Aperture fixed when 35mm lenses used on small CCD's??

I've often heard it said that a 100mm f/2.8 lens (for example) designed
for a 35mm camera if put on a smaller CCD sensor will become a 160mm
f/2.8 i.e. the 'effective' focal length gets multiplied by some factor
(1.6) in my example and the aperture remains constant. I'm not so
convinced the latter is true.

I make a few observations.

1) A 100mm f/2.8 lens put on a small digital sensor remains a 100mm
f/2.8 lens. The lens remains the same.

2) The smaller CCD sensor means the focal length (as compared to 35mm)
is longer, as everyone agrees.

3) The aperture whilst still f/2.8 is "effectively" larger, as much of
the light is thrown away, missing the sides of the sensor. So the
viewfinder will be darker than if fitted with a f/2.8 lens which filled
the sensor and no more.

At first the digital format would seem to allow long focal, fast
telephotos. i.e. my 70-200 f/2.8 would become a 112-320 f/2.8, which
would be a very nice fast lens indeed. But I'm not so sure the lens
would have the light gathering power of a real f/2.8 lens, but instead
be effectively an f3.5 (I think). I suspect if the focal length is
multipled by 1.6, the apeture will be multipled by sqrt(1.6), although I
might be wrong on the exact calculation.


PS,
Does anyone know if Nikon are developing a full frame (35mm) digital SLR
like Canon and Kodak?? It seems such a move would have a lot of
technical advantages (lower noise) and people with expensive 35mm lenses
would get the full benefit, and not throw much of the light away, which
is what I think would happen now.

Dave wrote in :

1) A 100mm f/2.8 lens put on a small digital sensor remains a 100mm
f/2.8 lens. The lens remains the same.

Yes

2) The smaller CCD sensor means the focal length (as compared to 35mm)
is longer, as everyone agrees.

That contradicts 1.

The smaller CCD equals cropping. Sometimes this is referred
to as a longer equivalent focal length. But it is still the same
focal length.

3) The aperture whilst still f/2.8 is "effectively" larger, as much of
the light is thrown away, missing the sides of the sensor. So the
viewfinder will be darker than if fitted with a f/2.8 lens which filled
the sensor and no more.

The aperture is still f/2.8.

But you are right with the finder being darker. Moreover,
if you want the same amount of pixels, the actual sensors
must be smaller, thus collecting less light, thus you cannot
have as high ISO - making the system less sensitive.


/Roland

Dave wrote in :

1) A 100mm f/2.8 lens put on a small digital sensor remains a 100mm
f/2.8 lens. The lens remains the same.

Yes

2) The smaller CCD sensor means the focal length (as compared to 35mm)
is longer, as everyone agrees.

That contradicts 1.

The smaller CCD equals cropping. Sometimes this is referred
to as a longer equivalent focal length. But it is still the same
focal length.

3) The aperture whilst still f/2.8 is "effectively" larger, as much of
the light is thrown away, missing the sides of the sensor. So the
viewfinder will be darker than if fitted with a f/2.8 lens which filled
the sensor and no more.

The aperture is still f/2.8.

But you are right with the finder being darker. Moreover,
if you want the same amount of pixels, the actual sensors
must be smaller, thus collecting less light, thus you cannot
have as high ISO - making the system less sensitive.


/Roland

lets put it this way. My Nikon 70-200 on my D70 remains at 2.8 all the way
through. My exposures show this to be true. If your theory is true then the
shutter has to lie or the daylight must intensify as I throw the switch!
"Dave" wrote in message ...
I've often heard it said that a 100mm f/2.8 lens (for example) designed
for a 35mm camera if put on a smaller CCD sensor will become a 160mm f/2.8
i.e. the 'effective' focal length gets multiplied by some factor (1.6) in
my example and the aperture remains constant. I'm not so convinced the
latter is true.

I make a few observations.

1) A 100mm f/2.8 lens put on a small digital sensor remains a 100mm f/2.8
lens. The lens remains the same.

2) The smaller CCD sensor means the focal length (as compared to 35mm) is
longer, as everyone agrees.

3) The aperture whilst still f/2.8 is "effectively" larger, as much of the
light is thrown away, missing the sides of the sensor. So the viewfinder
will be darker than if fitted with a f/2.8 lens which filled the sensor
and no more.

At first the digital format would seem to allow long focal, fast
telephotos. i.e. my 70-200 f/2.8 would become a 112-320 f/2.8, which would
be a very nice fast lens indeed. But I'm not so sure the lens would have
the light gathering power of a real f/2.8 lens, but instead be effectively
an f3.5 (I think). I suspect if the focal length is multipled by 1.6, the
apeture will be multipled by sqrt(1.6), although I might be wrong on the
exact calculation.


