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?

  • \$\begingroup\$ 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. \$\endgroup\$
    – jrista
    Commented Dec 10, 2012 at 20:49
  • \$\begingroup\$ 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? \$\endgroup\$
    – Jono
    Commented Dec 10, 2012 at 23:32
  • \$\begingroup\$ 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. \$\endgroup\$
    – jrista
    Commented Dec 11, 2012 at 0:28
  • \$\begingroup\$ @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?) \$\endgroup\$
    – Jono
    Commented Dec 14, 2012 at 0:40
  • 1
    \$\begingroup\$ 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 results...it 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... \$\endgroup\$
    – jrista
    Commented Dec 14, 2012 at 2:42

4 Answers 4


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.

  • \$\begingroup\$ 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. \$\endgroup\$
    – Jono
    Commented Dec 13, 2012 at 23:58
  • \$\begingroup\$ 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). \$\endgroup\$
    – Jono
    Commented Dec 14, 2012 at 0:17
  • \$\begingroup\$ 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. \$\endgroup\$
    – jrista
    Commented Dec 14, 2012 at 4:09
  • \$\begingroup\$ 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. \$\endgroup\$
    – jrista
    Commented Dec 14, 2012 at 4:15
  • \$\begingroup\$ 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... \$\endgroup\$
    – jrista
    Commented Dec 14, 2012 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

  • \$\begingroup\$ Yeah, I saw that. Not sure what the constant of 3438 arcsec represents (i.e. 57.3 arcmin or 0.955 degrees). \$\endgroup\$
    – Jono
    Commented Dec 10, 2012 at 19:06

I think, the root problem was ... the distance between the Barlow and the DLSR. I think, the Barlow-to-DSLR distance was excessive and the barlow was actually operating at 3X ( or more ) instead of the expected 2X. The imaging system had way too much effective focal length. And that is why the image of the moon would not fit with the image sensor of the DSLR. Shortening the distance between the barlow and the DSLR, would allow the image of the moon to fit per the math formulas.


3458 is to convert from radians to arc-minutes 180/pi (deg/radian) * 60 minutes/degree = 3,437.7467707849398 3438 is the common rounded used number.

23.6mmx15.8mm sensor use smaller for circle to fit is 15.8 mm divide by two for the angle is 7.9 mm Tan-1, Arctan(7.9/1300) = 0.3481778° multiply by 2 is now 0.6963555°

alternatively working in arc-mins ( the calculator, avoids trig functions based on sin Ø ~= Ø (rads) when angles are under 1°, (cosØ~=1) this is the assumption the calculator works with to get approx close calc)

sensor height x convert radians to arcmins (180/pi*60) / focal length gives 1300 mm x ~3438 / 1300mm = 41.78184536800157 arc mins ( calc rounded to 41.78492307692308 arc min (close enough)) 41.7818 / 60 = 0.69636408946669° ~same as working with trig

you are missing a factor of approx 2x somewhere. It is probably in the barlow now is acting as a lens ( telescope is lens #1, barlow #2, and so your distance from this second lens becomes a magnifying effect, like in a view camera where you move the camera lens by extending the bellows and can macro/magnify what the lens does at infinity.

here is the math: F = focal length of objective or primary f = focal length of Barlow [1] J = joint focal length (effective focal length) d = separation of Barlow and original focal plane (objective focal plane) x = separation of barlow and new focal plane (eyepiece focal plane) M = amplification of Barlow J = (F×f)/(f-d) ...(1) (combined lens formula) *********** M = J/F ...(2) (by definition) = f/(f-d) The separation of the Barlow and the new focal plane can be calculated from M and f: x = f×(M-1) ...(3) ...from which we get : M = 1 + (x/f) ok math ( I am assuming you amount adapter is going to cause a 2x magnification BEYOND the 2x of the barlow itself)

  1. Based on Separation of Eyepiece Focal Plane and Barlow

Let us take a 75mm focal length x2 (nominal) Barlow used at its designed amplification. (f = 75mm, M = 2)

M = 1 + (x/f) δx = f(M - 1) = 75(2 - 1) mm = 75mm

This relationship (the separation of Barlow and the new focal plane is equal to the focal length of the Barlow) holds for any x2 Barlow.

Let us now use the old trick of increasing Barlow amplification by inserting an extension tube to the camera, the t-mount, (working here much like a star diagonal between the eyepiece and Barlow), is between the camera and the barlow. Assume that the extensions adds 150mm to the optical path. ('guessed' from 75*3)

M = 1 + (x/f) = 1 + (75 + 150)/75 = 4.0

i.e. a nominal x2 Barlow has become an x4 Barlow.

  1. Based on Separation of Objective Focal Plane and Barlow.

Let's take a 130mm f/10 objective (F = 1300mm) with a 75mm focal length Barlow (f) placed 56.25mm inside focus (d). ('guess' 56.25 from 75/4 = 18.75, 75-18.75 = 56.25)

Substituting in equation (1): J = (F×f)/(f-d) = (1300 × 75)/(75-56.25) mm = 5200mm

Substituting in equation (2): M = J/F = 5200/1500 = 4

Hence we have an amplification factor of ×4.

Substituting in equation (3): x = f×(M-1) = 75 × (4 - 1) mm = 225mm

I hope that explains how the t-mount tube has changed your math your barlow is effectively 225mm by this 56.25mm (2") extension.

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    \$\begingroup\$ I'm sorry, but this near unreadable, and I know the maths involved. \$\endgroup\$
    – scottbb
    Commented Nov 21, 2016 at 21:01

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