) is calculated by the focal-length divided by the diameter of the aperture; same diameter means same amount of light. The 50mm has a diameter of 35.7mm at f1.4 (50/1.4). For the 100mm we get an aperture of 100mm/35.7mm = 2.8. So 50mm f1.4 on a MFT equals 100mm f2.8 on a FF, in terms of amount of light. But the FF sensor has 4 times the area of an MFT sensor. This means the amount of light per area is 4 times smaller on a FF sensor. On the other hand, with the same pixel count, the light per pixel would be the same.

I'm using APS-C, and think about switching to FF. My feeling is, that consumer lenses on a FF will be sharper than semi pro lenses on APS-C, because the pixel on the FF are bigger. I will have to do some internet research on that when the time comes for a new camera.
The information presented here is absolutely true in terms of DOF equivalence. It cannot be true in terms of light gathering ability per pixel because otherwise if I mounted a APS-C lens to my a7riv, I should expect it to darker and it is not. The sensor is not compensating, the aperture sets the amount of light per pixel with everything else held equal.

Can you please post a link to where you looked this up?
 

Travel_Photographer

Travel, Landscape, Architecture
Aug 30, 2019
94
125
Difficult it is... (I had to look it up myself to understand it better)

Let's say you shoot with an MFT (crop factor 2) with a 50mm f1.4. To get the same angle of view (the same frame) with a FF you need a 100mm lens. The aperture size (f-number) is calculated by the focal-length divided by the diameter of the aperture; same diameter means same amount of light. The 50mm has a diameter of 35.7mm at f1.4 (50/1.4). For the 100mm we get an aperture of 100mm/35.7mm = 2.8. So 50mm f1.4 on a MFT equals 100mm f2.8 on a FF, in terms of amount of light. But the FF sensor has 4 times the area of an MFT sensor. This means the amount of light per area is 4 times smaller on a FF sensor. On the other hand, with the same pixel count, the light per pixel would be the same.

I'm using APS-C, and think about switching to FF. My feeling is, that consumer lenses on a FF will be sharper than semi pro lenses on APS-C, because the pixel on the FF are bigger. I will have to do some internet research on that when the time comes for a new camera.
I think you're getting caught up in the "physics" and engineering part of this. The beauty of the "F-stop" is that it is universal across all sensor sizes and lenses. Manufacturers design the size of the lens elements and aperture to produce consistent exposure at the same exposure settings across all sensor sizes and formats. As others have stated in terms of exposure, F2.8 is F2.8 regardless of sensor size or lens. In practical use for taking a photo, it doesn't matter physically how much light is collected in photons etc etc. The exposure is the same. If three people are next to each other, one with a full-frame 5D Mark IV, one with an APS-C 90D, and one with a tiny Canon G7 X Mark II, and they all set their cameras to F2.8 and same shutter speed and ISO, they're all getting the exact same exposure. The physics of how much light is going through the lens and how it is focused on what size sensor area etc etc is complicated. That's why it's so nice to be able to just know what exposure you're going to get at F5.6, shutter 1/1000th, ISO 100 regardless of camera, lens, or sensor.

