Well, there you go. What character did you mean to place between the quote marks - * ? ;-)
--
Its RKM
-
Welcome to the new forums! Please read this first. For known issues we are working to resolve, click here.
Article about light loss on various sensors
- Thread starter Horshack
- Start date
--
Its RKM
But were they all measured the same way?Nice test. So we know have "independent" numbers for the 7D, 5D and 5D mk II.
I was making a, perhaps invalid, assumption that since the gain they show for the 50D (and the Nikon D200) is a similar magnitude to the claimed sensitivity loss that this would also be true for the 5D. They show a loss of 0.25EV at f/1.4 for the 5D, whilst I only calculate a gain of 0.060EV is applied - more than a factor of 4 lower.
Expressed in the same terms (EV) as the 3rd chart on the LL article, the gain of the EOS 5D is much lower than any of the cameras shown, and also much lower than the "T-stop loss" they claim for the sensor.
I don't find it hard to believe that at all. Its a very common misconception that the light delivered by the lens is inversely proportional to the square of f/#, and the lens manufaccturers play along, marking their lenses as f/1.2 when they are really delivering only f/1.3's worth of light.
Also, these "T-stop losses" are similar to what we accepted without question for over a century with film. Film is a near lambertian surface, very diffuse, so its absorbtion, and hence sensitivity, varies as a cos^2 law as a function of incident angle. Similarly, the apparent area of the lens reduces as cos^2, giving an overall cos^4 fall off. So, going from f/2 to f/1.4 may seem to be a full stop increase, film would only receive 0.75 stops more light - doubling the shutter speed to compensate for the increased lens stop would result in under-exposure by a quarter stop. From f/1.4 to f/1.2 you expect an extra half stop, but film would only deliver an extra tenth of a stop!
Compared to that baseline, all these microlensed digital cameras aren't doing too bad at all!
--
Its RKM
The "stack" isn't the "well", and the paper doesn't suggest it is.I said "deeper wells relative to surface", which means either that they are effectively deeper, or just that theyir surfaces are smaller, and therefre their relative depth is bigger. If we really want to be pedantic, I should have said have smaller surface to stack height ratio, but I was aiming for a more intuitive explanation.
Anyway in the article they suggest that usually what happens when technology advances is that stack height increases (the wells get deeper).
--
Its RKM
Yes - however there is no indication of it being digitally compensated for in RAW.
Once you are in the macro region the dependence of exposure on focus distance becomes significant - certainly a lot more significant than the trivial changes being discussed here.
For example, a 100mm lens focussing on an object at 200mm distance is 2 stops slower than focussing on an object at infinity. (Image distance from lens' principle point is 100mm for infinity focus, and 200mm for 200mm focus - twice as far, so a factor of 4x less light on the image, ie. 2 stops.).
Remember extension tube corrections?
--
Its RKM
David J. Littleboy
Senior Member
Yes, I'm aware of all that stuff (which is there because of simple lens geometry); been using ext. tubes since the early 70s. (OK, correction: I used ext. tubes in the 70s; took a long break from photography, and nowadays they come built in to the macro lenses (or the body if I hadn't wimped out with a Mamiya 7 instead of an RZ67).)
But what I was asking was: is the exprosure different for an in-focus image vs. an out of focus one? And how would an individual pixel tell if it was seeing an in-focus point with a bad-bokeh halo (which many fast lenses don't have) or anything else for that matter.
Whatever, the paper still reeks of batguano.
--
David J. Littleboy
Tokyo, Japan
Bogdan_M
Senior Member
You tell me, you are certainly in a better position to discern that. I think John Sheehy's method of apturing black frames is slightly more reliable, but my knowledge in the field is limited.But were they all measured the same way?Nice test. So we know have "independent" numbers for the 7D, 5D and 5D mk II
IMO, your assumption is indeed invalid.
The LL post claims that the sensitivity losses are very different from sensor to sensor. It also claims that the manufacturers apply gain to make up for that. So the applied gain must be inversely proportional to the light loss due to pixel vignetting.
To elaborate a bit, the LL article presents the light loss for the 550D at f 1.4 as 0.63 stops (second graph). The manufacturer is applying a gain of about 0.35 stops to compensate the light loss due to pixel vignetting (third graph, the red line). The rest of 0.28 stops is the loss due to absorbtion/reflection by lens elements (I assume) and therefore is constant at all apêrtures, no need to make up for it.
