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## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

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Re: Eyesight dangers using telescopes
From: George Huxtable
Date: 2009 Jun 28, 11:39 +0100

```I will try again, about energy density falling on the retina, but without
much hope of convincing Douglas Denny.

Douglas Denny wrote, in [8850], (re: Star-star distances for arc error)
"In optics the apparent luminance of an object using a telescope, divided by
the apparent luminance of the object viewed directly,  is proportional to A
/ M^2  where a is the aperture area and M is magnification. ( N.B.  I use as
a source for my references:  'Geometrical and Physical Optics' by R.S.
Longhurst ).
So the maths is clear, whatever the luminance of the object (L), ( and with
the Sun it is very high indeed) the luminance seen through a telescope is
equal to L times a constant, times the Aperture area, divided by the
Magnification squared.   So "brightness" of the image is a function of two
quantities: aperture and magnification."

agreed (though unsure about the "constant", which ought to equal 1 under the
assumptions we have made, I believe)

"Increasing aperture does indeed increase luminance of the image in direct
proportion.". Yes. To the AREA of that aperture, note, in the way that
Longhurst framed it. There is confusion throughout Douglas' posting, between
aperture-as-diameter, and aperture-as-area.

"Increasing magnification decreases luminance squared." I don't understand
that statement, which appears to be written backwards. If Douglas had
written instead "luminance decreases as the square of the increase in
magnification", I would go along with that.

"What happens with aperture changes with constant magnification?
If a telescope is used of unity magnification, the maths says increase the
aperture area and the luminance of the image increases. If aperture is
doubled then L doubles.  If aperture is quadrupled, then L is quadrupled.
Use a telescope of unity magnification but aperture area double that of the
pupil diameter of say 3mm and the luminance must double to 2L."

No. That's where Douglas has it wrong. He has assumed the impossible.

I think in a previous posting, Douglas eventually agreed with me (and with
Johnson's book) that the ratio of entrance pupil diameter to exit pupil
diameter of a telescope was equal to its magnification. (if apertures are
consided as aperture-areas rather than diameters, then the ratio of those
areas is the magnification squared). If he is still unconvinced, an
astronomical telescope with a magnification of 1 (which does nothing but
invert) has two lenses of equal focal length l, spaced 2l apart. It's
completely symmetrical, and from that symmetry alone, it's obvious that the
exit pupil must be exactly the same size as the entrance pupil.

So, with such a telescope, provided its entrance aperture was no more that
the 3mm which Douglas has assumed for the exit pupil, then all the light it
collected in that tiny objective would enter the eye.

But double that diameter, and the diameter of its exit pupil is also
doubled. However, now most of the light in that larger 6mm exit diameter
pencil of  now misses the hole in the iris, which remains at 3mm diameter.
Only a quarter of the collected light now enters the eye. The  lenses can be
made as big as desired, but the effective aperture of the system remains as
3mm, the stop created by the iris.

The only way to get more light into the eye, and take advantage of a larger
objective, is by introducing magnification, which allows the light pencil
diameter to be compressed, with diameter inversely proportional to the
magnification. In that case, the useful diameter of the objective can be
increased in proportion to the magnification, but not usefully more; its
area then increases in proportion to the magnification squared, as does the
energy entering the eye. Magnification and useful aperture are inseparably

"So it is clear in my mind, for brightness of image in a telescope there are
two parameters working in opposition: aperture increasing directly and
magnification decreasing and by a square function."

No, the possible objective area and the total input energy increase by the
magnification squared, as explained, and as the resulting image is spread
over a retinal area also proportional to the magnification squared, the
energy density remains exactly the same, according to Longhurst as well as
me.

George.

contact George Huxtable, at  george{at}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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