I have a question. Consider two lenses with same f-ratio but one has a larger focal length than the other (say 200 mm vs 50mm). Which lens would give better image quality when shooting portraits in low light considering the same composition, same lighting conditions and same f-ratio?
Not from the front pupil to the sensor, but from the scene to the entrance pupil. All the light passing through the entrance pupil is projected on the sensor. Explained in more detail here.
Re: Longer focal length or shorter for good IQ for low light portrait photography?
Get another lens so you can have Long FL (over 100mm) and another one of short FL (about 50mm). Ensure you have constant light source.
with both lenses. Repeat two or three more time to get multiple samples. alternating lens for each pose.
Try to maintain perspective of your model and the size of the photo. This means being farther with longer lens.
I did not understand your example. The aperture for the 100 mm lens will be larger than the 50 mm lens for the same f/stop value. Then, wouldn't the sensor collect more light for the 100 mm lens?
And it has to travel farther from the front pupil to the sensor, so your inverse square law comes into effect, and is exactly cancelled by the greater area.
Not from the front pupil to the sensor, but from the scene to the entrance pupil. All the light passing through the entrance pupil is projected on the sensor. Explained in more detail here.
Well, strictly speaking all the light collected by the entrance pupil is delivered to the exit pupil.
All the light from the exit pupil is not necessarily delivered to the sensor.
Generally until you stop down some light is lost (fall off) towards the corners.
So yes, on axis, but not necessarily over the entire sensor.
Hi. Thanks for the advice. I am planning to do it over the weekend. As you said, I have to constrain the framing, lighting etc so that the results aren't influenced by them.
Hi. Thanks for the advice. I am planning to do it over the weekend. As you said, I have to constrain the framing, lighting etc so that the results aren't influenced by them.
While there are pros and cons for Jpeg and raw, make it easy on yourself by adhering to either one throughout your entire experiment.
Use the "Off the camera" to eliminate the influence of your computer post processor
Other than these two, asked yourself which factor (or factors) affect the validity of the comparison of Image Quality.
a) f/stop
b) shutter speed
c) ISO
d) light intensity
e) placement of light source or sources
f) perspective (placement of camera and subject)
You decide which of this factors need to be constant and which may (or need) be altered. to have valid comparison.
I did not understand your example. The aperture for the 100 mm lens will be larger than the 50 mm lens for the same f/stop value. Then, wouldn't the sensor collect more light for the 100 mm lens?
And it has to travel farther from the front pupil to the sensor, so your inverse square law comes into effect, and is exactly cancelled by the greater area.
Not from the front pupil to the sensor, but from the scene to the entrance pupil. All the light passing through the entrance pupil is projected on the sensor. Explained in more detail here.
Well, strictly speaking all the light collected by the entrance pupil is delivered to the exit pupil.
All the light from the exit pupil is not necessarily delivered to the sensor.
Generally until you stop down some light is lost (fall off) towards the corners.
So yes, on axis, but not necessarily over the entire sensor.
...that is simply one of the better corrections to something I've said that I can remember -- thanks! If I may ask, does this, then, have a lot to do with why fast lenses have so much vignetting wide open?
I did not understand your example. The aperture for the 100 mm lens will be larger than the 50 mm lens for the same f/stop value. Then, wouldn't the sensor collect more light for the 100 mm lens?
And it has to travel farther from the front pupil to the sensor, so your inverse square law comes into effect, and is exactly cancelled by the greater area.
Not from the front pupil to the sensor, but from the scene to the entrance pupil. All the light passing through the entrance pupil is projected on the sensor. Explained in more detail here.
Well, strictly speaking all the light collected by the entrance pupil is delivered to the exit pupil.
All the light from the exit pupil is not necessarily delivered to the sensor.
Generally until you stop down some light is lost (fall off) towards the corners.
So yes, on axis, but not necessarily over the entire sensor.
