Stupid questions for the maths-minded... with way too much time on their hands.

[FrenchyArnaud]

I did not attend the lessons in Optics at school. Maybe the sun was shining, or Gremlins II was on TV or I really needed to be elsewhere – wherever this wonderful place is.  Anyways, I did not attend the lessons in optics and now I can tell you what – I really regret it.

So here’s a  maths problem for those who DID attend school unlike me and who are maths-minded. This maths problem has no direct practical use – it’s just about understanding the physics. Because I did not attend school, by I still want to understand weird and wonderful stuff.

Considering two reflectors, with mirrors of 114 and 150mm in diameter. They have surfaces of 10207 and 17672 square millimetres thanks to PiR2 (yes I DID attend school this far). Therefore the bigger one has 85% extra surface compared to the smaller one.

Stupid question 1) Does the bigger collect 85% extra light ? (all things being otherwise equivalent) or is there something to take into account AFTER the square law ?

Stupid question 2) Does it mean it can theoretically put in sight objects that are theoretically 85% fainter ? (read slowly : this does not necessarily follow question 1!)

Stupid question 3) Does it mean that even with that 85% difference the gain in brightness is only around .35 magnitude ? (based on approximation x2.512 per order)

Bonus for those willing to win a  used wooden spoon and second-hand pair of pink socks : is there a way to calculate the actual limit in “faintness” or “positive magnitude” of a particular mirror ?

And if someone has a good resource to explain the difference between parabolic and hemispheric – I’ll take that too !

[CraigT82]

If you're willing to give yourself a headache... have a read of this website http://www.telescope-optics.net/functions.htm#1.4.1._Light_gathering_power

[Riemann]

1 hour ago, FrenchyArnaud said:

Question 1) Does the bigger collect 85% extra light ? (all things being otherwise equivalent) or is there something to take into account AFTER the square law ?

Question 2) Does it mean it can theoretically put in sight objects that are theoretically 85% fainter ? (read slowly : this does not necessarily follow question 1!)

Question 3) Does it mean that even with that 85% difference the gain in brightness is only around .35 magnitude ? (based on approximation x2.512 per order)

Bonus for those willing to win a  used wooden spoon and second-hand pair of pink socks : is there a way to calculate the actual limit in “faintness” or “positive magnitude” of a particular mirror ?

And if someone has a good resource to explain the difference between parabolic and hemispheric – I’ll take that too !

 

 

Hi - these are not stupid questions      Some thoughts that may help....

Question 1:   I am not sure where you get 85% from.   17672/10207 = roughly 1.73 i.e. 73% extra.    But leaving the arithmetic aside, the principle is basically correct.   The only other thing you have to worry about in a reflector is subtracting off the impact of the central obstructions before you compare the areas.    If the central obstruction is roughly the same size then you might actually end up with somewhere around 85% extra after all, but that would depend upon how big it was.

Question 2:  If thing A is 85% bigger than thing B it does not mean that thing B is 85% smaller than thing A.   So as an example, thing B is 100, thing A is 185, then thing thing A is 85/100 bigger than thing B but B is 85/185 smaller than thing A, i.e. A is 85% bigger than thing B but thing B is roughly 46% smaller than thing A.  

Question 3: You can't do the maths in this way for a couple of reasons.   If we stick with 85% then the number you need to be working with is 1.85 as the ratio of the amount of light.   A very crude estimate would be that 1.85 is about 1.85/2.512 = 0.86 of a magnitude.   However, this is not really right either because magnitude is a logarithmic scale.   When you subtract 1 from the magnitude you multiply the apparent brightness by 2.512     So to get the right answer you need log base 2.512 of 1.85 = 0.67 of a magnitude.   If you use 1.73 instead you get log base 2.512 of 1.73 = 0.60 of a magnitude.

Bonus question:  There are two things that affect your ability to see faint objects.   The first is the ability of your eyes to see faint things against a perfectly dark background.   This varies from person to person.    More telescope aperture helps in this regard.     The second factor is how dark the background is.   If an object's surface brightness is no brighter than the background then it does not matter how much aperture you chuck at it you still won't be able to see it.      Both factors come into play when you get objects that are only just brighter than the background.    I have never seen a formula for the limiting magnitude of a mirror and I think it would be very difficult to produce a useful one because it would depend on who was using it and where it was being used and whether the moon was out etc.

I'm not really sure how best to answer your last question.   Is there something in particular you want to know?    You might like to have a look at https://en.wikipedia.org/wiki/Parabolic_reflector

https://en.wikipedia.org/wiki/Parabola

and

https://en.wikipedia.org/wiki/Curved_mirror

 

Hope that helps a bit

[ronin]

Not sure where 85% came from as the ratio is 1: 1.7145 so just 71% more area. If the 2 values given are right.

1) The bigger one will collect 71% more light as given by the ratio's of their area.

2) Maybe, if the magnification is the same then the image size is the same and so the light collected get dumped/squeezed into the same size image and so is or should be brighter. Hiccup is twice the light is not twice as bright at the eye. So settle for "brighter".

3) If a mag increase of 0.35 corresponds to the collection are then yes, but I never really got on with the magnitude scale. Each step of 1 isd about 2.5 times as bright in lumen terms I think.

2) Has the problem that brightness is measured a bit odd: M31 is Mag 3.44, but is a lot dimmer then say a Mag 4 star. The brightness is the light given off by the whole area, M31 being big has a little light from it but over a large area. In effect a unit area of M31 is dim, but there are lots of units. A "unit" is any arbitory size you feel like.

