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Whilst I know it is extremely dangerous to view the sun using a telescope (or even just looking directly at it for prolong periods) can you do permanent damage to your eyes looking at Venus through a telescope? I was out with my telescope last night looking at Jupiter and moved over to Venus as it's right next door at the moment. At the moment it seems really really bright and for a quite while afterwards I felt as if I had arc eye. I've not really used my telescope for a couple of years but I remember looking at Venus previously and whilst it was bright it seemed perfectly OK to view.

Hopefully it doesn't seem like a stupid question, I was a bit surprised because it did feel like I'd just deliberately watched a welding spark. It was quite early in the evening; but the sun had fully set and the sky was dark. Is the brightness of Venus to us affected by it's position - it's probably in quite a good position to reflect sunlight as it's quite low in the west at the moment.

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Of course not. It all depends where you see it. –  Garmen1778 Mar 11 '12 at 23:06
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up vote 8 down vote accepted

No, there is absolutely no possibility of eye damage from looking at Venus. Like the Moon, it appears extremely bright when observed from a dark location in a dark sky, but when observed in full daylight (which is the way I usually observe it) it is quite dim, only slightly brighter than the background sky.

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Additionally, the OP probably only felt like they had arc-eye as their eyes were accustomed to the dark, and the contrast from Jupiter to Venus that night may have been a bit of a jump. It takes your eyes a while to adjust to changes in light level at night as your pupils are wide open. –  Rory Alsop Mar 11 '12 at 21:35
    
Thanks, I didn't think you could but thought I'd check. –  Paul Hadfield Mar 12 '12 at 8:56
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A telescope can never increase the surface brightness (brightness per unit of apparent area) of an extended object like a planet or nebula (as opposed to a point-object, like a star). This can easily be demonstrated using the theorems of optics.

So, if you look through a telescope at any object that is larger than a mere dot - let's say the Moon, or any of the major planets, or a nebula, etc. - the surface brightness you see through the scope is actually not more intense than what you see naked-eye. It could be less bright (and, in fact, it always is), or it could be equal (or, rather, pretty close); but it can never be more bright than the naked-eye brightness of that surface.

This is very counter-intuitive, but it's just one of the many paradoxes of optics. Most people, when looking at the Moon through a telescope, go "wow, that's too bright!" In fact, it's not brighter (per unit of apparent surface) than the naked eye Moon, and in fact it's almost always less bright. But it's bigger. That's all.

So, when you look at a large image of Moon (or even Venus) in a scope, what you see is a bright sunlit object, more or less as bright as Earth's ground at noon, and it's made big. But your eyes are dark-adapted. That's why it hurts. If your eyes were not dark-adapted, it would not hurt.

Here's a trick: when observing the Moon, or some of the bigger planets, don't get dark-adapted. Observe from the front porch, all lights turned on around you. Does the Moon or Venus bother you anymore? Nope, or much less so.


Here's an even more counter-intuitive fact: the surface brightness of an extended object in a scope does not depend on the size (aperture) of the scope, but only on the exit pupil. Big scopes, small scopes - the Moon, or a nebula, look equally bright (per unit of surface) if the exit pupil is the same.

The surface brightness of an extended object in a scope is equal to its naked-eye surface brightness when the exit pupil of the scope is equal to your eye's flesh-and-blood pupil. It becomes smaller when the scope's exit pupil is smaller than your eye's pupil.


Some math:

M = the magnification of the scope

D = the aperture (the diameter of the primary mirror or lens)

d = the exit pupil (the thickness of the "pencil of light" exiting the scope)

F = the focal length of the primary lens or mirror

f = the focal length of the eyepiece

Everyone knows the basic formula for magnification:

M = F / f

But few people know that magnification is also related to the aperture and exit pupil:

M = D / d

So then the exit pupil is:

d = D / M

E.g., a 200 mm (8") F/6 dobsonian, using a 12 mm eyepiece.

D = 200 mm; F = 1200 mm; f = 12 mm

M = F / f = 1200 / 12 = 100x

So then the exit pupil is:

d = D / M = 200 / 100 = 2 mm

Most people, when fully dark-adapted, have an eye pupil of about 6 mm. In this telescope, a 6 mm exit pupil happens when magnification is:

M = D / d = 200 / 6 = 33x

The eyepiece needed to achieve that is:

f = F / M = 1200 / 33 = 36 mm

The 36 mm eyepiece is the longest one that makes sense in this telescope, and the 33x magnification is the lowest useful magnification. If you use an eyepiece even longer, the exit pupil becomes greater than your eye's pupil and you're wasting light and aperture.


Note: All of the above is true for extended objects, like the Moon, the big planets, nebulae, etc. But for point-objects, like most stars, the story is different.

A telescope can and does increase the brightness of point-objects like stars. The increase is directly proportional to the scope's aperture, D.

Let's say a scope has an aperture D = 200 mm. Your eye has an aperture of 6 mm. How much brighter will a star look through this scope?

(200 / 6)^2 = 33.3^2 = 1111x

Finally, why a star looks brighter through a scope, while the Moon's surface does not? After all, the scope collects more light no matter what object you're looking at.

This is because it collects more light from the Moon, but the Moon also appears bigger in the scope. So all that extra light is "diluted" over a larger apparent area. It can be shown that the two effects (collecting more light, and making the object appear bigger) compensate exactly when the exit pupil is equal to your eye's pupil.

However, a star is a star. It always looks like a dot. So all that extra light, instead of being spread out in a bigger image, is concentrated in the same tiny dot - which appears brighter.

Does that make sense?

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+1 for such a detailed response, thanks. –  Paul Hadfield Mar 12 '12 at 18:59
    
+1 epic answer! –  Mark Mayo Mar 13 '12 at 22:28
    
I don't exactly know the applicability of this theory, but in the case of the Sun, it would seem to be trumped by observation. Examples, youtu.be/tPhXwk1nvCg, youtu.be/7O4Gl5lBKf4 –  Sonia Mar 17 '12 at 18:17
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@Sonia: The bigger the image of the sun, the greater the total luminous flux. Trust me, if the sun appeared, say, 100 times bigger across in the sky, you could easily light paper with it even without a telescope. That is, if you could somehow survive without burning to a crisp yourself. –  Ilmari Karonen Mar 17 '12 at 19:25
    
@IlmariKaronen - you are correct. The Sun delivers 1 kW / m^2 of radiation at ground level. If it appeared 100x bigger, even without being more bright (which is the case in a telescope) then its radiation would be 10 MW / m^2, enough to scorch, melt or ignite most materials - that would be like orbiting the Sun not far from its surface, waaay inside the orbit of Mercury; that's a very hot place. Moreover, if a telescope could increase the surface brightness of any object, then perpetual motion machines would become feasible. Just one more way physics is non-intuitive. –  Florin Andrei Mar 19 '12 at 18:43
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You description sounds more like an afterimage than arc eye. Venus through a telescope is easily bright enough to leave a distracting afterimage for a moment. Permanent damage though? No.

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