Hi Bill. In the ray tracing images how does one understand light fall off? I am sorry if my question sounds dumb. Also, is the right most thin rectangle (outside the lens) the sensor?
I did not understand your example. The aperture for the 100 mm lens will be larger than the 50 mm lens for the same f/stop value. Then, wouldn't the sensor collect more light for the 100 mm lens?
And it has to travel farther from the front pupil to the sensor, so your inverse square law comes into effect, and is exactly cancelled by the greater area.
Not from the front pupil to the sensor, but from the scene to the entrance pupil. All the light passing through the entrance pupil is projected on the sensor. Explained in more detail here.
Well, strictly speaking all the light collected by the entrance pupil is delivered to the exit pupil.
All the light from the exit pupil is not necessarily delivered to the sensor.
Generally until you stop down some light is lost (fall off) towards the corners.
So yes, on axis, but not necessarily over the entire sensor.
The black tick marks near the middle are the aperture.
Notice how they limit the blue on-axis rays.
Notice whether the green off-axis rays fill the aperture; f/1.8 no, f/3.3 yes.
Good question. The gray rectangle to the left of the Image plane (labeled 'I') is the cover glass. The cover glass is normally included as part of the optical prescription for a lens that's designed for a mirrorless camera.
--
Bill ( Your trusted source for independent sensor data at PhotonsToPhotos )
I have a question. Consider two lenses with same f-ratio but one has a larger focal length than the other (say 200 mm vs 50mm). Which lens would give better image quality when shooting portraits in low light considering the same composition, same lighting conditions and same f-ratio?
to get similar composition you will need to stand further back with the longer fl.
The further away, the less light will hit the sensor. With the inverse-square law', which applies to light, that means if you stand 2x further back the sensor will only get 1/4 as much light
since you mentioned 'low light' that could mean a more blurred and/or noisy image, since to get the same IQ you will need to either shoot at a slower shutter speed, or increase the ISO which will result in more noise, esp in the shadows
I was thinking the same too. I initially thought the longer focal length lens will give lower noise as the exit-pupil diameter is larger. However, if the same composition has to be maintained then one has to stand farther away for this lens, and, the light reaching the sensor will be reduced in intensity due to inverse square law (?). So, I do not know if there is a generic answer.
Let's consider a 50mm lens at f/2 (aperture diameter = 50mm / 2 = 25mm) and a 100mm lens at f/2 (aperture diameter = 100mm / 2 = 50mm). Since the aperture diameter of the 100mm lens at f/2 is twice as large as the aperture diameter of the 50mm lens at f/2, it will have 4x the area, and thus pass 4x as much light through the lens onto the sensor.
However, to maintain the same framing, you would be twice as far back with the 100mm lens, so the intensity of the light reaching the aperture will be 1/4 as much. Thus, 1/4 the intensity of light passing through an aperture with 4x the area results in the same total amount of light being projected on the sensor.
In my experience, there's only one kind of stupid question: when a question is asked, an answer is given, the answer is ignored, and the exact same question is asked again. That second question is a stupid question.
But when one doesn't understand something, asks, gets an answer, understands that answer and learns something (as opposed to mindlessly accepting the answer), then the question was not only not stupid, but the very best kind of question!
I think my best days of math are way in my past, (lol)
but I gave the math a try...if I screwed it all up please explain my errors.
So the big question is, if we keep the framing for both lenses (50ml and 100ml lenses)
the same, we would need to 'zoom with our feet' moving the 100ml lenses further from
the target than the 50ml lens. If we do that, will the light falling on the front of
the 100ml lens be the same as the light falling on the 50ml lens?