Consider it like this M31, Andromeda is Mag 3.44 and difficult to find and see, Polaris in Mag 9.2 (hugely "dimmer") but easy to find and see. At a guess M31 is defined as about 10 times brighter then Polaris is if you work in Magnitude. But this much "brighter!" object is almost invisible, whereas people navigated the world by looking at Polaris.

[billyharris72]

Hi Arnaud:

Interesting questions, so here's my take.

1) All other things remaining equal, in terms of the number of photons from a continuous source entering the tube, yes. The mirror that has 1.85 times the area will collect 1.85 times the light. The light from any object you are observing more or less meets this criterion (as the distance means the light waves are parallel and entering evenly across the whole aperture. (Though credit to Riemann for being less lazy than me, it's 1.73.

2) All the photons from the wider area collected are being focussed to a point and entering the eye. So the image in the larger tube would be 1.73 times brighter, and the smaller tube would be 0.58 (1/1.73) times as bright. Another way to think about this is in terms of exit pupi, which varies with aperture when magnification is held constant.

3) I make x1.85 increase in brightness equal to a 0.60 magnitude difference: log_2.512_(1.73). Again, Reinamm got there before me and as he notes there are lots of other factors that affect what can and can't be seen. For most of us contract with the sky background is probably the limiting factor, and bigger scopes, while better, do seem to produce diminishing returns.

Billy.

 

[FrenchyArnaud]

Thanks to all, I will study your thoughts and come back to you. 

85% : yes it's actually 73% but I retain 85% because I estimated that on my 114, there is close to 12% obscuration due to big spider (and I mean BIG. It's ridiculously big, when you think it's supposed to hold such a little mirror)  while I estimated that on the 150 there is only 6% obscuration. So I retained  (17672*.94) / (10207*.88) which is give or take... 85%.  I just forgot to explain this from my notes or to get back to the initial calculation...

Eventually the actual value is irrelevant - the logic of the physics is much more interesting

[Riemann]

17 minutes ago, ronin said:

Has the problem that brightness is measured a bit odd: M31 is Mag 3.44, but is a lot dimmer then say a Mag 4 star. The brightness is the light given off by the whole area, M31 being big has a little light from it but over a large area. In effect a unit area of M31 is dim, but there are lots of units. A "unit" is any arbitory size you feel like.

Consider it like this M31, Andromeda is Mag 3.44 and difficult to find and see, Polaris in Mag 9.2 (hugely "dimmer") but easy to find and see. At a guess M31 is defined as about 10 times brighter then Polaris is if you work in Magnitude. But this much "brighter!" object is almost invisible, whereas people navigated the world by looking at Polaris.

Whilst my points in reply to the bonus question were correct, this point is really more important.

[cloudsweeper]

Rays parallel to the axis of a mirror with a circular section come to different foci the further out from the axis they are, whereas with a parabolic section, they all come to a fixed focus, ensuring sharp images.

Doug.

[FrenchyArnaud]

Ok, Rieman clearly wins the pink socks - but sorry, I will ask you to share the wooden spoon with ronin and billyharris72

While I assume I have now the whole logic right, it actually enlights ( enlights... I am soooo funny ) an assumption I had - before I read and thought about your answers.

Basically, the bigger the better but it's mostly dependant on who looks at what, where. Ok, I can figure that out. However there is this idea of light constrast that bugs me : 

1 hour ago, billyharris72 said:

For most of us contract with the sky background is probably the limiting factor, and bigger scopes, while better, do seem to produce diminishing returns.

 

1 hour ago, ronin said:

Consider it like this M31, Andromeda is Mag 3.44 and difficult to find and see, Polaris in Mag 9.2 (hugely "dimmer") but easy to find and see. At a guess M31 is defined as about 10 times brighter then Polaris is if you work in Magnitude. But this much "brighter!" object is almost invisible, whereas people navigated the world by looking at Polaris.

 

1 hour ago, Riemann said:

The second factor is how dark the background is.   If an object's surface brightness is no brighter than the background then it does not matter how much aperture you chuck at it you still won't be able to see it.      Both factors come into play when you get objects that are only just brighter than the background.   

Those three remarks are actually expressing the same concept : if your object is not brighter than the background - it's invisible, whatever the gear. Which makes kind of sense... And all of the sudden I realize that the smaller 114mm will probably show me more than the 150, because I can physically carry it farer away, in darker places. My hopes to spot DSOs are vanishing at lightspeed ( seriously, am I not hilarious? )

So if magnitude is not a valid indicator of the actual perceived brightness through the EP, and you can't really oppose the catalogue of brightnesses to your aperture, how do you select the easiest galaxies to actually see ? trial and error until you actual spot a smudge ?

 

[Sirius Starwatcher]

I've solved the problem and seen the light !.   Forget the math and just by the biggest aperture you can afford. It works for me.

[billyharris72]

Trial and error is definitely good. Surface brightness data can also be found, which can be more useful. Magnitude expresses the total light output from an object, which may be spread over a large or a small area, and the larger the area for a given magnitude the dimmer the object (stars being easy to see for a given magnitude as they are point sources). Surfact brighness normalises for size and so can be a useful guide (though I find it does not always work, which brings us back to trial and error).

[FrenchyArnaud]

I did actually try the  150... See here.