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I've shanged the threadname to make it more relevant.

Frank Reed has perceptively addressed these matters in  recent postings.

Douglas Denny has written, in [8760]-

"I wrote the following in a private coversation about this to a friend on
this forum:

===============
Pupil diameter of an eye is about 6mm - an area of approx 28 sq mm.
An ordinary binocular aperture is about 50mm diam,  or  1963 sq mm.  which
is 70 times greater.  So neglecting inefficiencies in the optics, if all
light is transferred through the exit pupil of the telescope to the eye, the
energy entering the eye due to the telescope is 70 times that entering the
naked eye.

Worse ....... this 70 times increase of illuminance is then focussed down in
the  eye to the foveal area  which has a diam of  0.3mm  which is 0.28 sq
mm. Compare   the ratio 0.28 to 1963
which is 1 to 7010

i.e.  the telscope of 50 mm diam increases the flux at the fovea by 7000
times."
-------------



However, let's examine Douglas' statement above, bit by bit, in the context
of potential eye damage by sunlight

"Pupil diameter of an eye is about 6mm - an area of approx 28 sq mm.". Fair
enough for the dark-adapted eye, as occurs at night, when there's no danger
from Sun exposure.. But unrealistic in daytime observation, the only time
when Sun exposure matters, when the pupil will be closed down, to 2mm dia or
even less.

"An ordinary binocular aperture is about 50mm diam,  or  1963 sq mm.  which
is 70 times greater.  So neglecting inefficiencies in the optics, if all
light is transferred through the exit pupil of the telescope to the eye, the
energy entering the eye due to the telescope is 70 times that entering the
naked eye. "

Well, the size of the exit pupil, the pencil of parallel light from a
distant point-source, that leaves the telescope, is the aperture divided by
the magnification.; simple as that. So to get all the light from such a
source which passes through a 50mm aperture to enter the 6mm pupil of the
eye that Douglas presumes, that could happen only if the magnification was
at least 8.3x; an unlikely (but conceivable) value for anyone to be using at
sea without special stabilisation.. If a more-plausible 2mm pupil is
assumed, then the magnification would have to be all of 25x; hard to imagine
in a marine environment. If it was less than that, much of the light falling
on the objective would fail to enter the eye pupil.

But still, let's take it that the eye really does have a 6mm pupil and the
magnification really is 8.3, to humour  Douglas by bending over backwards in
meeting his prior conditions. In that case, the total energy entering the
eye pupil would indeed be increased by 70x by using a (lossless) telescope,
compared with direct view, just as Douglas claims.

What about image size at the retina? Assuming that the eyeball is 25mm dia,
and taking the Sun's angular diameter as 32 arc-minutes, without a telescope
direct sunlight paints a bright disc on the retina that's 0.23mm dia. All
the energy entering the pupil falls on that disc. It's not far short of the
size of the fovea, the high-resolution spot that we should be especially
careful to protect.

Next, interpose our telescope. Now the angular divergence of the light
entering the pupil is 8.3x greater. That's the meaning of angular
magnification. So the Sun now points a disc on the retina 8.3x greater in
diameter, or 1.93mm. The area of that disc, to nobody's surprise, is 70x the
area of the disc we would get without a telescope, into which (we
calculated) 70x the energy would fall. Result: surface illuminance of the
bright disc is exactly the same as it was without the telescope.

Where Douglas went wrong was in stating- "Worse ....... this 70 times
increase of illuminance is then focussed down in the  eye to the foveal area
which has a diam of  0.3mm  which is 0.28 sq mm. " This is wrong on several
counts. The Sun image ISN'T focussed down to the foveal area, it's focussed
to a disc many times larger than the foveal area.  It's the total entering
energy that increases by 70x, not the illuminance, which is the energy per
unit area. And by the way, (though it's not very relevant) taking Douglas'
stated diameter of the fovea as 0.3mm, its area would be 0.07 sq. mm, not as
he states 0.28 sq.mm.

I do not wish to detract from the warnings about the dangers of looking at
the Sun, either through a telescope or directly, without one. The energy
density, causing local damage to the retina, can be no greater when a
telescope is used. The area of retina which could be damaged by such an
event could be much greater when a telescope is in use, as would also be the
total amount of heat deposited within the eye. I have discussed only damage
to the retina, but use of a telescope will greatly increase energy density
falling on the corneal surface of the eye, and that, and the iris, may be
vulnerable to such damage.

It's refrettable that, when trying to get that across the message about
potential dangers, the physical mechanisms that give rise to those dangers
are so often misidentified.  I hope that Frank and I have between us done
our bit to put these matters into perspective.

George.

contact George Huxtable, at  george@hux.me.uk
or at +44 1865 820222 (from UK, 01865 820222)
or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.


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