Question

A standard 14.16-inch (0.360-meter) computer monitor is 1024 pixels wide and 768 pixels tall. Each pixel is a square approximately 281 micrometers on each side. Up close, you can see the individual pixels, but from a distance they appear to blend together and form the image on the screen.

Assuming that the screen is sufficiently bright, at what distance can you no longer resolve two pixels on diagonally opposite corners of the screen, so that the entire screen looks like a single spot? Note that the size (0.360 meters) quoted for a monitor is the length of the diagonal.

lambda = 550 nanometer
pupil diameter = 3.10 mm

Answer 1

arc length = R * central angle
0.360 = R x (0.7)(291)(10^-6)

= R x 2 x 10^-4
R = 0.360 / (2 x 10^-4)

= 0.18 x 10^4 = 1800
@ = wavelength / diameter

wave length = 550 x 10^-9

diameter = 3.1 x 10^-3

therefore  @ = 177.4 x 10^-6 radians 

Answer 2

So using arc length = R x central angle, we have:

0.360 = R x (0.7)(291)(10^-6) = R x 2 x 10^-4
Figure about 0.7 arc minute for the resolution of the human eye:
one arc minute is about 291 micro-radians:

so R = 0.360 / (2 x 10^-4) = 0.18 x 10^4 = 1800

You can argue about a factor of 2 (one cycle vs 3/2 cycle, etc.) but this should be the right order of magnitude.

For Rayleigh criteria, I assume you used the formula:
@ = wavelength / diameter

for wave length = 550 x 10^-9 and diameter = 3.1 x 10^-3
this gives @ = 177.4 x 10^-6 radians

Answer 3

So using arc length = R x central angle, we have:

0.360 = R x (0.7)(291)(10^-6) = R x 2 x 10^-4

so R = 0.360 / (2 x 10^-4) = 0.18 x 10^4 = 1800

You can argue about a factor of 2 (one cycle vs 3/2 cycle, etc.) but this should be the right order of magnitude.

For Rayleigh criteria, I assume you used the formula:
@ = wavelength / diameter

for wave length = 550 x 10^-9 and diameter = 3.1 x 10^-3
this gives @ = 177.4 x 10^-6 radians
Since this is very close to the 180 micro-radian figure I used, it is clear that your problem is not the formula but your application of it.