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How can I determine the diameter of the image circle (i.e. the diagonal size of CCD required to fully utilise a telescope's optics) at near-infinity focus? I have looked around online and found this calculator, but I am more interested in how the figure was calculated than knowing the figure itself.

Background: I am new to astrophotography and recently bought a Celestron Nexstar 130 SLT telescope with a Barlow lens, T-adaptor and T-ring to connect it with my Nikon D80 to take photographs of the night sky in prime focus. The telescope dimensions are 130mm aperture and 650mm focal length. With the 2x Barlow lens (required to diverge the focus enough to mount the camera outside the optical tube assembly) it effectively becomes a telescope with 1300mm focal length. I can measure a picture of an object with known angular size (e.g. the moon) and use crop factors to establish a ratio, but this method doesn't help me understand how a bigger/newer telescope would fare on the same CCD.

Edit: Since this question now has a bounty, I would like to emphasize the real question is How to calculate the size of the image circle at infinity focus?, and not the follow-up question of Would the image circle size decrease if I was able to ditch the Barlow lens?

share|improve this question
A barlow lens is really just a magnifying lens. It takes a smaller portion of the center of the image circle of whatever the telescope is resolving, and magnifies it by its order (2x, 3x, etc.) It could be thought of as akin to a teleconverter (TC) in normal photography...same basic effect, and the image circle does not change when adding or removing a TC. – jrista Dec 10 '12 at 20:49
Is that so? Having never owned a full frame camera I've not had the chance to play. I know with a DX lens on an FX camera the image circle is reasonably well defined. Are you saying that if I added a TC to that mixture that the circle would remain the same size, just the contents would be 2x as big? – Jono Dec 10 '12 at 23:32
Yes, exactly. You need not restrict yourself to using a TC on FF, though. I use both 1.4x and 2x TC's on my 7D's APS-C sensor. A TC is pretty much the same concept as a barlow lens...they are just a magnifying glass. They bend the light produced by the base apparatus, be that a telescope or a camera lens, in such a way that a smaller circular area within the original image circle is enlarged. There isn't really any magickery about it, and I don't know of any inherent trait of either a barlow or TC that would cause them to reduce image circle size. – jrista Dec 11 '12 at 0:28
@jrista, I wasn't trying to say that the Barlow lens reduced the image circle size, rather the opposite. I do understand your point of enlarging a smaller central portion of the image circle by 2x, but don't see why it's necessary to crop the field of view by 1/2 (surely you'd just add a 2x bigger CCD i.e. 4x area and you'd be back to capturing your original field of view?) – Jono Dec 14 '12 at 0:40
Oh, I'm sure there are a few reasons why it's not practical. For one, to enlarge by two fold with such a small device, and not lose your FoV, you would be bending light a LOT in the corners. It would be pretty tough to maintain clean rectilinear is probably possible to maintain the full FoV...but they are going to be blurry, probably exhibit a lot of optical aberrations, and generally not be very usable. You have to trade something off for the increased magnification... – jrista Dec 14 '12 at 2:42
up vote 1 down vote accepted

You can think of a lens focused at inifinity as simply a pinhole for the purpose of figuring out what gets projected where. Consider this simplified diagram of a camera:

The box is the camera, the fat line on the left the image plane (where the sensor or film is), and the small hole at right represents the effective point thru which the lens projects the outside view onto the image plane. In this drawing, F is the focal length and S is the size of some projection onto the image plane. From basic high school geometry, the tangent of the view angle is S/F:

  tan(Ang) = S/F

Therefore obviously

  Ang = arctan(S/F)

Yes, it really is that simple. You didn't say how big the sensor is in your camera, so I'll use a 35mm frame, which is 36x24mm. The largest circle it can fit is therefore 24mm in diameter or 12mm in radius. You say your final effective focal length is 1300mm. Arctan(12mm/1300mm) = 0.529°. That was the angle from the center to the edge, so double that to make the total view angle of the largest circle on your sensor, which is 1.06°.

