Should 4/3 lenses be half the size of full frame lenses?

it exists.

there you go again, picking only one item.

 

Only if the pint glass has perpendicular sides, and I've never seen
one like that. Beer glasses are either tapered, bowed out, or
schooner-shaped so cutting the height in half would result in a
container that would hold significantly less than a half-pint.

 

[]

... and the hole falls out of the argument. <G>

David

 

Pity the guy who got the top half ...

.... a no-pint glass.

 

It is indeed, and there's no need for you to try to complicate it.

ISO simply has to do with the sensitivity of the sensor or film to light.

That's all.

For any given subject and light conditions, ISO defines the necessary
exposure to get a properly exposed photo. Higher ISO = less exposure, etc.,
etc.

The properly exposed photo may have more or less noise, depending on things
such as sensor size, which does not change the ISO. There is no "normalize
it for equivalency," whatever that is supposed to mean.

ISO is ISO.

Yes, "it's really quite simple."

 

Yes. Stop right there.

You didn't stop where you should have stopped.

 

That depends on how you define "in half." No one said anything about
"cutting the height in half."

 

Just so.

 

if you take a lens without a camera, there is no exposure. it's just a
lens, sitting on the table.

or more accurately, you're stopping prematurely, ignoring things that
matter. why is that?

 

i'm not complicating it at all. for something to be equivalent, it must
be equivalent in all aspects, not only the ones you feel like
comparing.

if a photo from one camera has more or less noise than a photo from a
different camera, then the two cameras didn't produce equivalent
images, did they?

 

I take your point, but in trying to make something "equivalent in all
aspects" here, you are attempting the impossible.

ISO is ISO. It has to do with exposure, that's all. Small sensors will have
more noise than large sensors at the same ISO. There is no way you can make
them "equivalent in all aspects."

Obviously not, and the difference obviously has nothing to do with ISO.

What you're trying to do is something like putting the same 300-hp engine in
a small sports car and a big truck, and then saying the horsepower must be
"normalized" (as you call it) because the two don't accelerate the same. The
difference in performance has nothing to do with, and does not change, the
fact that the engine is still the same 300 hp regardless of what vehicle
it's in.

It's the same with ISO. The fact that two different cameras produce
different amounts of noise at the same ISO does not mean the ISO has to be
"normalized" or otherwise tinkered with in either case.

 

Because they don't in fact matter in this connection.

The f-number is the f-number. Period.

 

it's not impossible at all.

yes you can, simply by raising the iso of the larger sensor which
increases its noise to match that of the smaller and noisier sensor.

nope. if you want to use a car analogy, put a 150hp motor in the
smaller and lighter car and a 300hp motor in the larger and heavier
car, and they both have the same 0-60 times.

an engine without a car isn't particularly useful, just as a lens
without a camera.

it does if you want to call something equivalent.

 

only without a camera or photo.

 

Of course it does.

Try this:
To spread the light from a 1x1 area onto a 2x2 area quadruples
the area lighted, quarters the light per area. Since the lens
is unchanged (and since the transmission didn't change (much)),
the f-stop must have doubled.

Or, try this:
To have the same angle of view (which you keep, after all) with a
sensor of 2 times the linear dimension, (which the 2x teleconverter
does) you need a lens twice the focal length. The teleconverter
does that without changing the lens, so the aperture stays the
same, but the effective focal length must have doubled. Since the
f/stop is effective focal length/effective aperture diameter,
the new f/stop value is doubled (as the focal length doubles).

Same thing, different words. But I like the spreading and less
light per area better.

I must correct myself: s/80mm/70mm/

The NASA had some built to photograph the dark side of the moon.
Later, Kubrick used a lens for Barry Lyndon (where some scenes
were lit only by candles).

The 70mm was a German wartime night lens, btw., though the design
is prewar.

-Wolfgang

 

OK, then noone can help you

OK, but then you are not looking at the photos, but at tiny details
you need larger to properly edit them. And if it wasn't for the
12 MPix, you'd need 141% and 283%.

-Wolfgang

 

Changing the focal length while keeping the maximum aperture the same is
what changes the f-number. For example, f/2.8 means the aperture is 1/2.8 of
the focal length. If you double the focal length while keeping the same
physical aperture, it is then f/5.6. That's what a 2x tele converter does.

What you like better doesn't change the definition. The f-number is f (focal
length) divided by aperture. That's the way it's written and that's what it
means. Yes, obviously this does mean less light per unit area, just as
anything else than increases the f-number does.

 

Fair enough. It's not, basically, a photographic problem,
and hence outside the remit of this newsgroup.

For both dust on the sensor, and actual image details,
they'll be twice as big (in pixels; same size in physical
on-sensor image dimensions) on the 6MP sensor, so I won't
need to enlarge more to get my usual on-screen working
size.

 

Yes they do give the lens a faster real f-stop.

Equivalent DOF.

 

Then "equivalent DOF" should be stated. I don't think "equivalent f/2.8"
conveys anything about DOF to most readers.