Just for one example of something that is more complex than it might seem, is your example of the aperture in millimeters. What you wrote about the aperture in millimeters is not exactly accurate. The size of the "hole" in the lens in regards to exposure and especially depth of field, is not technically the physical aperture. It's the "entrance pupil". You may be thinking of the "aperture" as the size of the "physical" hole that is created by the blades of the diaphragm, but that's not what's important, nor is it what should be "measured" in your calculations. It is the size of the "entrance pupil" that is calculated when you divide the focal length by the F-stop. And the entrance pupil is not even a "real" or "physical" thing. It is, for lack of a better description, and "optical phenomenon", an "illusion". To put it in real terms like your example, a 100mm lens with the aperture set to the F4 would have an "entrance pupil" of 25mm (100/4). A 50mm lens with an aperture of F4 would have an entrance pupil of 12.5mm (50/4). This is regardless of sensor size. You can see this "illusion" in real life and get a better sense of it with the following example. Let's say you have a Canon 70-200mm F4L lens. You set the aperture to F4 and leave it there, so no matter what you zoom the lens to, the aperture will stay at F4. If you zoom the lens to 100mm, and you take off the front and back lens caps, and look through the lens from the front with F-stop set to F4, you will see the "entrance pupil" hole and it will "appear" to be 25mm large to your eye. That hole is an illusion, it is the "appearance" of the aperture blades as affected by the focal length you selected. While continuing to look through the lens, zoom to 200mm and you will see the "entrance pupil" grow from 25mm to 50mm. The illusion of the hole will have doubled in size as you doubled the focal length from 100mm to 200mm. Now the physical aperture blades haven't moved or changed size at all. The physical hole is the same at both focal lengths. They're still creating a hole however many millimeters in diamater. It's not relevant how many millimeters the physical hole is. What's relevant is the size of the entrance pupil.

Anyway, you get the idea of what I'm trying to say. It's much more complicated than just a few simple calculations of aperture sizes and sensor sizes. Manufacturers create the lenses with specific sizes of the elements and the diameter of the aperture blades to produce the exact amount of light required for that sensor size so that the F-stops work universally. That's the whole beauty. The system allows us to just enjoy photography with a universal system that works across all the different formats. Depth of field is a whole different story and has it's own calculations, but for now I'm just talking about exposure.

(For those interested in a brief comment about DoF, it is the size of the entrance pupil in millimeters that has a direct effect on Depth of Field) That's why full-frame has more depth of field than Micro Four Thirds. The larger the entrance pupil, the narrower the DoF. On a full frame camera, if you take a photo with a 100mm lens at F4, the entrance pupil is 25mm. The exact same photo and exposure on Micro Four Thirds would be taken with a 50mm at F4 (you would use 50mm to get FF equivalent focal length). So you would have same exact framing and exposure BUT the Micro Four Thirds photo would be taken with an entrance pupil of only 12.5mm, so not as much background blur as the Full Frame with an entrance pupil of 25mm).
 

aj1575

EOS RP
Dec 15, 2010
202
0
I think you're getting caught up in the "physics" and engineering part of this. The beauty of the "F-stop" is that it is universal across all sensor sizes and lenses. Manufacturers design the size of the lens elements and aperture to produce consistent exposure at the same exposure settings across all sensor sizes and formats. As others have stated in terms of exposure, F2.8 is F2.8 regardless of sensor size or lens. In practical use for taking a photo, it doesn't matter physically how much light is collected in photons etc etc. The exposure is the same. If three people are next to each other, one with a full-frame 5D Mark IV, one with an APS-C 90D, and one with a tiny Canon G7 X Mark II, and they all set their cameras to F2.8 and same shutter speed and ISO, they're all getting the exact same exposure. The physics of how much light is going through the lens and how it is focused on what size sensor area etc etc is complicated. That's why it's so nice to be able to just know what exposure you're going to get at F5.6, shutter 1/1000th, ISO 100 regardless of camera, lens, or sensor.
The problem with your example is, that the people with the 3 different cameras need to work at different focal length to get the same picture. Changing your focal length also changes the size of the aperture in the equation.
Just answer this question: If I take a 50mm f1.4 on a MFT camera, and like to take the same picture from the same spot with a FF camera, what Focal length / aperture combination do I have to choose to get the same exposure with the same shutter speed at the same ISO?
 

Travel_Photographer

Travel, Landscape, Architecture
Aug 30, 2019
94
125
Just answer this question: If I take a 50mm f1.4 on a MFT camera, and like to take the same picture from the same spot with a FF camera, what Focal length / aperture combination do I have to choose to get the same exposure with the same shutter speed at the same ISO?
Answer: 100mm lens, aperture F1.4, same shutter and ISO.

