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University of Wisconsin - Milwaukee

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Problem 61 Hard Difficulty

An American standard analog television picture (non-HDTV), also known as NTSC, is composed of approximately 485 visible horizontal lines of varying light intensity. Assume your ability to resolve the lines is limited only by the Rayleigh criterion, the pupils of your eyes are 5.00 mm in diameter, and the average wavelength of the light coming from the screen is 550. nm. Calculate the ratio of the minimum viewing distance to the vertical dimension of the picture such that you will not be able to resolve the lines.

Answer

15.4

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Video Transcript

so we have the number of visible horizontal lines in the NTSC. TV is a nickel's for 85. Let's say H is the height of the TV. Then the spacing between the lines in the TV would be equals h over M, which is gonna be 80/45 Now when the screen is viewed from a distance off our let's say our is this tens between the screen and do you observer All right, so this is the distance between the screen and the observer. Then the angular separation between the two lines those two hoods and alliance would be did, uh, because g over our all right. And if the lines are to be unresolved than there must be less than the minimum angle of resolution required by the rally kept Criterion. So what that means is data has to be less than that. I mean, that is required by the criterion to be able to resolve something else. So this actually means d over our has to be less than 1.2 to Lambda over the here lam raise the wording of the light being used. He's the fritter diameter, and in this case, the purser diameter is five millimeter. It is given we're going to write down in terms of meter five times 10 to the negative three meter, and the Lambda is also given. So Lamda is also given, which is 515 at a meter. 5 50 Nana meter. This is three and that is 10 to the negative night. All right, so let's go to next pitch. So from here, from the first piece, we can write down our over H has to be greater than D over 4 85 times 1.22 times lambda. So this valley here is fairly Here is equality five times 10 to the negative three, which is a capital D and 4 85 times one point 22 times 10 to the negative nine. Sorry, 1.22 And this is 5 50 times into negative. Nine, 5 50 times tended. They get nine, and that means are over. Age has to be this fat crater than this value. The value is 15.4

University of Wisconsin - Milwaukee
Top Physics 103 Educators
Marshall S.

University of Washington

Farnaz M.

Other Schools

Aspen F.

University of Sheffield

Meghan M.

McMaster University