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deeprony7
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in YDSE, 10 fringes are observed to be formed in a certain segment of the screen when light of wavelength 400nm is used. if the wave length of light is changed to 200nm, the number of fringes observed in the same segment of the screen is?
 3 years ago
 3 years ago
deeprony7 Group Title
in YDSE, 10 fringes are observed to be formed in a certain segment of the screen when light of wavelength 400nm is used. if the wave length of light is changed to 200nm, the number of fringes observed in the same segment of the screen is?
 3 years ago
 3 years ago

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BrilliantMinds Group TitleBest ResponseYou've already chosen the best response.0
d sin(theta) = n(l), where l = wavelength d (y/x) = n(l), where y is distance between central fringe and nth order fringe y = (x/d)n(l) You can see that the y is proportional to (l). So when the wavelength used is halved, the fringe pattern will be 'squeezed' to the center, so the number of observed fringes doubles! Since you can squeeze twice the amount of fringes into the same distance. This is also not surprising and intuitive because as the wavelength decreases, the magnitude of optical path difference n(l) will be of a smaller value!
 3 years ago

ac7qz Group TitleBest ResponseYou've already chosen the best response.0
Since we start measuring (x = 0) at the first fringe, n = 9, not 10. If the wavelength is halved, and we still start our measurement at the first fringe, the last fringe in our space will be number 19.
 3 years ago
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