The best way to think about it is this. In *any* camera, the exposure is based on *only* three things: shutter speed, ISO, and F-Stop. It is a "standard" and "universal" exposure across all cameras for those three settings. Focal length, sensor size, etc. are not factored into the equation at all. It is just the shutter speed, ISO, and F-Stop across every camera, every manufacturer, every lens, every sensor. Pick the same three numbers for those three settings, and you will get the same exposure on any camera with any lens at any focal length.

Here's a handy calculator that I sometimes use. It shows the exposure values for any combination of the three settings. Notice how focal length and sensor size are not part of the calculator, because they are not part of the equation. This calculator works for all cameras, lenses, and sensors.

 
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SteveC

M50 & T6i
Sep 3, 2019
475
341
Answer: 100mm lens, aperture F1.4, same shutter and ISO.

The best way to think about it is this. In *any* camera, the exposure is based on *only* three things: shutter speed, ISO, and F-Stop. It is a "standard" and "universal" exposure across all cameras for those three settings. Focal length, sensor size, etc. are not factored into the equation at all. It is just the shutter speed, ISO, and F-Stop across every camera, every manufacturer, every lens, every sensor. Pick the same three numbers for those three settings, and you will get the same exposure on any camera with any lens at any focal length.

Here's a handy calculator that I sometimes use. It shows the exposure values for any combination of the three settings. Notice how focal length and sensor size are not part of the calculator, because they are not part of the equation. This calculator works for all cameras, lenses, and sensors.

It sounds like aj1575 doesn't have the same concept as to what "exposure" is that you do (I believe yours is correct). You've reduced yourself to simply asserting that X, Y, and Z don't affect exposure--the problem being that if he understands the term differently, you might be coming across nonsensically.

So with that I'll try to clarify what "exposure" actually means. In the end, exposure simply amounts to "how bright will the photo look." (OK, that's not quite the meaning yet--and *I* am new enough to possibly be off base.)

If your sensor doubles in size, sure, twice as much light will hit it (all else being the same; no fair switching out sensors then waiting for nighttime to take the comparison shot), BUT that light will be spread out over twice the area.

So more technically, exposure is the light per unit area, or you could even call it flux--except that you also have to factor in how sensitive the sensor is.

Because of this, as you've said, the only things that can affect it are shutter speed, ISO and F stop, and I'll explain why:

shutter speed is obvious...double the amount of time the shutter is open, double the amount of light that hits a unit area of the sensor.

ISO is actually compensating for light level by making the sensor more or less sensitive. Double the sensitivity, twice the exposure.

Aperture obviously will affect the exposure too--if light gets to come in through twice as wide an aperture, then four times (not twice) as much will hit the sensor. Four times because you're doubling the height and width of the aperture simultaneously and that means you are quadrupling the area of the aperture. The other wrinkle with aperture is that it must be given as a ratio of the focal length, because if you double the focal length, you will quarter the "cone" of light that makes it to the sensor with the same diameter aperture. So photography always speaks of f/2.0 (for instance) which, for a 100 mm lens = 100/2 = 50 mm. If you double the focal length, you're basically taking an image of something only twice as wide, therefore you're only collecting 1/4 of the light in the first place, so to compensate, you'd need to double your aperture. Rather than deal with this, when making comparisons it's easier to give the aperture as a fraction of the focal length, so you can compare what two lenses will do for exposure. F/2.0 will give the same exposure no matter the focal length. (Note: it's technically focal length, f, divided by a number, so it SHOULD be written with the slash or a colon, though many don't bother with that.)

Those, plus the ambient brightness (which is outside of the camera, and not controllable in the camera) are the only things that affect how much light hits the sensor per unit area and how that sensor will respond to the light hitting it, which together govern how bright the picture will be.
 

Kit.

EOR R
Apr 25, 2011
1,626
967
And the entrance pupil is not even a "real" or "physical" thing. It is, for lack of a better description, and "optical phenomenon", an "illusion".
It might be called an "optical illusion", but the fact that you, as an object in the scene, see it when you look onto/into the lens, means that this "optical illusion" is what determines which rays of light from the scene are allowed to pass through the lens toward the sensor

Which affects both the depth of field and the number of photons registered by the sensor per unit of time.
 

