Field of view (Have i got this right?)

[astoc]

I've been trying to properly understand optics in the context of astronomy, and I think I have everything pretty much licked apart from 'field of view'.

Everyone seems to know that a faster telescope gives a richer field of view, but I can't seem to find a decent explanation as to why. Heres what I think is going on, please correct me if i'm wrong:

A shorter focal length necessarily alters the area over which the image is produced at prime focus. We're all used to seeing images of parallel rays at infinity focusing on the axis of the system, but it is equally valid to consider parallel rays at infinity striking the lens (or indeed mirror) at a non zero angle to the axis. Take two stars for example; one at the 'top' and one at the 'bottom' of the scope as seen by the unaided eye. If the separation between the two stars is sufficient, the angle at which rays from each star reach the objective (angle to the axis) will no longer be negligible and the image of each star will form at different vertical positions in the focal plane.

This is very important for imaging, because a ccd/cmos chip has a finite area. All other things being equal, a shorter focal length (and so a smaller f/ratio) will 'squash' the image in the focal plane to a smaller area. Stars in the image that may not have covered the chip with a longer focal length may now cover the chip thereby increasing the field of view as seen by the camera. This also has the effect of increasing the photon count per unit area of the image. Conversely a longer focal length (with the same size aperture) will spread the image out at the focal plane, therefore increasing the magnification of the image. For small circular objects such planets this is particularly of benefit because the greater the magnification at the focal plane the more detail can be teased out by the ccd/cmos chip.

For eyepieces, a much larger EP focal length would be required to see more of the larger image of the longer focal length scope. Put the same large focal length EP into a faster scope with a smaller prime focus image and much much more of the image will 'fit' into the apparent field of view of the EP.

So to sum up, (briefly because I should be getting ready to go out ), every telescope irrespective of focal length will produce an image of the same area of sky - the field over which the objective will receive light is naturally the same. What differs is the size of the image produced at prime focus, and in turn this affects the size and focal length of the EP needed for the observer to 'fit' this image into the exit pupil of the EP.

I have of course ignored distortions such as coma etc arising as a result of the greater curvature of shorter focal length optics.

Please, please tell me this is correct, haha.

John

[FraserClarke]

Yep, sounds like you've got it.

The image scale vs focal length is just simple geometry. Going back to your example of parallel rays of light. There is a 'special' ray which goes exactly through the middle of the lens/mirror. This is known as the 'Chief Ray'. Because it goes through the middle of the optics, the optic does not affect it in any way (i.e. a concave mirror is flat in exactly the middle, so just bounces the Chief ray off as a flat mirror would). So, the Chief rays from two stars separated by a given angle go out of the mirror at the same angle (relative to each other) that they came into it at. The mirror focuses the rays from all other parts of the mirror into a point. The shorter the focal length, the 'quicker' it does this. The quicker the focus is formed, the less distance there is for the Chief rays to move apart, so the smaller the image scale is (see; simple geometry!). So, the image size depends entirely on focal length. The longer the focal length, the larger the image scale.

The amount of light in the image (in total) just depends on the size of the mirror (bigger mirrors collect more photons). But brightness is amount of light per unit area. So the bigger the image, the less light there is per unit area (for the same aperture size). If you work that out, you come out with the 'brightness' only dependent on F/ratio

[astoc]

Fantastic. Cheers for that! And now I see where the term 'fast scope' comes from, too.

Glad I don't have it backward.

John

[narrowbandpaul]

your last sentence teadwarf is spot on!...trying getting folk to believe you that aperture aint king or imaging an extended object.

going back to the first comment....think about an object, size D a distance f away from . what angle does it subtend.....?

well in the small angle approx angle= D/f, but if f is the focal length and D is the aperture, then angle=D/f = 1/F I(focal ratio)

if you like telescopes have a kind of fundamental field of view. in radians it is 1/F

i noticed this relationship a while back, but wasnt too sure of its meaning. Im still not, but the ratio 1/F is a unit of angle.....