# How to estimate my eyepiece FOV?

As a follow-on to my question about seeing Jupiter and Saturn at the same time in my scope, I'd like to ask how I might determine the Field of View (FOV) of my eyepieces. More specifically, I'd like to determine their aFOV (Apparent Field of View?) because that is a configuration setting in Stellarium when you are using ocular view and would like to specify the specs of your eyepiece.

I have two eyepieces that came with my AstroMaster114 telescope. One is 10mm and the other is 20mm. Beyond that, I have no idea what type they are (Plössl? Ortho? Erfle?).

I'd also like to understand how this FOV/aFOV might be impacted by the use of my Celestron #93640 2x Barlow adapter.

Here are pictures of the eyepieces:

Celestron 20mm Erecting Eyepiece

Celestron 10mm eyepiece

• @uhoh you mentioned you might have an answer for me. – S. Imp Dec 19 '20 at 2:12
• yes I did, writing now, thanks for the other comment, will take about 20 minutes... – uhoh Dec 19 '20 at 2:20
• okay, 57 minutes :-) – uhoh Dec 19 '20 at 3:18

Random references from the internet:

As long as we need moderate accuracy only, there are several ways to do this.

The stated focal length of your eyepiece is approximate, 10 mm might really be 9.6 mm or 10.4 and they simply didn't bother with the details, each little lens' focal length inside the eyepiece will have some manufacturing tolerance. If your telescope's focal length is stated to be 1000 mm it might be 960 mm or 1040 mm and there'd be no reason for them to measure every mirror and give you it's own focal length to several digits of accuracy. So let's call these methods estimates +/-10%.

### Examination of the eyepiece itself.

At the "input" end of the eyepiece there will be a field stop, an aperture that defines the your field of view (FOV) in the sky. It's likely a metal ring with a circular hole in it.

If the ring has a diameger $$d$$ of 10 mm and your telescope's focal length $$f_T$$ is 1000 mm, then you can ignore all of the other math and say that your FOV is $$d/f_T$$ = 10/1000 = 0.01 radians or 0.57°. Technically it is

$$FOV = 2 \arctan \left(\frac{r}{f_T} \right)$$

where $$r=d/2$$ is the radius of the hole. That doesn't matter much here but it does next.

Since your question asks for the apparent FOV of the eyepiece, you can get it by remembering that the plane of the aperture is what the eyepiece of focal length $$f_E$$ focuses on with a conjugate at infinity (for your eye to refocus to your retina), so the eyepiece's apparent FOV (aFOV) is given by

$$aFOV = 2 \arctan \left(\frac{r}{f_E} \right).$$

IF for some reason your eyepiece didn't have an aperture stop, you could bring a ruler into focus when looking through it and trying to focus your eye at infinity (perhaps by keeping both eyes open and focusing across the room with the other eye) and count the number of millimeters visible across the diameter. This doesn't sound easy though.

### Looking through the telescope

At night you can measure the time it takes for a star to move across the FOV of your eyepiece and get the real FOV of the system, then divide by the telescope's focal length $$f_T$$ to get the apparent FOV (aFOV). The Earth rotates 360° degrees on its axis every 23 hours 56 minutes 4 seconds and change (86164 seconds), so at a declination of 0 stars move 0.00418°/sec or 0.251°/minute. If you use something at a different declination, just multiply those speeds by $$\cos(dec)$$ to slow it down.

This gives you the real FOV of your system, again multiply by the magnification ($$f_T/f_E$$) to get aFOV.

You can also just look at something of a known or measurable or estimatable size far away, like a house or yardstick or meterstick to estimate the system FOV and multiply by magnification to get eyepiece FOV.

• Thanks for this. It will take some digesting! – S. Imp Dec 21 '20 at 21:47