Does anyone know how the acuity of your vision translates to a difference in limiting magnitude?

e.g., the kind of answer I'm looking for would be "For each factor of 2 improvement in your vision (20/80 to 20/40, 20/40 to 20/20, 20/20 to 20/10, etc., your personal limiting magnitude for point sources increases by 1, independent of seeing or sky brightness."

(except I just made those numbers and conditions up out of thin air)

I also implicitly introduced the ansatz of a smooth logarithmic dependence. Yes, no, approximately so?


There are two causes of limiting magnitude. First is pure sensitivity, where no matter how dark the sky is, one can't see stars fainter than about 6.5. If your vision loses acuity and blurs things out, it may not at first have much effect on limiting magnitude, especially if you use averted vision where things are not seen that sharply anyway. Of course, enough blurring spreads the signal over too many of your retina's rods and cones causing you to no longer be able to identify the star due to 'eyenoise'. If you have light pollution either natural (Moon, twilight) or artificial, things get bad much faster when you blur things and your eye can no longer see the signals because of low contrast.

I know it's not giving you a good answer because the situation can be complex. But you can easily do an experiment yourself. Get drugstore eyeglasses in various strengths (+1, +1.5 diopters, etc.) and look at a part of the sky where the magnitudes are well known, e.g. the bowl of the Little Dipper has stars of magnitude +2, +3, +4, and +5.

Translating to the values on an eyechart can be ambiguous because making out letters is a completely different process than seeing limited magnitudes. I have noticed several different conversion tables while searching. You might want to do an experiment with your own eyes by looking at standard eyecharts with the same glasses you used to estimate the limiting visual magnitudes.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.