Crop sensor effect on apertures

[Peter 13]

[...] but my statement was that croped body differs from ff body only in FOV.

And you are wrong. First, bodies do not have FOV, they need lenses for that. Second, no matter how hard you try to spin the discussion, comparing at the same size is the only sane thing to do. Let's say that you found out that Joe, a.k.a. GB, bought a 16" laptop to view his great 5D images in full screen mode. Are you going to run to BB to buy a 10" notebook because, after all, you own a crop body, and you should view your images on screen that is 1.6 times smaller? When he prints 4x6, do you print smaller?

There is one essential difference between FF and crop bodies - what you (Joe, actually ) can do with the 50/1.2 on FF, no crop body can do, with any existing lens in the world. Not even with a hypothetical one (the mirror box is an obstacle, and the image would be terribly soft). You are right in one thing only - if you view Joe's images in thumbnail size, then - yes, you can take such images with your cell phone.

 

[Great Bustard]

For example, consider the following two scenarios:

7D: 50mm f/2.8 1/100
5D: 80mm f/4.5 1/100

I'm sure you'll agree that, from the same position (persepctive), they'll both result in the same framing (50mm x 1.6 = 80mm). I'm sure you'll also agree that the aperture diameters are the same (50mm / 2.8 = 80mm / 4.5 = 17.9mm).

That's right.

Excellent.

Thus, with the same shutter speed, the same amount of light ( the same number of photons) will fall on each sensor.

Nope. This is where you get confused. You assume that if 100 photons enter each lens, all 100 photons will fall on each sensor. This is incorrect.

That's where you're wrong. Same scene, same distance (perspective), same framing, same aperture diameter, same shutter speed means same number of photons fall on the sensor.

Until you understand that point, you won't understand the rest.

Like I pointed out - please verify your assumption that if 100 photons enter a lens, all 100 make it to a sensor. This is very important.

It's not an assumption, it's a fact. And yes, it's not only "very important" -- it's central to the concept of Equivalence.

And I don't mean (minor) light losses here caused by the optics.

Sure, sure.

Fundametally, make sure you understand very well how light travels through a shorter vs longer lens lens - and how that affects photon counts reaching the sensor.

You read my mind.

 

[x-vision]

Same scene, same distance (perspective), same framing, same aperture diameter, same shutter speed means same number of photons fall on the sensor .

OK. I guess we have to agree to disagree here.

 

[carlk]

Hmmm... I don't think FOV have any relationship with output size.

Lens, or angle of view of the lens, does decide the FOV covered by the image circle but sensor size, which covers only a portiong of the image circle will decide the FOV of the final image.

iljai, Lee and few others (including myself) are correct about the format and FOV. You need to get that part straight before you can understand how DOF was affected by the format size.

[...] but my statement was that croped body differs from ff body only in FOV.

And you are wrong. First, bodies do not have FOV, they need lenses for that. Second, no matter how hard you try to spin the discussion, comparing at the same size is the only sane thing to do. Let's say that you found out that Joe, a.k.a. GB, bought a 16" laptop to view his great 5D images in full screen mode. Are you going to run to BB to buy a 10" notebook because, after all, you own a crop body, and you should view your images on screen that is 1.6 times smaller? When he prints 4x6, do you print smaller?

There is one essential difference between FF and crop bodies - what you (Joe, actually ) can do with the 50/1.2 on FF, no crop body can do, with any existing lens in the world. Not even with a hypothetical one (the mirror box is an obstacle, and the image would be terribly soft). You are right in one thing only - if you view Joe's images in thumbnail size, then - yes, you can take such images with your cell phone.

 

[Great Bustard]

Same scene, same distance (perspective), same framing, same aperture diameter, same shutter speed means same number of photons fall on the sensor .

OK. I guess we have to agree to disagree here.

Well, that is one option. The other is to understand why what I'm saying is a fact. Did you think it was just a coincidence that sensors with the same efficiency had the same noise for Equivalent images? For example:

http://ezstrobesphoto.blogspot.com/2009/01/more-d300-and-d700-equivalent-images_25.html

Perhaps you don't understand how noise relates to the total amount of light and the sensor efficiency? Fear not, it's all here:

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

 

[dsjtecserv]

Same scene, same distance (perspective), same framing, same aperture diameter, same shutter speed means same number of photons fall on the sensor .

OK. I guess we have to agree to disagree here.

I'm honestly curious: What percentage of the 100 photos do you suppose do fall on the sensor? And where to the truant photons go? There must be some some explanation for that, especially if you are suggesting that the answer is something other than 100.

Dave
--
http://www.pbase.com/dsjtecserv

 

[x-vision]

Perhaps you don't understand how noise relates to the total amount of light and the sensor efficiency? Fear not, it's all here:

You have it all figured out, don't you ().

Like I said, clarify for yourself how the focal length factors in the photon counts. This will shed some light (hehe) on the relationship between light intensity and amount of light.

Btw, what if someone that you really trust tells you that amount of light and light density/intensity are in fact the same thing. You've been assuming all along that they are different. But what if they are the same?

That was it for me on this thread.

 

[Great Bustard]

Perhaps you don't understand how noise relates to the total amount of light and the sensor efficiency? Fear not, it's all here:

You have it all figured out, don't you ().

Yeah, I do, actually. The thing is, I'm not alone, and a lot of people in this thread have been trying to explain it to you. I'm surprised that Lee Jay (ljfinger) didn't make an effort, as he's the one who taught me, and he's in this thread.

Like I said, clarify for yourself how the focal length factors in the photon counts.

The focal length, along with the sensor size, merely determines how much of the scene is captured.

This will shed some light (hehe) on the relationship between light intensity and amount of light.

It's just part of the equation, as I've told you many times. We also need to know the aperture diameter and the shutter speed. Well, we need to know the scene luminance as well, but that's only necessary for an absolute count of the total light -- not necessary in a comparison between formats since we are talking about the same scene.

Btw, what if someone that you really trust tells you that amount of light and light density/intensity are in fact the same thing.

Then they would be wrong, and I would explain it to them. That's just such a silly thing to say. It's like saying the density of water and the amount of water are the same thing.

You've been assuming all along that they are different. But what if they are the same?

What if Santa Claus were real? What if this were all a dream? What if Jessica Alba returned my emails?

That was it for me on this thread.

Well, sure -- the thread is short now, anyway. But, seriously, do yourself a favor and learn the difference between the total amount of light and the density of light (exposure). I'll give you a head start -- it's spelled out, in detail , here:

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

I believe you know the link. Here's a quick fact: there's a difference between photons and photons / mm².

 

[Lee Jay]

Perhaps you don't understand how noise relates to the total amount of light and the sensor efficiency? Fear not, it's all here:

You have it all figured out, don't you ().

Yeah, I do, actually. The thing is, I'm not alone, and a lot of people in this thread have been trying to explain it to you. I'm surprised that Lee Jay (ljfinger) didn't make an effort, as he's the one who taught me, and he's in this thread.

You don't need me, you're doing just fine.

--
Lee Jay
(see profile for equipment)