f ratio = focal length/diameter (mirror or objective).
eg. A telescope of 600mm focal length with an objective of 80mm Dia will give an f ratio of 7.5
= focal length of the objective/focal length of the eyepiece.
So, our 600mm telescope with a 20mm eyepiece give 600/20 = 30 times.
A 2x barlow will double the focal length to 1200mm making the magnification 1200/20 = 60 times.
Focal reducers reduce the focal length, of course. The popular .63 FR will reduce the focal length to 378mm, so the 20mm eyepiece will give 378/20 = 19 times (18.9 actually)
In photography the focal length controls the image size.
The formula is Pixel size (microns) /focal length(mm) x 206 = Field of View/pixel (arcsec).
This gives you the amount of sky that each pixel sees, eg. For our 600 mm telescope and a digital camera with square 7.8 micron pixels, the formula will give us
7.8 x 206/600 = 2.678 arcsec/pixel.
Multiplying this by the number of pixels in the array gives the size of the camera’s Field Of View in arcsec. So for a camera with an array of 3024 x 2016 pixels,
the result is 2.678 x 3024 = 2.25° and 2.678 x 2016 = 1.5°
Giving a Field of View of 2.25° x 1.5°
A critical job in astrophotography, but how close do you have to get to be “in focus”?
The formula is,
In-Focus Zone = Focal Ratio² x 2.2 in microns.
for our example telescope it is 7.5 x 7.5 x 2.2 = 124 microns, and with 1000 microns in a millimetre, it means .124 of a mm, say 1/8th of a mm.
For a Meade SCT, it is 10 x 10 x 2.2 = .22 of a mm.
It’s all in the photographic speed of the telescope.
A fast f/2 system will have to be within (2 x 2 x 2.2 = 8.8 microns) = less than a 10th of a mm! Not easy with a rack and pinion focuser, and varying ambient temperature making the telescope expand and contract.
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