Notes on eyepiece projection for enthusiastic insomniacs

Ford Prefect asks:   (With apologies to Douglas Adams)
What is the magnification and effective focal ratio of a telescope when I project the 
image of Jupiter into my camera with a 6mm eyepiece,  and can I then work out the expected 
size of Jupiter on my film ?

Marvin replies
The magnification is  ( v - f ) / f 

Where  	v = image distance.   In practice the distance of the eyepiece from the back of the camera.
        f = focal length of the eyepiece.

The image size on the film is the primary image size multiplied by magnification.
The focal ratio of the system is the ratio of the instrument multiplied by magnification.

Proof : 
1/f = 1/v  + 1/u  (the lens formula )
1/u = 1/f - 1/v
1/u = ( v - f ) / fv
but Magnification = v/u 
so M = v ( v -f ) / fv
or            (v - f ) / f 	QED
( or v/f  - 1 )

In use: 
I have a 6mm eyepiece and its about 50mm away from the film.  The instrument is  300mm f/6 reflector
Magnification = 7.3
Focal ratio = 44
Effective focal length is  13 m.

Did it work in practice ?
Yes !  The image size of Jupiter at the prime focus of the 300 F/6 is  given by   ( Theta . F / 206265 ) 
where Theta = apparent size of Jupiter ie 60 “arc  F = 1800mm,  so  Jupiter is 0.5 mm in diameter.

After eyepiece projection, the image should be magnified up to 7.3 x 0.5 mm

The measured diameter on the processed film was 3.8mm  !!