To put this in perspective, the view angle of the moon is about 0.5° (it varies, but this is within its variation). That means for a 1300mm lens and a "35mm" sensor, the moon will fill about half the height of the image.

For a cropped sensor, it would appear bigger. Actually, the projection of the moon onto the sensor is the same regardless of how big the sensor is, but for a smaller sensor it takes up a proportionately bigger portion of the frame. This would be reflected in the equations above in that the 12mm value would be smaller, since this was the radius of the largest circle that could fit into the sensor frame.

Also note that the simplified geometry of a camera diagrammed above works for the effective focal length. This will be the actual focal length when the lens is focused at infinity, but can be different when the lens is focused close, depending on the lens design. However, you are asking about astrophotography, so the focus will always be at infinity and I won't go into that further.

share|improve this answer
OK, now that's the same formula that's in Wikipedia's angle-of-view entry (just rearranged). Celestron give 1.7 degrees as this scope's FOV in the specs, which when plugged into 2*TAN(a/2)*f gives me 38.6mm (a number smaller than the 35mm FX diagonal of 43.3mm, indicating vignetting on full frame CCDs). I tried shooting the moon with my own DX (23.6mmx15.8mm) and expected it to fill only 76% of the height but it was actually too big to capture completely. Somewhere, the formula and figures aren't matching reality. – Jono Dec 13 '12 at 23:58
Another thing, @jrista's comment states that a Barlow lens doesn't affect image circle size. However, that would imply we must use the shorter focal length of 650mm in these formulas, which gives us an image circle small enough to cause vignetting on even my DX CCD (and I didn't see any, even when shooting in daylight). – Jono Dec 14 '12 at 0:17
Keep in mind, telescopes can usually be used with focal reducers as well. The only way that is possible is if the entrance pupil is capturing a much larger AoV of the sky than would be indicated without using a focal reducer. I believe most quality telescopes offer a lot of latitude when it comes to "content" to fill up the image circle, either focal reduced or focal extended. And in specific terms, even at "infinity" (which is a concept that is difficult to actually achieve in reality), a telescope is not actually a pinhole camera. – jrista Dec 14 '12 at 4:09
The NexStar telescope you are using actually has a 130mm aperture, which is quite large...orders of magnitude larger than the 1mm or smaller aperture you would have with an actual pinhole. There are plenty of blogs and articles on the net, specifically with astrophotography, that clearly demonstrate that even with the immense distances of stars in the night sky, light from each star is NOT parallel, so the math is only an approximation with a certain degree of error. – jrista Dec 14 '12 at 4:15
The optics, angles, and math with a telescope are a little different than when simply using a camera. Remember that with a reflector, the telescope is not actually a lens...its just a tube with a given aperture that gathers light and moves it around. Reflecting telescopes ...project, rather than bend like a lens. An actual lens only comes into play at the eyepiece end... – jrista Dec 14 '12 at 4:17

I am not sure about the answer to Would the image circle size decrease if I was able to ditch the Barlow lens?

But if you look at the JavaScript of that page you will see

var sensorw = "Sensor Width"
var sensorh = "Sensor Height"
var maxres  = "Max Res"
var focleng = "Focal Length"

var thisF = sensorw * 3438/focleng;
var thisF2 = sensorh * 3438/focleng;
var thisF3 = sensorw * 3438/focleng * 60/maxres;
var thisF4 = focleng/Math.sqrt(sensorw * sensorw + sensorh * sensorh);


  • thisF = Arc Min of Sky- Width:
  • thisF2 = Arc Min of Sky- Height:
  • thisF3 = Arc Seconds/Pixel:
  • thisF4 = Magnification (X):

This is how that page calculates it

share|improve this answer
Yeah, I saw that. Not sure what the constant of 3438 arcsec represents (i.e. 57.3 arcmin or 0.955 degrees). – Jono Dec 10 '12 at 19:06

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