Finally...I did it.

The question is:

If I focus a pair of binoculars at something very far away (~infinity),

how close can I look and still see something this is sharp?

In the perfect infinity case, the objective makes an image at the focal length,

the eyepiece's focal plane mates up to that, and you see a virtual image

at infinity (your eyes relaxed)... So...calculate two relays of the object..

It does depend on your eyes what happens next, but I chose the standard

of 0.5 meters for the closest virtual image I focus clearly without eye strain...

So...my program being written,

here is a table of the effect of power (mostly) on

the 'near-point' of a binoculars' range:

(assume: Fl(obj) = 3.7*Dia , 0.5m closest tolerable final image, focus set at infinity, relaxed eyes)

POWERxObjective Near-Range Limit (0.5m virtual image distance)

-------------------------- -------------------------

6x30 17meters

7x35 , 7x50 23m

8x30 31m

8x40 30m

10x50 48m

15x70 108m

20x80 194m

If you want a less claustrophobic, more-exacting standard,

here are some ranges for a 0.7m virtual image standard:

POWERxObjective Near-Range Limit (0.7m virtual image distance)

-------------------------- -------------------------

6x30 24

7x50 33m

10x50 86m

20x80 274m

Other than figuring two imaging sets,

the real tricky part was solving for the desired virtual distance

(of the final image). Discrete math to the rescue! I just tried

every distance from 1000meters down to 1 until the virtual image

popped below 0.5 (or 0.7).... voila!

A bit brute-force, but no cumulative errors, and the PC has lots of power to

repeat things.

**Edited by MartinPond, 12 July 2018 - 12:07 PM.**