TV resolution and the Rayleigh criterion with 485 lines

[Ownaginatious]

Homework Statement

I'm not sure what I'm messing up here:

"An American standard television picture is composed of about 485 horizontal lines of varying light intensity. Assume that your ability to resolve the lines is limited only by the Rayleigh criterion and that the pupils of your eyes are 5.13 mm in diameter. Calculate the ratio of minimum viewing distance to the vertical dimension of the picture such that you will not be able to resolve the lines. Assume that the average wavelength of the light coming from the screen is 570 nm."

Homework Equations

diameter of pupils = d = 5.13*10^-3 m
wavelength = r = 570 * 10^-9 m

Now I think I'm supposed to use this equation:

(theta) = 1.22 * (wavelength)/(Diameter)

The Attempt at a Solution

I'm not really sure where to go from here, and can't figure out how the 485 horizontal lines plays into this. Can someone guide me through this one? I've never had a problem in my homework like this before.

Thanks!

- Dillon

 

[rl.bhat]

Vertical dimension of the picture tube / numbers of lines = the distance between two lines which is to be resolved.

 

[Ownaginatious]

Vertical dimension of the picture tube / numbers of lines = the distance between two lines which is to be resolved.

But there is no vertical dimension of the screen :s

 

[rl.bhat]

The answer may be in the form of d/D

 

[Ownaginatious]

The answer may be in the form of d/D

But what would the upper-case D represent?

 

[rl.bhat]

Vertical dimension of the picture tube.

 

[Ownaginatious]

I'm still unsure of how to solve the problem...

 

[rl.bhat]

In this problem theta = (D/485)/d = 1.22(wavelength/diameter)

 

[Ownaginatious]

Okay, great, that yields a correct answer :)

If you don't mind rl.bhat, could you show me how you derived that (D/485)/d?

EDIT:

Oh wait, never mind, I just realized that you just used s = (radius)*(theta). Thanks again!