How to calculate the size of the image circle at infinity focus?

Question

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?

Answer 1

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.

Answer 2

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);

Values:

• 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