Random tip: Light the ground not the sky (or your neighbour's telescope). MAS FAQ

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### (-) 1036 How do I calculate FOV for Eyepiece projection ?

`Notes on eyepiece projection for enthusiastic insomniacs` `Eyepiece projection is when the camera sensor (without it's lens) is positioned above the eyepiece. The image is 'projected' by the eyepiece onto the sensor. Typically, eyepiece to sensor distance is fixed by the camera adaptor, however this distance results in additional magnification` `Worked example. What is the magnification and effective focal ratio of my 300mm f/6 telescope when I project the image of Jupiter into my camera with a 6mm eyepiece (so I can work out the expected size of Jupiter on my film / DSLR sensor) ?` `Out goal is to get an image of Jupiter 'filling the frame' i.e. covering as many pixels as possible, to give the best resolution` `The magnification formula for a simple lens (eyepiece) is M = ( v - f ) / f ` `v = is the 'image distance' i.e. the distance of the eyepiece from the position of the camera film/sensor.` `f = focal length of the eyepiece (eg 6mm)` `The image size on the film is then 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 )` `Back to our example, the distance from the 6mm eyepiece to the 'film' plane (image sensor) was estimated as 50mm.` `(NB. The telescope is a 300mm f/6 reflector).` `So Eyepiece Magnification is thus = (50 - 6) / 6 = 7.3` `This gives us a combined focal ratio f/6 X 7.3 = 44, and an effective focal length of 44 x .3 = 13 m.` `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 i.e. 60" arc, F = 1800mm, so Jupiter is 0.5 mm in diameter.` `After eyepiece projection, the image would be magnified to 7.3 x 0.5 mm = 3.65mm` `The actual measured diameter on the processed film was 3.8mm  !!` `(the difference is likely due to the estimate of eyepiece to film distance of 50mm)`

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