(I phrased the problem this way to bypass practicles like F and the T-stop, since each lense
could have different light-transmitting capabilities, and for this we wouldn't need to
consider the F-stop, which might need to be different for artistic purposes, such as depth of field)
First, let's get data for real-life lenses. I selected the following data for the canon ef lenses:
50ml:
50ml lens data from web:
angle of view, diagonal: 46 degrees
tan(46 degrees): 1.03553031379
100ml lens data from web:
angle of view, diagonal: 23.4 degrees
tan(23.4 degrees): 0.43273864224
Now let's remember out trig formula:
Trigonometry. ... In a right triangle, the tangent of an angle
is the length of the opposite side divided by the length of the
adjacent side
Let's give our model some more data, specifying the target frame size:
So lets say the length of the target is 2 meters (about 6 feet, say for a head-to-foot-shot).
And since we are using trig math we want to use 1/2 of the target length as the opposite
side of our triangle...
our Tr (radius of target) is 1 meter. This will be the same for both lenses, since we want
the same framing.
let lenXXmm be the length from the target to lens XX so len50mm is the length from the target to the front of the 50mm lens
len100mm is the length from the target to the front of the 100mm lens
so we have the following formulas:
tan(angle of view) = opposite / adjacent which would be:
rewrite the math to make solving it easier:
len50mm = Tr / tan(46 degrees) # 1.03553031379
len100mm = Tr / tan(23.4 degrees) # 0.43273864224
Let's use the actual values for the tan(lenXXXmm) to make it cleaner:
len50mm = Tr / 1.03553031379
len100mm = Tr / 0.43273864224
since we are using Tr value of 1 meter that simplifies to:
len50mm = 1/1.03553031379 => 0.9656887748
len100mm = 1/0.43273864224 => 2.31086365392
or we would need to stand approx 2.4x (2.39296936469) as far with the 100mm lens to get the same framing as the 50mm lens
so know that we know the radius of the target (Tr) and the length each lens is from the target (to get the same framing) let's calculate how much less light will reach the 100ml lens, compared to the 50ml lens:
using the inverse square law we first square the length ratio:
2.39296936469 ** 2 => 5.34009082701
then, using the inverse square law, we would have light level at:
1/5.34009082701
=> 0.18726273248
so the front of the 100mm lens would need to be 5.34009082724 times the area of the 50mm lens
area of a cirlce is pi * r ** 2
a50ml = pi * r50ml ** 2
a100ml = pi * r100ml ** 2
now let's check if the front of each lens gets the same amount of light from the target.
We do this by assuming it's true, calculating the values, and checking if the results match
the physical dimensions of the lenses.
since a100mm = a50mm * 5.34009082724
pi * r100mm ** 2 = a50mm * 5.34009082724
substituting for a50ml (pi * r50ml ** 2):
pi * r100mm ** 2 = pi * r50mml ** 2 * 5.34009082724
(remove pi from both sides)
r100mm ** 2 = r50mm ** 2 * 5.34009082724
take the square root of both sides:
r100mm = sqrt(r50mm ** 2 * 5.34009082724)
factoring right side: (I think I remember this factoring right...lol it's been a looong time for me...)
r100mm = sqrt(r50mm ** 2) * sqrt(5.34009082724)
factoring by squaring both sides of the equation:
r100mm = r50mm * sqrt(5.34009082724)
using calculator:
r100mm = r50mm * 2.31086365397
so.... the front opening of the 100mm lens would need to be 2.3x the front opening of the 50ml lens
I don't have the data for the front openings of the lenses...
from the canon specs for the physical dimensions of each lens:
50mm Manufacturer Spec Size (DxL): 2.72 x 1.55
so, doing the math (I'm not sure but I think the D is the diameter we would want
but I did the math for both dimensions)
which gives us:
1.55 * 2.3 => 3.565
2.72 * 2.3 => 6.256
Neither seemed a good match for the 100ml dimensions...
100mm Manufacturer Spec Size (DxL): 3.06 x 4.84
So, in conclusion, the front of each lenses would receive different amounts of light...*sigh*
Or, my math could be wrong. It's been a long while since I did much math...or maybe my
dyslexia strikes again...or the actual size of the front of each lenses might be different
than what I pulled up from the data sheets (the actual front opening is never the exact
size of the outer diameter of the lens, right?)