Travel_Photographer

Travel, Landscape, Architecture
Aug 30, 2019
94
125
It might be called an "optical illusion", but the fact that you, as an object in the scene, see it when you look onto/into the lens, means that this "optical illusion" is what determines which rays of light from the scene are allowed to pass through the lens toward the sensor

Which affects both the depth of field and the number of photons registered by the sensor per unit of time.
Agreed.
 

Czardoom

I'm New Here
Jan 27, 2020
21
47
besides the macro and the pricepoint... maybe the weight, why would one want this VS the other 24-105 RF's available? seems dumb IMO, even for the RP.
Because it is much lighter. And also considerably smaller. And MUCH CHEAPER. There. Three good reasons.

It seems like Canon has given us a really good choice between the two 24-105 RF lenses. We have smaller, lighter and much more affordable - and we have the higher quality and cost L lens. Choose the one you want. Making them too similar apparently is what people here on the forum wanted. Makes no sense, but why is that not surprising.
 

Travel_Photographer

Travel, Landscape, Architecture
Aug 30, 2019
94
125
It sounds like aj1575 doesn't have the same concept as to what "exposure" is that you do (I believe yours is correct). You've reduced yourself to simply asserting that X, Y, and Z don't affect exposure--the problem being that if he understands the term differently, you might be coming across nonsensically.

So with that I'll try to clarify what "exposure" actually means. In the end, exposure simply amounts to "how bright will the photo look." (OK, that's not quite the meaning yet--and *I* am new enough to possibly be off base.)

If your sensor doubles in size, sure, twice as much light will hit it (all else being the same; no fair switching out sensors then waiting for nighttime to take the comparison shot), BUT that light will be spread out over twice the area.

So more technically, exposure is the light per unit area, or you could even call it flux--except that you also have to factor in how sensitive the sensor is.

Because of this, as you've said, the only things that can affect it are shutter speed, ISO and F stop, and I'll explain why:

shutter speed is obvious...double the amount of time the shutter is open, double the amount of light that hits a unit area of the sensor.

ISO is actually compensating for light level by making the sensor more or less sensitive. Double the sensitivity, twice the exposure.

Aperture obviously will affect the exposure too--if light gets to come in through twice as wide an aperture, then four times (not twice) as much will hit the sensor. Four times because you're doubling the height and width of the aperture simultaneously and that means you are quadrupling the area of the aperture. The other wrinkle with aperture is that it must be given as a ratio of the focal length, because if you double the focal length, you will quarter the "cone" of light that makes it to the sensor with the same diameter aperture. So photography always speaks of f/2.0 (for instance) which, for a 100 mm lens = 100/2 = 50 mm. If you double the focal length, you're basically taking an image of something only twice as wide, therefore you're only collecting 1/4 of the light in the first place, so to compensate, you'd need to double your aperture. Rather than deal with this, when making comparisons it's easier to give the aperture as a fraction of the focal length, so you can compare what two lenses will do for exposure. F/2.0 will give the same exposure no matter the focal length. (Note: it's technically focal length, f, divided by a number, so it SHOULD be written with the slash or a colon, though many don't bother with that.)

Those, plus the ambient brightness (which is outside of the camera, and not controllable in the camera) are the only things that affect how much light hits the sensor per unit area and how that sensor will respond to the light hitting it, which together govern how bright the picture will be.
I think the disconnect may be because of the mixing and matching of terms from two different areas. There's the "photography" terms that we're all used to (and most just use only those) and then there are the "engineering" and "physical terms." They're not easily mixed and matched. For example,

Changing your focal length also changes the size of the aperture in the equation.
From a "photography terminology" perspective, that is completely false. So I think that's where the confusion is. Changing the focal length does not change the size of the aperture. If it did, every time you changed lenses or zoomed a zoom lens your exposure would change based on the focal length. It does not. Pick an aperture, and change lenses and zoom all you want, the exposure will be identical.

