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# Observed brightness of the moon

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I thought I knew the answer to this question but I think I've talked myself out of it! It relates to the "A" level physics syllabus and I'll use the moon as an example but any non-extended object could be the subject.

The question is, when viewed through a telescope, is the moon brighter than when viewed naked eye? Does the brightness of the moon change with increasing aperture? What about with increasing magnification?

My knee-jerk answer, everything else being equal, is yes (scope has bigger collecting area), yes (that's why big scopes are usually stopped down) and no (image will dim with increasing magnification).

The further question was posed as to whether my answer is simply a result of the observer's perception of brightness rather than actual measured luminosity per square millimetre ...? The text that started this stressed a difference between extended and non-extended objects and suggested the image brightness compared to the naked eye did not change due to the magnification of the image by the optics. Then I started to think too hard about it ...

I still think the obvious answers are correct simply based on photons per unit of collecting area but that also implies no difference between extended and non-extended objects ...

Am I missing something obvious?

AndyG

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I have seen arguements and reasoning both ways on this. But like yourself I would go for objects being brighter with a scope - on the condition that the magnification is not so great as to render the image disproportionally bigger and so reduce brightness.

If you collect 80x light but magnify it to 100x area then it makes sense that it is dimmer. However a factor is what is 80x the light gather? Sensibly it means a collection area 80x more then your fairly standard everyday pupil. But if my pupil opens to 6.5mm and yours to 7.2mm then on the same scope we are collecting different multiples.

This comes out on CN about once or twice a years where several are adamant that the naked eye gives the brightest. My thought being simply why do we use a scope then? Find it easier/better/safer to stay out of the arguement.

Extended and point objects are best considered differently. (Sorry no real idea why I say that) But a point is considered as zero size, and zero in any maths is generally a problem. On an extended object the brightness is the sum of all the light off of the whole object. Likely they take an "average" point then integrate over the full area to arrive at a value. Hence something like M31 is the "brightest" object in the sky, but so dim it is hard to see.

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Bear in mind the moon is an extended object which only has reflected light which is fixed. Look at the moon naked eye it is as bright as it can be - the reflected light is to all intents and purposes fixed. But note that as the moon equates to a small part of the sky your pupil does not really contract. However look with one eye and and scope with the moon filling all the field and and your pupil contracts a lot. Does this mean it Is brighter?  No, it just takes up more field so feels brighter.  This Is the difference and source of confusion I reckon. Just because something seems brighter does not mean it is. Actual brightness and apparent brightness are not the same.  Someone else said previously that if you have a candle 100m away and bring it to 6" away, is it brighter? No, just closer.

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Someone else said previously that if you have a candle 100m away and bring it to 6" away, is it brighter? No, just closer."

I might regret this, but it is apparently brighter but not intrinsically brighter. It's brighter because it's closer and therefore by the 1/r^2 rule it must be brighter, there is more energy pouring into the same unit solid angle of the observers eye. If unresolved, the flame will just appear brighter, if resolved, the light might be smeared over more rods , so now you have 2 rods firing not just one. But the flame isn't burning any harder.

for a telescope, the brightness increases by the light gathering area but decreases by the f/ratio, compared to the eye. That's because the image size and inversely , the brightness per unit area,scales by the f/ratio.

cheers

Mike

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5 hours ago, 101nut said:

I still think the obvious answers are correct simply based on photons per unit of collecting area but that also implies no difference between extended and non-extended objects ...

Am I missing something obvious?

AndyG

Perhaps it might help to think of a telescope as an instrument for mapping circles of 'sky' to circles on your retina.

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