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I keep reading the general opinion that the improvement is just not worth the risk if you attempt to clean the mirror on a Newtonian.
Clearly, there must be a point at which it really is an important thing to do. I thought I would try a back of fag packet calculation and see where it took me. This is my argument and I would be grateful if someone would cast an eye and comment whether it's conclusion is realistic.
I'll take the example of a 200p (mine). Diameter 200mm.
Main mirror area : A1 = πR12
Distance D between the two mirrors is about 1m
Secondary mirror area: about A2 = πR22
Assume that the surface of the main mirror is covered by 1% with dust and that the dust has reflectivity of 20%. (I had to start somewhere). Pointing the scope at a single star, there will be E Watts of power entering the tube and 0.002E scattered randomly from the dust. The reduction in image brightness is negligible but a proportion of the light scattered from the dust will find its way to the secondary mirror and will uniformly (?) illuminate the image that the eyepiece sees and hiding any object of similar brightness. (i.e contrast is affected)
The proportion of scattered light hitting the secondary mirror will be (assuming a point source at the main reflector, spread over a hemisphere) approximately equal to
A2/2πD2 = 10-4π/2π = 0.5X 10-4 π.
The energy will therefore be 0.002E X 0.5 X 10-4 π = about 3x10-7 E
That ratio looks to me to be a relative magnitude of 15! Can that be right?
But that's for just one star. A hundred, of varying magnitudes could make a difference - as could the Moon tonight,
Can someone find the flaw in my argument please? It seems to imply that a really grotty mirror cannot hurt.
But, otoh, regular photographers spend all their time keeping their lenses spotlessly clean - and they don't even suffer from the back scattering that my calculation deals with.
Clearly, there must be a point at which it really is an important thing to do. I thought I would try a back of fag packet calculation and see where it took me. This is my argument and I would be grateful if someone would cast an eye and comment whether it's conclusion is realistic.
I'll take the example of a 200p (mine). Diameter 200mm.
Main mirror area : A1 = πR12
Distance D between the two mirrors is about 1m
Secondary mirror area: about A2 = πR22
Assume that the surface of the main mirror is covered by 1% with dust and that the dust has reflectivity of 20%. (I had to start somewhere). Pointing the scope at a single star, there will be E Watts of power entering the tube and 0.002E scattered randomly from the dust. The reduction in image brightness is negligible but a proportion of the light scattered from the dust will find its way to the secondary mirror and will uniformly (?) illuminate the image that the eyepiece sees and hiding any object of similar brightness. (i.e contrast is affected)
The proportion of scattered light hitting the secondary mirror will be (assuming a point source at the main reflector, spread over a hemisphere) approximately equal to
A2/2πD2 = 10-4π/2π = 0.5X 10-4 π.
The energy will therefore be 0.002E X 0.5 X 10-4 π = about 3x10-7 E
That ratio looks to me to be a relative magnitude of 15! Can that be right?
But that's for just one star. A hundred, of varying magnitudes could make a difference - as could the Moon tonight,
Can someone find the flaw in my argument please? It seems to imply that a really grotty mirror cannot hurt.
But, otoh, regular photographers spend all their time keeping their lenses spotlessly clean - and they don't even suffer from the back scattering that my calculation deals with.