Now from an "engineering" and "physics" perspective, the size of the "entrance pupil" (which remember is not even a "physical" thing,) does change "behind the scenes". But that is not the same as saying the "aperture" has changed, when discussing photography. If the aperture is set to F5.6, it is F5.6 regardless of what lens or focal length is being used.

The aperture size (f-number) is calculated by the focal-length divided by the diameter of the aperture
I'm not sure what that means, but it isn't accurate. The appropriate formula is:

Focal length / F-stop = Size of the Entrance Pupil in millimeters

You start with the focal length and F-stop, and you derive the size of the entrance pupil in millimeters.

Once you select the F-stop, the entrance pupil will work its magic behind the scenes, widening and narrowing as it needs to in order to provide consistent exposure regardless of what focal length is selected. So if F5.6 produces the correct exposure for the scene, you can zoom all you want, or switch to your backup camera which may be a Micro Four Thirds, and use the exact aperture, shutter speed, and ISO, and you'll get the same exposure.

To summarize:

"Exposure" is set by selecting:
* Shutter Speed
* ISO
* F-number

(Notice there is NO mention of the physical size of the aperture opening or entrance pupil in the above. It is only those three things that combine to set exposure.)

Separately, there is the formula:

Entrance Pupil Size = Focal Length / F-number

(Notice there is no mention of exposure in that formula. It is purely a physical calculation with no reference to a resulting exposure).

Perhaps most importantly, In neither of the two relevant formulas is the word "aperture". If we talk about the whole thing using only the words shutter speed, ISO, F-number, and entrance pupil, it would make the discussion much more straightforward.
 
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jd7

EOS 7D MK II
Feb 3, 2013
782
151
Difficult it is... (I had to look it up myself to understand it better)

Let's say you shoot with an MFT (crop factor 2) with a 50mm f1.4. To get the same angle of view (the same frame) with a FF you need a 100mm lens. The aperture size (f-number) is calculated by the focal-length divided by the diameter of the aperture; same diameter means same amount of light. The 50mm has a diameter of 35.7mm at f1.4 (50/1.4). For the 100mm we get an aperture of 100mm/35.7mm = 2.8. So 50mm f1.4 on a MFT equals 100mm f2.8 on a FF, in terms of amount of light. But the FF sensor has 4 times the area of an MFT sensor. This means the amount of light per area is 4 times smaller on a FF sensor. On the other hand, with the same pixel count, the light per pixel would be the same.
I believe this is not altogether correct.

Firstly, a 50mm f/1.4 on MFT is equivalent to a 100mm f/2.8 on full frame in terms of trying to produce images which are "equivalent" (at least very similar) in terms of framing and depth of field (assuming you are take photos of the same subject from the same spot, ie same perspective). However, from an "exposure" point of view, f/2.8 is still two stops "slower" than f/1.4, so if you use the same ISO, the full frame shot will be two stops dimmer than the MFT shot. (Let's leave aside any potential difference in quantum efficiency between two sensors!)

So, what does "exposure" mean? In what I've said above, I've used exposure in the way photographers often do, which is to refer to image brightness, ie light density on the sensor (light per unit area). However, exposure can also refer to the total light gathered by the sensor.

Going back to your example, from an "exposure" point of view, 50mm f/1.4 on MFT and 100m f/1.4 on full frame would give the same image brightness (at same ISO), ie the same light density on each part of the sensor. However, the full frame sensor has an area which is four times the area of the MFT sensor, so the full frame sensor gathers four times as much total light (total light exposure, if you like) than the MFT sensor. If you then view the images at the same output size (on a screen or in a print), you can think of the full frame image having four times as much total light packed into the same area you are viewing, which is important in giving full frame sensors a noise advantage.