The loss of light for the 5D, according to the same graph, is only 0.35 stops. So by the same reasoning, the gain applied to the 5D would be about 0.7 stops. Of course the graphs aren't perfectly readable, so I may be wrong on the exact numbrs, but I don't think I am wrong on the overall tendency.
I think it is pretty in line with your test results.
Marking them, I understand, it's marketing after all. But actually falling for it, is a different measure. They must have a few good engineers in their payroll, and maybe a few Ph D's too. I would be surprised if everybody would be into the "common misconception". But this is less on topic.calculate a gain of 0.060EV is applied - more than a factor of 4 lower.
Expressed in the same terms (EV) as the 3rd chart on the LL article, the gain of the EOS 5D is much lower than any of the cameras shown, and also much lower than the "T-stop loss" they claim for the sensor.
I don't find it hard to believe that at all. Its a very common misconception that the light delivered by the lens is inversely proportional to the square of f/#, and the lens manufaccturers play along, marking their lenses as f/1.2 when they are really delivering only f/1.3's worth of light.
--
http://www.flickr.com/photos/bogdanmoisuc/
No "must" about it! That is an even more precise assumption than the one I made - I just reckonned that the gain would be a similar magnitude to the claimed light loss, whilst you are suggesting it must be the inverse. If my assumption ins invalid them yours certainly is.
No, its more than an order of magnitude higher than I measure from the RAW data, as shown below: at f/1.4 the gain applied is only 0.06EV, not 0.7EV.
--
Its RKM
YES ;-)
For exactly the same reason. Here's a simple test you can try yourself. Put a 50/1.4 lens on your camera on a tripod and point it at a uniform scene - a white wall or even a grey card - at the minimum focus distance (0.45m). Set maximum aperture (f/1.4) and switch the lens to manual focus. Then rack the focus from infinity to 0.45m. The exposure is 1/3rd of a stop less at infinity focus (out of focus) than it is at 0.45m (in focus).
The pixel doesn't have to. All it is seeing is less (or more) light from the lens, depending on whether the out of focus is due to the principle plane of the lens being further (or closer) to the image plane than ideal focus. The pixel (and TTL light meter) are just responding to the amount of light passed by the lens, which does change as a function of focus.
In the case of the lens focus being fixed and the subject being moved from out of focus in front, through focus to out of focus behind, the exposure also changes. Two effects come into play here:
1. Inverse square law - as the subject moves further away less of the light it reflects is captured by the lens aperture (the lens aperture subtends a smaller fraction of the total angle that the light is reflected into)
2. Blurring - the light captured by the lens is focussed onto fewer pixels when in focus than when out, resulting in a higher light density in focus.
In most cases, other objects in the scene compensate for both of these effects - as the subject moves further away, other objects come into view, adding their light; the light density lost when moving out of focus is added to that of other objects in the scene. However, if you try this with a single lit object in an otherwise black background then you will see both effects in action and the exposure will indeed change.
--
Its RKM
oscarvdvelde
Senior Member
Quote from the site:
" 2. If the camera is automatically going to increase the ISO due to a significant light loss at the sensor, does it make sense to buy bigger, heavier and much more expensive large aperture lenses? "
Larger aperture lenses probably still make sense, because at a given f-stop (say f/2.0) these lenses tend to show less optical vignetting than smaller base-aperture lenses. But for crop-sensor cameras the digital compensation is almost a full stop and vignetting is less of a problem, so in that case this question becomes more relevant.
If marginal light rays were the problem, it is remarkable the darkening is worse for APS-C cameras than full frame, since the angle of light from the lens axis to pixels near the sensor edges should not be as problematic as for a full frame sensor! So Luminous Landscape has a wrong explanation here: the problem has probably nothing to do with marginal light rays, but instead with sensor vs readout electonics fill factor. Then we see that later models with microlenses keep the light level approximately equal while pixel size is reduced.
" In fact, is not even clear that large aperture lenses will deliver a shallower depth of field as intended. The DxO measurements to date prove that the marginal light rays just don’t hit the sensor. The point regarding depth of field is that these rays are also responsible for a larger blur spot when out of focus. If they are lost, they not only don’t contribute to the light intensity at the sensor, but they also don’t blur the out of focus planes as much as you would expect at wide apertures. "
Since marginal light rays are most likely not the problem, I doubt this statement is valid. Anyway, this could be easily tested. Just shoot the same wide aperture lens open on a 5D, then a 5DII or future 1DsIV and compare the bokeh and depth of field. Or test between 40D and 7D.
