Re: A Modest Proposal

One would think that the use of 50 mm instead of 43 mm comes from a
combination of 3 things. First, it's just a heck of a lot easier to
use/remember/etc. For example, would you then got to a 150 mm or a 129
mm. Second, you probably want a little extra light around the outside
of the frame, so you round up a bit. Third, and most likely, when you
are designing a lens, there's a lot of math going into the angles, etc.
Back in the early days when the designers were using slide rules, it
was probably easier to stick 50 into a formula than 43. But these are
just guesses.

Pat.
Dave Martindale wrote:
"ih@xxxxxxxxxxxx" <ih@xxxxxxxxxxxx> writes:


Somehow
some manufacturer introduced the 50 mm as the normal lens and now it is
general
accepted as the normal lens. I believe that when people refers to 35
mm
equivalent is based on 50 mm. So I used that as the reference.

When you see lenses marked with a "35 mm equivalent" focal length, that
calculation depends only on the sensor dimensions. What's considered a
"normal" lens for the focal length has no part in it.

For example, if you have a digital camera whose sensor is 7.2 mm wide,
that is 1/5 the width of a 35 film frame, and so all of the "35
equivalent focal length" specification are 5 times the true focal
length. If the lens is actually 7-21 mm, it will be quoted as "35-105
equivalent". It simply doesn't matter whether you think "normal" is 40
or 43 or 50 or 55 mm on a 35 camera.

I am referring to using the digital camera to take a picture through
the
eyepiece of a telescope or microscope. If the focal length of the
eyepiece
is 32 mm and using the maximum zoom of the CP-995 (which is also 32
mm),
the image you got is the same as formed by the objective of the
telescope,
i.e., the final magnification is 1. If the focal length of the eyepiece
is 16, the
final magnification is 2. Of course you have to make sure that the
images
match.

That doesn't make sense to me at all. The magnfication of the
*telescope* is an angular magnification, determined by the ratio of the
objective and eyepiece focal lengths. The resulting virtual image is
projected at infinity (or so we assume, anyway).

Then the camera uses its lens to capture that image. On a camera, the
lens focal length primarily determines angle of view, and you generally
want a rough match between the camera angle of view and the eyepiece
image size. That might mean using a 50 mm lens on a full-frame camera,
or a 10 mm lens on a P&S digicam, but the angle of view is the same for
both of these. And it doesn't matter whether the telescope eyepiece is
a 55 mm or a 9 mm.

If you want to calculate the magnification of the digital image you get,
you do need to know the true focal length of the camera lens as well as
the overall magnification of the telescope. But you'll get the same
final magnification using a telescope with a 2000 mm objective and a 40
mm eyepiece as you will with a 500 mm objective and a 10 mm eyepiece.

Having the focal length of the camera lens match the focal length of
the eyepiece has no value at all that I'm aware of. They do two very
different jobs, and have entrance and exit pupils in very different
places.

You *would* like to match the exit pupil of the eyepiece with the entrance
pupil of the camera lens, but that doesn't depend on the focal length,
and can vary greatly between zoom lenses that have the same focal
length.

Dave

.

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