Longer focal length or shorter for good IQ for low light portrait photography?

foot said:

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:
https://www.the-digital-picture.com/Reviews/Lens-Specifications.aspx?Lens=989
100ml:
https://www.the-digital-picture.com/Reviews/Lens-Specifications.aspx?Lens=674

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:

tan(46 degrees) = Tr / len50mm # 1.03553031379
tan(23.4 degrees) = Tr / len100mm # 0.43273864224

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?)

any thoughts?

Click to expand...

Moti said:

Suman Vajjala said:

Hi,

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?

Thanks and regards

Suman

Click to expand...

Why don't you just buy a decent flash and learn how to use it. You'll always get better quality than with any lens at low light.

moti

Click to expand...

foot said:

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:
https://www.the-digital-picture.com/Reviews/Lens-Specifications.aspx?Lens=989
100ml:
https://www.the-digital-picture.com/Reviews/Lens-Specifications.aspx?Lens=674

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:

tan(46 degrees) = Tr / len50mm # 1.03553031379
tan(23.4 degrees) = Tr / len100mm # 0.43273864224

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

Click to expand...

• 50mm lens: [diagonal] AOV = 46° on FF (this is actually a focal length of 51mm)
• 100mm lens: [diagonal] AOV = 23.4° on FF (this is actually a focal length of 104.5mm)

• 50mm (51mm) lens --> x = 10 / [2 · tan (46°/2)] = 11.8m.
• 100mm (104.5mm) lens --> x = 10 / [2 · tan (23.4°/2)] = 24.1m.

foot said:

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?)

any thoughts?

Click to expand...

J A C S said:

Why same f-stop?

Click to expand...

kiwi2 said:

Answer yourself this question. Do you only see a bubble of light around you with the outside world fading into darkens at further distances?

I know I don't.

Click to expand...

bclaff said:

Suman Vajjala said:

bclaff said:

...

Click to expand...

Hello Bill. Thank you very much for the explanation. Isn't aperture the diameter of the exit pupil?

Click to expand...

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.

bclaff said:

Also, what are the lens elements to the right of I-I?

Click to expand...

I don't know what you mean by I-I

Perhaps you're referring to the tick marks.

The first pair of tick marks is the aperture/diaphragm/iris

The second set is something called a flare stop.

bclaff said:

Shouldn't the exit pupil shrink as we increase the f-stop? I am again sorry if the question sounds silly.

Click to expand...

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 .


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.

Click to expand...

BAK said:

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.

Click to expand...

BAK said:

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.

Click to expand...

BAK said:

(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

Click to expand...

Dr JLW said:

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.

Click to expand...

Suman Vajjala said:

J A C S said:

Why same f-stop?

Click to expand...

It was to even out the comparison. I thought initially (now I know it is incorrect) that longer focal length, faster lenses will capture more light than a shorter one (keeping the same composition etc).

Regards

Suman

Click to expand...

Great Bustard said:

foot said:

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:
https://www.the-digital-picture.com/Reviews/Lens-Specifications.aspx?Lens=989
100ml:
https://www.the-digital-picture.com/Reviews/Lens-Specifications.aspx?Lens=674

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:

tan(46 degrees) = Tr / len50mm # 1.03553031379
tan(23.4 degrees) = Tr / len100mm # 0.43273864224

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

Click to expand...

First of all, similar triangles is the way, way, way easier way about things. Secondly, the focal length (and thus angle-of-view) is for infinity focus. At closer distances, the effective focal length (effective angle-of-view) changes, and then there's focus breathing to take into account, as well.

OK, that all said, let's do the math with your figures:

• 50mm lens: [diagonal] AOV = 46° on FF (this is actually a focal length of 51mm)
• 100mm lens: [diagonal] AOV = 23.4° on FF (this is actually a focal length of 104.5mm)

So, let's say we want the frame diagonal on the focal plane to be x. Then the distance we need to be from the center of the frame is given by tan (AOV/2) = (w/2) / x. Solving for x, we get x = w / [2 · tan (AOV/2)]. Working this out for a [diagonal] width on the focal plane of 10m for a:

• 50mm (51mm) lens --> x = 10 / [2 · tan (46°/2)] = 11.8m.
• 100mm (104.5mm) lens --> x = 10 / [2 · tan (23.4°/2)] = 24.1m.

This is right in line with the ratio of the focal lengths: 104.5mm / 51mm = 24.1 / 11.8 = 2.05. So, with that correction to the 2.4x figure that you got, the rest of the below:

Great Bustard said:

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?)

any thoughts?

Click to expand...

...will follow.

Click to expand...

Suman Vajjala said:

... I thought initially (now I know it is incorrect) that longer focal length, faster lenses will capture more light than a shorter one (keeping the same composition etc).

Click to expand...

Suman Vajjala said:

Moti said:

Suman Vajjala said:

Hi,

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?

Thanks and regards

Suman

Click to expand...

Why don't you just buy a decent flash and learn how to use it. You'll always get better quality than with any lens at low light.

moti

Click to expand...

Hi,

I have a flash and I use it sometimes. I asked the question as I (wrongly) assumed that longer focal length, faster lenses would be better than the shorter focal length lenses. But, I learned from the forum members that it isn't true.

Regards

Suman

Click to expand...

kiwi2 said:

foot said:

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:
https://www.the-digital-picture.com/Reviews/Lens-Specifications.aspx?Lens=989
100ml:
https://www.the-digital-picture.com/Reviews/Lens-Specifications.aspx?Lens=674

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:

tan(46 degrees) = Tr / len50mm # 1.03553031379
tan(23.4 degrees) = Tr / len100mm # 0.43273864224

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?)

any thoughts?

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Here's my thoughts. A simpler way to do the math...

www.dpreview.com/forums/post/62524790

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foot said:

foot said:

...

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And to calculate the 'light gathering area' we would use the formula for the area of a circle, which would be (aperture/2) ** 2 * pi

...

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bclaff said:

Suman Vajjala said:

... I thought initially (now I know it is incorrect) that longer focal length, faster lenses will capture more light than a shorter one (keeping the same composition etc).

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Light gathered is the product (multiplication) of the entrance pupil area and the solid angle. So, it's easy to think of it this way ...

Holding f# constant if you double the focal length you double the diameter of the entrance pupil (and get 4x the area).
But you also halve (1/2) the angle of view (and as it works out the solid angle is a quarter (1/4)).

So the 4x and the 1/4x cancel out.

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J A C S said:

Suman Vajjala said:

J A C S said:

Why same f-stop?

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It was to even out the comparison. I thought initially (now I know it is incorrect) that longer focal length, faster lenses will capture more light than a shorter one (keeping the same composition etc).

Regards

Suman

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The point of my remark is that shorter lenses are typically faster (say, 50/1.4 vs. 200/2.8), so one could exploit this, with all the implication about the DOF.

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Suman Vajjala said:

bclaff said:

Suman Vajjala said:

... I thought initially (now I know it is incorrect) that longer focal length, faster lenses will capture more light than a shorter one (keeping the same composition etc).

Click to expand...

Light gathered is the product (multiplication) of the entrance pupil area and the solid angle. So, it's easy to think of it this way ...

Holding f# constant if you double the focal length you double the diameter of the entrance pupil (and get 4x the area).
But you also halve (1/2) the angle of view (and as it works out the solid angle is a quarter (1/4)).

So the 4x and the 1/4x cancel out.

Click to expand...

Thanks, Bill. May I know why one uses solid angle instead of area?

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