PS,
Does anyone know if Nikon are developing a full frame (35mm) digital SLR
like Canon and Kodak?? It seems such a move would have a lot of technical
advantages (lower noise) and people with expensive 35mm lenses would get
the full benefit, and not throw much of the light away, which is what I
think would happen now.

lets put it this way. My Nikon 70-200 on my D70 remains at 2.8 all the way
through. My exposures show this to be true. If your theory is true then the
shutter has to lie or the daylight must intensify as I throw the switch!
"Dave" wrote in message ...
I've often heard it said that a 100mm f/2.8 lens (for example) designed
for a 35mm camera if put on a smaller CCD sensor will become a 160mm f/2.8
i.e. the 'effective' focal length gets multiplied by some factor (1.6) in
my example and the aperture remains constant. I'm not so convinced the
latter is true.

I make a few observations.

1) A 100mm f/2.8 lens put on a small digital sensor remains a 100mm f/2.8
lens. The lens remains the same.

2) The smaller CCD sensor means the focal length (as compared to 35mm) is
longer, as everyone agrees.

3) The aperture whilst still f/2.8 is "effectively" larger, as much of the
light is thrown away, missing the sides of the sensor. So the viewfinder
will be darker than if fitted with a f/2.8 lens which filled the sensor
and no more.

At first the digital format would seem to allow long focal, fast
telephotos. i.e. my 70-200 f/2.8 would become a 112-320 f/2.8, which would
be a very nice fast lens indeed. But I'm not so sure the lens would have
the light gathering power of a real f/2.8 lens, but instead be effectively
an f3.5 (I think). I suspect if the focal length is multipled by 1.6, the
apeture will be multipled by sqrt(1.6), although I might be wrong on the
exact calculation.


PS,
Does anyone know if Nikon are developing a full frame (35mm) digital SLR
like Canon and Kodak?? It seems such a move would have a lot of technical
advantages (lower noise) and people with expensive 35mm lenses would get
the full benefit, and not throw much of the light away, which is what I
think would happen now.

Dave writes:

I've often heard it said that a 100mm f/2.8 lens (for example)
designed for a 35mm camera if put on a smaller CCD sensor will become
a 160mm f/2.8 i.e. the 'effective' focal length gets multiplied by
some factor (1.6) in my example and the aperture remains constant. I'm
not so convinced the latter is true.

I make a few observations.

1) A 100mm f/2.8 lens put on a small digital sensor remains a 100mm
f/2.8 lens. The lens remains the same.

2) The smaller CCD sensor means the focal length (as compared to 35mm)
is longer, as everyone agrees.

3) The aperture whilst still f/2.8 is "effectively" larger, as much of
the light is thrown away, missing the sides of the sensor. So the
viewfinder will be darker than if fitted with a f/2.8 lens which
filled the sensor and no more.

So if you crop the center out of a 35mm shot, does it get darker?

Really, the F number correlates to what irradiance will hit the image
plane for a given light field entering the lens. Irradiance is the
power density in watts per square meter; varying the number of square
meters you catch doesn't change this at all.

snip

--
-Stephen H. Westin
Any information or opinions in this message are mine: they do not
represent the position of Cornell University or any of its sponsors.

Dave writes:

I've often heard it said that a 100mm f/2.8 lens (for example)
designed for a 35mm camera if put on a smaller CCD sensor will become
a 160mm f/2.8 i.e. the 'effective' focal length gets multiplied by
some factor (1.6) in my example and the aperture remains constant. I'm
not so convinced the latter is true.

I make a few observations.

1) A 100mm f/2.8 lens put on a small digital sensor remains a 100mm
f/2.8 lens. The lens remains the same.

2) The smaller CCD sensor means the focal length (as compared to 35mm)
is longer, as everyone agrees.

3) The aperture whilst still f/2.8 is "effectively" larger, as much of
the light is thrown away, missing the sides of the sensor. So the
viewfinder will be darker than if fitted with a f/2.8 lens which
filled the sensor and no more.

So if you crop the center out of a 35mm shot, does it get darker?

Really, the F number correlates to what irradiance will hit the image
plane for a given light field entering the lens. Irradiance is the
power density in watts per square meter; varying the number of square
meters you catch doesn't change this at all.

snip

--
-Stephen H. Westin
Any information or opinions in this message are mine: they do not
represent the position of Cornell University or any of its sponsors.

wrote:

Dave writes:

I've often heard it said that a 100mm f/2.8 lens (for example)
designed for a 35mm camera if put on a smaller CCD sensor will become
a 160mm f/2.8 i.e. the 'effective' focal length gets multiplied by
some factor (1.6) in my example and the aperture remains constant. I'm
not so convinced the latter is true.

