The visual acuity page on wikipedia suggests 1.8 arcmin @ 1km away (but says nothing about the relationship expected between distance and angular resolution) whilst the page on angular resolution suggests that 1 arcmin is perfectly normal, and I have not been able to find many other sources.

I'd like to know what the value is.

As an aside, is there any great source of freely available information on eyes and ophthalmology that could have avoided me asking the question at all?

Thanks in advance.


In general, resolution varies with distance, but when expressed as angular resolution this takes distance into account; the minimum distance between two points that can be resolved will increase in proportion to distance for a given angle.

The retina is the light sensitive part of the eye. It is packed with photoreceptor cells known as rods (more sensitive and useful in low-light conditions but of lower resolution) and cones (less sensitive so need greater light intensity, but higher resolution). Only cones confer colour vision, whereas rods confer achromatic vision. This is why colour is difficult to perceive in low light intensities.

The fovea is the highest resolution part of the retina. It is packed with cone cells, offering maximum resolution. Interestingly, their requirement for higher light intensity is why it is easier to see faint stars at night by looking slightly to the side of them, as the fovea is not sensitive enough.

This diagram demonstrates the resolution of the fovea.

enter image description here

When a normal eye can just resolve two points, the angle θ is 1/60th degree (or 1 arcminute). This corresponds to being able to read line 8 on a Snellen Chart, which is also known as 6/6 vision, or 20/20 vision in the USA (this refers to 6 meters or 20 feet, which are equivalent).

So angular resolution does not change with distance, but is best at the fovea and less at other parts of the retina. The maximum angular resolution of 1 arcminute corresponds to about 29cm at 1km distance.

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.