True Field Of View

I was calculating the true field of view for each of my eyepieces using this formula.

True Field of View = Magnification / Apparent Field of view

Example 32mm Plossl with 50 degrees of apparent field of view
Scope’s focal length 1250mm.
1250 / 32 yields a magnification of 39.06. Plug the magnification into the formula above.
1.28 = 39.06 / 50 True field of view is 1.28

And then I realized does this formula apply to 2 inch eyepieces? Because if you had 2 32mm eyepieces and both had the same exact apparent field of view but one was 2 inches and the other 1 1/4” they both give you the same true field of view. But shouldn’t the 2 inch eyepiece have a great true field of view? I know the above formula is not precise for calculating this value but it should be close. If I am missing something please excuse the ignorant question. Thanks

Justin

Justin
From what little I know, I think the only thing that determines TFOV is the field stop diameter of ep and focal length of scope?
Craig

Your formula applies equally well to 2 inch and 1.25 inch eyepieces. It should be stated that your formula is technically only roughly correct, but is usually within a few % of the true value (most eyepieces have some degree of curvature or distortion of the field that renders your calculation not quite right...that said, it's what I use).

What really determines the field of view is the "field stop diameter", as explained in more detail by Clyde. If you have a 32mm Plossl with 50 degree apparent field, it's going to have the same field stop regardless of the barrel size. But you'll notice that 1.25 inch 40mm Plossls typically have about a 42degree apparent field of view, instead of the usual 50-52. That's because the 1.25inch barrel itself is acting as the field stop. You can get a 50 degree field in a 40mm Plossl, but you have to increase the barrel size.

Maybe that clears things up. Not my clearest explanation ever, but I gave it a shot.

Justin De Guire said:

I was calculating the true field of view for each of my eyepieces using this formula.

True Field of View = Magnification / Apparent Field of view

Example 32mm Plossl with 50 degrees of apparent field of view
Scope’s focal length 1250mm.
1250 / 32 yields a magnification of 39.06. Plug the magnification into the formula above.
1.28 = 39.06 / 50 True field of view is 1.28

I think that this is inverted. The way you have it, the higher the magnification, the larger the true FOV. It should be:

True Field of View = Apparent Field of view / Magnification

You were actually doing this in your calculations and didn't realize it. 8)

Best regards,
Doug