However, if you take photos of the same subject from the same spot with a 50mm f/1.4 on MFT and a 100m f/1.4 on full frame, you will have the same perspective and angle of view, but the full frame will have shallower depth of field. That is because the distance to subject is the same in both cases, but the full frame shot is taken with a larger "aperture" (OK, entrance pupil as someone else has point out, ie apparent aperture, but let's not get too bogged down on that here). The point to remember is that although photographers often say they set their aperture to 1.4 or some other f stop number, f stop actually means relative aperture (ie aperture relative to focal length), not physical aperture (really entrance pupil). So, when we use f/1.4 on a 50mm lens and f/1.4 on a 100mm lens, we use the same relative aperture, but not the same physical aperture. In that case, the physical aperture when you use a 50mm lens at f/1.4 is 50mm / 1.4 = 35 mm, but when you use a 100mm lens at f/1.4 is 100mm / 1.4 = 71.43 mm.

Referring to relative aperture (f stop) allows consistent reference to image brightness (exposure in the sense of light density on the sensor) as we change focal lengths (and since it is referring to light density on the sensor, it does not change as sensor size changes). However, you need to factor in the physical aperture (entrance pupil) if you want to compare depth of field. You also need to factor in sensor size if you want to compare total light gathered (total exposure, if you like).

You might find this an interesting read: http://www.josephjamesphotography.com/equivalence/

Hope that helps!
 
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SteveC

M50 & T6i
Sep 3, 2019
475
341
To summarize:

"Exposure" is set by selecting:
* Shutter Speed
* ISO
* F-number

(Notice there is NO mention of the physical size of the aperture opening or entrance pupil in the above. It is only those three things that combine to set exposure.)

Separately, there is the formula:

Entrance Pupil Size = Focal Length / F-number

(Notice there is no mention of exposure in that formula. It is purely a physical calculation with no reference to a resulting exposure).

Perhaps most importantly, In neither of the two relevant formulas is the word "aperture". If we talk about the whole thing using only the words shutter speed, ISO, F-number, and entrance pupil, it would make the discussion much more straightforward.
There is no mention of the physical size of the aperture, because it--or rather the entrance pupil which is what actually matters, is implicit in the F number. It's not that it doesn't matter, it's that it's being expressed in a different form. F number is a way of expressing the relative size of the entrance pupil, rather than expressing it directly.

The F number is simply the entrance pupil in terms of the focal length, because what really matters for exposure is the ratio of the entrance pupil width to the focal length. Rather than burden photographers with doing that division every single time--it's simply labeled that way in the first place. It allows photographers to assume the exposure equivalent when they switch lenses and both are set to the same F/#, whereas if the entrance pupil were labeled in millimeters, they'd have to think to themselves, "hmm, with the 100mm lens I had a 25mm entrance pupil, so I'd better set the entrance pupil on my 50mm to 12.5." Instead, they can just see f/4 is the setting on the one lens and apply that setting to the other lens.
 

stevelee

FT-QL
Jul 6, 2017
1,319
324
Davidson, NC
Canon seems to have fallen in love with 24-105 as the range for basic consumer normal zooms. I guess they have the marketing numbers to back it up. Being Canon, they seem unfazed by Nikon's 24-120. Apparently Canon thinks the good magic number 105 trumps the not so good magic number 7.1.
I got the non-L 24-105 as my kit lens with the 6D2. I’ve been pleased with it as my general-purpose lens. For my travel camera, I used the G7X II for three years, and found it’s 24-100 equivalence just as useful. For interiors and scenic vistas, I have needed something wider, and rarely missed a longer focal length. Perhaps that effective similarity of range has made me more comfortable going between the two cameras. For my fall trip, I upgraded to the G5X II, and did occasionally zoom in to the 120mm equivalent (and appreciated the popup EVF in bright sunlight). So I can see the 24-105 as a good range for a kit/starter lens one could live with. My 24-105mm is just f/5.6, so I don’t see the 2/3 stop loss of the 7.1 as that much of a drawback in real life. The noise from a modern full-frame sensor shouldn’t be noticeably worse at ISO 3200 than at 2000, for example.
 
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stevelee

FT-QL
Jul 6, 2017
1,319
324
Davidson, NC
+1 of course.