Besides, in order to profit from the shallow depth of field and bokeh of such lenses, one should use a focusing screen made for manual focusing. The standard ones are bright and do not project the light rays effectively enough on a plane. The eye can compensate its own focus through it, whereas it should be focused on the screen only. I was once surprised to see photos I carefully focused with a manual focus/aperture 50mm f/1.4 to be out of focus and with a totally different depth of field than in the viewfinder... the ground glass simply did not allow to visualize this.
--
Weather & Photography
http://www.lightningwizard.com
Steen Bay
Veteran Member
The graphs show the light loss in the center of the image/sensor.
zerozeronine
Leading Member
Here's what I got:
85mm f/1.2L II on a 20D.
Av mode, ISO 100. Pressed the shutter button halfway and recorded the suggested shutter speed. Then rotated the lens a bit to disconnect it from the mount and did the same:
f/5.6 0.2s (lens connected to the mount) becomes 1/30s (lens disconnected).
f/2.8 1/8s (connected to the mount) becomes 1/30s (disconnected).
f/2 1/15s (connected to the mount) becomes 1/30s (disconnected).
f/1.2 1/40s (connected to the mount) becomes 1/30s (disconnected).
Seems like it just wants to go to 1/30s whenever you disconnect it.
But then I tried it on a darker scene, and the value it becomes when disconnected is slower (1/20s).
f/5.6 0.8s (lens connected to the mount) becomes 1/20s (lens disconnected).
f/2.8 5s (connected to the mount) becomes 1/20s (disconnected).
f/2 1/10s (connected to the mount) becomes 1/20s (disconnected).
f/1.2 1/25s (connected to the mount) becomes 1/20s (disconnected).
So I don't think this test proves anything. OTOH, I'm still not sure what's going on here with the "consistent" disconnected value changing with different scenes... I suppose all bets are off if the lens is disconnected.
Kaz
85mm f/1.2L II on a 20D.
Av mode, ISO 100. Pressed the shutter button halfway and recorded the suggested shutter speed. Then rotated the lens a bit to disconnect it from the mount and did the same:
f/5.6 0.2s (lens connected to the mount) becomes 1/30s (lens disconnected).
f/2.8 1/8s (connected to the mount) becomes 1/30s (disconnected).
f/2 1/15s (connected to the mount) becomes 1/30s (disconnected).
f/1.2 1/40s (connected to the mount) becomes 1/30s (disconnected).
Seems like it just wants to go to 1/30s whenever you disconnect it.
But then I tried it on a darker scene, and the value it becomes when disconnected is slower (1/20s).
f/5.6 0.8s (lens connected to the mount) becomes 1/20s (lens disconnected).
f/2.8 5s (connected to the mount) becomes 1/20s (disconnected).
f/2 1/10s (connected to the mount) becomes 1/20s (disconnected).
f/1.2 1/25s (connected to the mount) becomes 1/20s (disconnected).
So I don't think this test proves anything. OTOH, I'm still not sure what's going on here with the "consistent" disconnected value changing with different scenes... I suppose all bets are off if the lens is disconnected.
Kaz
When you disconnect an EF lens from the sensor it operates at full aperture only, it cannot stop down. When the lens is connected the camera uses full aperture metering. However, when the lens disconnects the camera switches to stop down metering but only at that single, fully open aperture.
Consequently, of all of the measurements that you made above, only two are relevant:
Which works out at a loss of 0.42EV when disconnected (closest step to 1/3 stop)
Which works out at a loss of 0.32EV when disconnected (closest step to 1/3 stop)
Looks fairly consistent within the range of shutter speeds available, with a gain of around 1/3EV. The rest of your measurements are useless for comparison purposes.
--
Its RKM
zerozeronine
Leading Member
Thanks for the explanation!
zerozeronine
Leading Member
All of this would mean that the 5D should have less vignetting with the same lens than the 5D II, right?
Has anyone tried that?
Has someone also compared the bokeh quality between a 5D II and 5D?
According to all this, it would seem that the 5D II should have better bokeh (i.e., faster falloff at the edges) at the expense of more vignetting. Interesting "tradeoff."
I hope we'll get more transparency about these things from the companies in the upcoming year.
Kaz
Has anyone tried that?
Has someone also compared the bokeh quality between a 5D II and 5D?
According to all this, it would seem that the 5D II should have better bokeh (i.e., faster falloff at the edges) at the expense of more vignetting. Interesting "tradeoff."
I hope we'll get more transparency about these things from the companies in the upcoming year.
Kaz
Similar threads
Keyboard shortcuts
- f
- Forum