Eye Resolution: Calculating the Limit and Overcoming Obstacles

[physiks]

Homework Statement

The pupil of the human eye is normally around 5mm in diameter. A person with 20/20 vision
should be able to read letters 9mm high at a distance of 20m, while in Chapter two of
The Two Towers Legolas can resolve objects of around 10cm size at a distance of five leagues
(about 24km). Comment

Homework Equations

The Rayleigh resolution limit for a circular aperture (assuming small angles is) θ=1.22λ/w where w is the diameter of the aperture.

The Attempt at a Solution

Ignoring the second part for now, using the above equation, we can resolve two point objects by eye when their angular separation is 0.0001342rad.
I have assumed λ=550nm for visible light.

A letter 9mm high at a distance of 20m subtends an angle given by θ=9/20000=0.00045rad at the eye.

Now these angles are of the same order which seems to indicate I'm going in the right direction. However I can't quite work out what I should exactly do with them next because:
- I don't know the separation of two letters, only the height of a single letter.
- The letters aren't actually point objects.

Any clues? Thanks.

 

[BvU]

Your factor 3 is a nice clue: a person will have difficulty distinguishing R from B because that is of the order of a 5 mm detail, right?

Don't worry too much about specifics: resolving objects of 10 cm means you need a resolution that is better than 10 cm.

If you start worrying about details: point sources 10 cm apart require approximately the same resolution as 10 cm diameter discs 10 cm apart. The diffraction pattern isn't really as steeply declining, but I would ignore that difference. It doesn't look as if they want you to actually calculate that pattern.
Even if you give Lego the benefit of the doubt and use 20 cm, he'll need really big eyes...

 

[physiks]

Your factor 3 is a nice clue: a person will have difficulty distinguishing R from B because that is of the order of a 5 mm detail, right?

Don't worry too much about specifics: resolving objects of 10 cm means you need a resolution that is better than 10 cm.

If you start worrying about details: point sources 10 cm apart require approximately the same resolution as 10 cm diameter discs 10 cm apart. The diffraction pattern isn't really as steeply declining, but I would ignore that difference. It doesn't look as if they want you to actually calculate that pattern.
Even if you give Lego the benefit of the doubt and use 20 cm, he'll need really big eyes...

Oh I see - I need to worry about actually telling what the letters are rather than worrying about whether adjacent letters blend into one another. So because my resolution is roughly a third of the size of the letter, it pretty much covers any variance in a letter allowing the reader to tell which letter is which.

Dealing with the resolution of point sources makes it quite difficult for me to think about resolution of continuous objects. So 'resolving objects of 10 cm means you need a resolution that is better than 10 cm', how can I logically understand why that is the case considering point objects?

 

[BvU]

Going from pont sources to continuous objects means integrating. Integrating is subdividing into small chunks and adding up the contributions from all chunks at a point on the retina.

Integrating leads to a convolution. All a bit too involved for this exercise. If a point source ("delta function") gives an Airy pattern, fairly steeply falling off on two sides, can you imagine an extended source (a "step function") falling off with about the same slope on one side only ?