- #1
fubag
- 105
- 0
[SOLVED] Resolving pixels on a computer screen
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.
If the maximum distance between the screen and your eyes at which you can just barely resolve two adjacent pixels is 1.30 meters, what is the effective diameter d of your pupil? Assume that the resolvability is diffraction-limited. Furthermore, use lambda = 550 nanometers as a characteristic optical wavelength.
I used Rayleigh's criterion which states (theta_minimum) = (lambda/a), where a is the width of the slit.
Then (theta_min) = (1.22(lambda))/D, where D is the aperture diameter.
Using "a" as 281 micrometers, i tried solving and got D = .34mm which doesn't seem right nor work.
Homework Statement
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.
If the maximum distance between the screen and your eyes at which you can just barely resolve two adjacent pixels is 1.30 meters, what is the effective diameter d of your pupil? Assume that the resolvability is diffraction-limited. Furthermore, use lambda = 550 nanometers as a characteristic optical wavelength.
Homework Equations
I used Rayleigh's criterion which states (theta_minimum) = (lambda/a), where a is the width of the slit.
Then (theta_min) = (1.22(lambda))/D, where D is the aperture diameter.
The Attempt at a Solution
Using "a" as 281 micrometers, i tried solving and got D = .34mm which doesn't seem right nor work.