So I've added or interchanged 'hockey dads' into my soccer mom statements the last few years. When I've said soccer moms before, the point was never that women somehow uniquely require a simple / low-tech camera (which is a ridiculous notion to hold) -- the point is that some folks just want to take snaps of the fam and not require climbing up the mode dial or discussing gear in forums to do it. One gender does not have a monopoly on that user need.

People love their kids and want to be able to capture them doing things in places that cell phones do not excel -- concerts, sporting events, etc. -- even if they've never owned or used a dedicated camera before. That applies to all genders.

- A
My soccer mom stereotype has more to do with vehicles than cameras. They are usually the ones in vehicles that can chauffeur around half the team.

My sports parents image is of women and men (about equal numbers) shooting with 7Ds, or at least aspiring to one. That fits the ones I’ve known.
 
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Kit.

EOR R
Apr 25, 2011
1,626
967
So, what does "exposure" mean? In what I've said above, I've used exposure in the way photographers often do, which is to refer to image brightness, ie light density on the sensor (light per unit area). However, exposure can also refer to the total light gathered by the sensor.
I think that using terms in non-standard ways is what increases confusion, not reduces it.

Exposure, both in radiometry and in photometry, is energy per unit of area. Given that some important sensor quantities (full well capacity, dark current noise) are proportional to the sensor area for which they are measured, exposure is a quantity that is important in describing non-linear effects (highlights cutout and, partially, very deep shadow noise) of the sensor converting photons to electrons. But that's it.

When we talk about the innate quantum noise of the light itself, what is important is not energy per unit of area, but energy per sensor (for photographers) or energy per pixel (for pixel-peepers).

The result of full well capacity being proportional to the sensor area is that the modern sensors, irrespective of their size, have their native ISO (as measured by their full well capacity) slightly below ISO 100. Which means that we cannot always compensate higher amounts of light noise of the smaller sensor by increasing the exposure. Once we pass the exposure equivalent of ISO 100, we will start to lose highlights.
 

aj1575

EOS RP
Dec 15, 2010
202
0
If the aperture is set to F5.6, it is F5.6 regardless of what lens or focal length is being used.

I'm not sure what that means, but it isn't accurate. The appropriate formula is:

Focal length / F-stop = Size of the Entrance Pupil in millimeters

You start with the focal length and F-stop, and you derive the size of the entrance pupil in millimeters.
Sorry for mixing up the entrance pupil and the aperture (I have to translate everything).

I think the interesting point from an engineering standpoint is, that you actually need to start with the focal length and the entrance pupil, because they are given by physics; that's what you see when you draw the path of the light. The f-number is just a number, that can be calculated with those two physical properties of a lens. (you can't measure the entrance pupil on a lens, but you measure it when you draw the path of the light).

This means the f-number in itself is not a given it is determined by the focal length and the entrance pupil, like the acceleration of a car, that is determined by its engines torque and the cars weight.

So when we are talking about f-number, we actually always talk also about focal length and entrance pupil, because the f-number is defined by them.
 

aj1575

EOS RP
Dec 15, 2010
202
0
I think the disconnect may be because of the mixing and matching of terms from two different areas. There's the "photography" terms that we're all used to (and most just use only those) and then there are the "engineering" and "physical terms." They're not easily mixed and matched. For example,
I think I have found the problem. When we look back when we were still using film, the ISO showed us how sensitiv the film was, no matter how big the area of the film was; the same amount of light on a certain area gave the same exposure, and to expose a bigger film you needed more light.
With digital, the situation changed. Basically the amount of light on the whole sensor is relative to the quality. (I know that is not entirely correct, but close enough). The same amount of photons on a 20MP APS-C sensor, gives about the same IQ as on a 20MP FF sensor. Usually FF sensors will see more light, so they tend to have better IQ. Fewer photons per pixel mean more noise. The ISO on a digital camera is just a gain in the calculation of the data.

Conclusion, you were correct with the ISO, aperture, shutter speed numbers. But in terms of pure IQ (same amount of photons an a pixel), my view was no wrong (since the ISO on digital camera are not what they used to be on film).