Instead of using trigonometry, you can use simply the law of similar triangles.
Focal length divided by sensor width is equal to subject distance divided by subject width (or field of view)
Doubling the focal length requires doubling the subject distance in order to get the same field of view.
As f/stop is equal to the focal length divided by entrance pupil diameter, doubling the focal length means doubling the diameter, which quadruples the area of the entrance pupil. This lets in four times the amount of light, which exactly offsets the dimming of light from being twice as far away, due to the inverse square law.
I did not understand your example. The aperture for the 100 mm lens will be larger than the 50 mm lens for the same f/stop value. Then, wouldn't the sensor collect more light for the 100 mm lens?
And it has to travel farther from the front pupil to the sensor, so your inverse square law comes into effect, and is exactly cancelled by the greater area.
Not from the front pupil to the sensor, but from the scene to the entrance pupil. All the light passing through the entrance pupil is projected on the sensor. Explained in more detail here.
Well, strictly speaking all the light collected by the entrance pupil is delivered to the exit pupil.
All the light from the exit pupil is not necessarily delivered to the sensor.
Generally until you stop down some light is lost (fall off) towards the corners.
So yes, on axis, but not necessarily over the entire sensor.
The black tick marks near the middle are the aperture.
Notice how they limit the blue on-axis rays.
Notice whether the green off-axis rays fill the aperture; f/1.8 no, f/3.3 yes.
Good question. The gray rectangle to the left of the Image plane (labeled 'I') is the cover glass. The cover glass is normally included as part of the optical prescription for a lens that's designed for a mirrorless camera.
Hello Bill. Thank you very much for the explanation. Isn't aperture the diameter of the exit pupil? Also, what are the lens elements to the right of I-I? Shouldn't the exit pupil shrink as we increase the f-stop? I am again sorry if the question sounds silly.
Natural and all light is affected by the inverse square law.
The beauty of F/# is that, with longer focal lengths of the same F/#, the the lens area goes as the square of the focal length. Thus as you get further away the light goes down by the inverse square and the collection area goes up to match.
If we look a bit deeper longer lenses have a small advantage. The longer lenses typically have fewer elements, and thus very slightly better transmission so with the same f/# a longer lens will probably get more light to the sensor. With multi layer coated lenses made after about 1970 this is at most a few per cent.
No. The entrance and exit pupils are virtual images of the (physical) aperture as viewed from the front and rear of the lens respectively.
F-number (some people call this aperture) is the ratio of focal length to the entrance pupil diameter.
Here's a new pair of images showing the entrance and exit pupils.
The entrance pupil is the blue line labeled P and the exit pupil is the red line labeled P'
BTW. You could do this yourself using the buttons in the Optical Bench.
You can see the lens I'm using as an example here .
[ATTACH alt="50mm lens at f/1.8 best viewed "original size" "]2557218[/ATTACH]
50mm lens at f/1.8 best viewed "original size"
Same lens at f/3.3 best viewed "original size"
Note the pupils do get smaller.
--
Bill ( Your trusted source for independent sensor data at PhotonsToPhotos )
I have a question. Consider two lenses with same f-ratio but one has a larger focal length than the other (say 200 mm vs 50mm). Which lens would give better image quality when shooting portraits in low light considering the same composition, same lighting conditions and same f-ratio?
DAYLIGHT If you move your camera 500 feet closer to the castle at noon, the amount of changed distance from the light source (the sun or clouds at 7000 feet, is tiny.
IN THE LIVING ROOM with an artificial light. (flash or continuous) If the light is 8 feet from the painting on the wall, and you move the light to 4 feet from the painting on the wall, the painting will look brighter to your eye.
If you take a meter reading, the light will be two stops brighter when four feet from the painting, compared to two stops dimmer when the light is moved back to eight feet.
(move the light to 5.6 feet and there will be one stop more or one stop less, compared to 4 feet or 8 feet.
BAK
