Re: [APML] Petzval hot spot?

[mschex@verizon.net]

Thanks Chuck, your argument certainly makes sense.  I am certainly not an optical engineer.  I have never owned one of these lenses but I thought that I had seen / heard of a gradient filter that screwed into the front.   It sounds like it would have had to be close to the image plane in order to even out the illumination.    

Mike
> 
> From: Chuck Vaughn <aa6g@aa6g.org>
> Date: 2002/09/18 Wed PM 02:44:21 EDT
> To: <astro-photo@seds.org>
> Subject: Re: [APML] Petzval hot spot?
> 
> Mike,
> 
> > Many of the systems do place the
> > radial gradient in front of the lens, this does in fact slow down the entire
> > system however -- but you get an evenly illuminated field.
> 
> Maybe I'm unable to visualize it but I can't see how it could work.
> Perhaps someone else can set me straight but let me explain further the
> way I see it. Feel free to punch holes in my argument because that way
> I'll know where I err'd. :-)
> 
> Let's examine an ideal system where the center of the field is 100%
> illuminated and the edge of the image circle is 50% illuminated and the
> drop off is linear.
> 
> A radial gradient placed in front of the film would have be the opposite,
> i.e., drop the illumination in the center by 50% and the edge 0%. There's
> an important point to remember here. Each point on the image plane, with or
> without the filter, sees the sum of all the non-obstructed rays entering
> the objective. Realizing this, it is easy to understand how the density
> of the near film filter would have to vary to even out the illumination.
> 
> What I don't understand is how the same thing could be done with a filter
> in front of the objective. Going back to my ideal example, the total
> attenuation of the filter would have to equal 50% in order to reduce the
> center illumination to 50%. But now the illumination at the edge of the
> image circle would be 50% from obstructions plus attenuation of the filter,
> perhaps another 50% depending on just how the obstructions blocked the
> objective. So now we have 50% illumination in the center and 25% at the
> edge. We still have the same 50% drop off.
> 
> In order for this to work the attenuation at the edges of the image circle
> would have to be zero. This is impossible though because the edges will
> see rays from a portion of the attenuated objective.
> 
> Perhaps the easiest way to think about this is a secondary obstruction.
> That would be step filter in from of the objective, 0% transmission within
> the obstruction and 100% out of it, yet this has no impact on the
> illumination of the image plane from center to edge. No circular
> obstruction placed in the center in front of the objective can alter
> the illumination curve at the image plane.
> 
> Chuck <aa6g@aa6g.org>
> 
> 
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