How Close Can You Get to a 720p TV Before Seeing Individual Lines?

In summary, the American Television Systems Committee sets the standard for HDTV, with one of the approved formats being 720p. To find the minimum acceptable viewing distance for this format, the Rayleigh's Criterion is used. To calculate this, the wavelength of light from the screen (505 nm) is divided by the diffraction coefficient for the eye (1.36) and then plugged into the equation L = (y / tan theta min). However, there seems to be some discrepancy with the final answer and further calculations are needed.
  • #1
kmj9k
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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!
 
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  • #2
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.
 
  • #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...
 
  • #4
Are you sure about that diffraction coefficient you are using? Perhaps it is not necessary.
 
  • #5
I believe so. In the book it says to take into account in which medium the diffraction pattern is observed.
 

FAQ: How Close Can You Get to a 720p TV Before Seeing Individual Lines?

1. What is the resolution problem in optics?

The resolution problem in optics refers to the limitation of a lens or optical system to accurately distinguish between two closely spaced objects. This is due to the diffraction of light, which causes the objects to appear blurry and indistinguishable.

2. How does the size of the lens affect the resolution problem?

The size of the lens directly affects the resolution problem. A larger lens has a greater ability to collect more light and produce a clearer image, resulting in a higher resolution. However, even with a large lens, the resolution is still limited by the wavelength of light.

3. Can the resolution problem be overcome?

While it cannot be completely overcome, the resolution problem can be improved by using techniques such as increasing the lens size, using multiple lenses, and using different wavelengths of light. Additionally, advancements in technology have led to the development of instruments such as electron microscopes which have higher resolution than traditional optical microscopes.

4. How does the numerical aperture affect resolution?

The numerical aperture (NA) of a lens is a measure of its ability to gather light. A higher NA results in a smaller diffraction limit and therefore, better resolution. NA is dependent on both the refractive index of the medium and the angle of light entering the lens.

5. What is the difference between resolution and magnification?

Resolution and magnification are two different aspects of an optical system. Resolution is the ability to distinguish between two closely spaced objects, while magnification is the ability to make an object appear larger. A system can have high magnification but low resolution, or vice versa.

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