I make a few observations.

1) A 100mm f/2.8 lens put on a small digital sensor remains a 100mm
f/2.8 lens. The lens remains the same.

2) The smaller CCD sensor means the focal length (as compared to 35mm)
is longer, as everyone agrees.

3) The aperture whilst still f/2.8 is "effectively" larger, as much of
the light is thrown away, missing the sides of the sensor. So the
viewfinder will be darker than if fitted with a f/2.8 lens which
filled the sensor and no more.

without any factual basis,
flys in the face of physics and optics...


So if you crop the center out of a 35mm shot, does it get darker?

Really, the F number correlates to what irradiance will hit the image
plane for a given light field entering the lens. Irradiance is the
power density in watts per square meter; varying the number of square
meters you catch doesn't change this at all.

snip

--
-Stephen H. Westin
Any information or opinions in this message are mine: they do not
represent the position of Cornell University or any of its sponsors.

Dave writes:
I've often heard it said that a 100mm f/2.8 lens (for example) designed
for a 35mm camera if put on a smaller CCD sensor will become a 160mm
f/2.8 i.e. the 'effective' focal length gets multiplied by some factor
(1.6) in my example and the aperture remains constant. I'm not so
convinced the latter is true.

What actually happens is that the *field of view* is smaller, equivalent
to the FOV you'd get with a 160 mm lens on a full-frame camera.

1) A 100mm f/2.8 lens put on a small digital sensor remains a 100mm
f/2.8 lens. The lens remains the same.

Certainly

2) The smaller CCD sensor means the focal length (as compared to 35mm)
is longer, as everyone agrees.

No. The smaller sensor means that the field of view with the same lens
is narrower. This is sometimes *described* as having the same effect as
using a longer focal length on the full-frame camera.

3) The aperture whilst still f/2.8 is "effectively" larger, as much of
the light is thrown away, missing the sides of the sensor. So the
viewfinder will be darker than if fitted with a f/2.8 lens which filled
the sensor and no more.

This is nonsense. You should try tracing some ray diagrams through a
lens. A f/2.8 lens collects a certain amount of light from *a single
point* in the subject and focuses all that light at one point in the
image, if the subject is in focus. At the same time, and entirely in
parallel, the lens is also collecting light from all of the other points
in the subject, and transferring it to corresponding points in the
image.

Now, when you reduce the sensor size, some of the points that used to be
in the image no longer are. And in a sense the light that used to go to
those points is "wasted". But for all of the points that remain in the
smaller image, they are just as bright as they always were. The
f/number of the lens does not describe total light throughput, because
that depends on angle as well as f/number. The f/number determines
brightness per point, or per unit area in the image, and that is
unchanged when some portion of the sensor area is removed.

At first the digital format would seem to allow long focal, fast
telephotos. i.e. my 70-200 f/2.8 would become a 112-320 f/2.8, which
would be a very nice fast lens indeed. But I'm not so sure the lens
would have the light gathering power of a real f/2.8 lens, but instead
be effectively an f3.5 (I think). I suspect if the focal length is
multipled by 1.6, the apeture will be multipled by sqrt(1.6), although I
might be wrong on the exact calculation.

No, you're wrong. It remains a f/2.8 lens. However, there's no free
lunch. To produce the same quality of image from the smaller sensor, a
given size print needs to be enlarged a factor of 1.6X more. That means
that the lens needs to deliver 1.6 times the resolution to the sensor to
get the same quality print. Thus, lenses that are marginal in sharpness
for full-frame use may look just plain unsharp with the smaller sensor,
while very sharp lenses can stand the extra magnification without
strain.

Dave

Hi Dave,

AFAIK "magnification factor" is a marketing term rather than an optical
term. As has been already stated, it is a cropping factor, which results in
a reduction of the FOV. If I cut off a tiny square of a 35mm film, I get a
"magnification", in terms of modern digicam marketing ;-) As far as the
f-stop is concerned, this will remain the same. Focal length and focal ratio
are characteristics of the lens - not the focal plane or whatever lies on
that. BTW the f-number is not defined through terms of power denity or
whatever interpretation anyone could come up with. In classical optics, the
f-number=focal length/diaphragm aperture, nothing more, nothing less. Try
http://en.wikipedia.org/wiki/F-stop and http://www.celestron.com/tb-trms.htm
for some terms.

I hope this helps :-)

dimitris

P.S. I remember well that f-number used to be the inverse of F-number, i.e.
f=1/F, I couldn't find any information on this on the web...