Resolving pixels on a computer screen

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Homework Help Overview

The discussion revolves around a problem involving the resolution of pixels on a computer screen, specifically focusing on a 14.16-inch monitor with a resolution of 1024 by 768 pixels. The problem requires determining the effective diameter of the pupil based on the ability to resolve adjacent pixels at a certain distance, using principles of diffraction and Rayleigh's criterion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the application of Rayleigh's criterion for both slit and circular apertures, questioning the appropriate expressions to use for a circular pupil. There are attempts to calculate the diameter of the pupil based on angular resolution and the distance to the screen.

Discussion Status

Participants are actively engaging with the problem, sharing their calculations and questioning the validity of their approaches. Some have identified mistakes in their reasoning, while others are clarifying the relationship between angular resolution and distance to the screen. Guidance has been offered regarding the correct use of formulas and the interpretation of parameters.

Contextual Notes

There is ongoing confusion regarding the application of the formulas and the definitions of variables, particularly in transitioning from resolving adjacent pixels to resolving pixels at the corners of the screen. The discussion reflects the challenges of applying theoretical concepts to practical scenarios.

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[SOLVED] Resolving pixels on a computer screen

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.
 
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Hi fubag,

Your first expression (theta_minimum) = (lambda/a) is Rayleigh's criterion for a slit; your second expression is the same thing for a circular aperture.

Since your pupil is circular all you need to use is the circular aperture expression (theta_min) = (1.22(lambda))/D. Here theta_min is the angle that two adjacent pixels make in your field of view when you're a distance 1.3 meters away.
 
ok so I used theta_min = atan(281micrometers/1.3m) and solve for D?

This gave me D = 3.10 mm.

Is this correct?
 
ok I got that part

Part 2:

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.

do i take theta to be really really small and use D = 3.1mm to find that theta, or use the .360 meters as a part of atan for finding theta?
 
ok still stuck on this one...

I understand I am using Rayleigh's criterion formula where (theta_min) = lamda / a . So I think I have to take a = .36m, since it is now the width of the slit or something.

and using the properties that tan (a) = opposite/adjacent, i want to solve for opposite which will give me the distance, using adjacent as what though? this is bothering me.
 
fubag said:
ok still stuck on this one...

I understand I am using Rayleigh's criterion formula where (theta_min) = lamda / a . So I think I have to take a = .36m, since it is now the width of the slit or something.

and using the properties that tan (a) = opposite/adjacent, i want to solve for opposite which will give me the distance, using adjacent as what though? this is bothering me.

It looks to me like this is really the same type of problem as part a; just the distance between the pixels has changed. The diffraction would still be due to the pupil of the eye, so you'll need to use the circular version of Rayleigh's criterion formula.
 
ok so I find (theta_min) = (550 * 10^-9) / (.36m), using a as distance between the two points.

then solve for D = (1.22(550 * 10^-9)) / (theta_min) ?

This gives me .44m which makes no sense.
 
ok i actually solved it now

found my mistake

thanks!
 
You already know the diameter of the pupil D; it won't change. What you are looking for is L, the length from you to the screen. That means you also know theta_min.

What's the relationship between theta_min, the ditance to the screen, and the distance between the pixels?
 
  • #10
fubag said:
ok i actually solved it now

found my mistake

thanks!


i don't get it... how'd you solve for the distance ? arctan doesn't work ..
 
  • #11


yeah i keep getting thetamin = 0.36/L
and answer 2029. what am I doing wrong?
 
  • #12


What is the value of thetamin (use the information in the problem statement)?
 
  • #13


i got theta min is 1.774 x 10^-4 from 550x10-9/3.1mm I believe. What's my mistake?
 
  • #14


would the theta min from 1.22 x 550 x 10-9 / 3.1mm be the right theta min then? I wuold get L=1660m for this theta min?
 
  • #15


yep, the right answer from the machine said the 1.22 x 550 x 10-9/3.1 one was correct? Thanks!
 
  • #16


minidee said:
would the theta min from 1.22 x 550 x 10-9 / 3.1mm be the right theta min then? I wuold get L=1660m for this theta min?
Looks good.

minidee said:
... Thanks!
You're welcome :smile:
 

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