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Optics: Resolution Problem

  1. Apr 16, 2007 #1
    1. The American Television Systems Committee (ATSC) sets the standards for high-definition television (HDTV). One of the approved HDTV formats is 720p, which means 720 horizontal lines scanned progressively - one line after another in sequence. Suppose the 720 lines are displayed on a television with a screen that is 28 inches high, and that the light coming from the screen has a wavelength of 505 nm. If the pupils of your eyes have a diameter of 5.5 mm, what is the closest you can be to the TV before seeing the individual horizontal lines? (This is the minimum acceptable viewing distance.)

    2. Rayleigh''s Criterion: theta min = 1.22 (lambda/D); L = (y/ (tan theta min))

    3. To find theta min, I did 1.22 ((505*10^-9)/1.36)/0.0055) and got 8.237 * 10^-5 rad. I divided the wavelength by 1.36 since that is the index of diffraction for the eye. Then, I plugged it into the L = equation: L= 0.7112 m /tan (8.237*10^-5) and got a final answer of 8634, which is incorrect.

    Where did I go wrong? Thank you!
  2. jcsd
  3. Apr 16, 2007 #2


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    I don't think you should be using y = 0.7112 m. You want to find the separation of two of the horizontal lines, which you can figure out because you know how many lines are in the 28" screen.
  4. Apr 16, 2007 #3
    I tried dividing 0.7112 m by 720 and got 9.8778e-4. I then plugged that into the L = y/ (tan theta min)) and got 11.99, which is still incorrect...
  5. Apr 16, 2007 #4


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    Are you sure about that diffraction coefficient you are using? Perhaps it is not necessary.
  6. Apr 16, 2007 #5
    I believe so. In the book it says to take into account in which medium the diffraction pattern is observed.
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