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![[Post New]](/templates/default/images/icon_minipost_new.gif) 7 Dec 2007 13:59:53 IST
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The width of a slit is .012 mm Monochromatic light is incident on it The angular position of the bright line is 5.2 deggrees The wavelenth of the incident light is Ans 7248 A Please give detailed solution rates assured for best ans
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8 Dec 2007 20:12:17 IST
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i think you mean width b/n slits is 0.012*10^-3
ok, let y be dist of angular position from centre point.
now dsin@ = n*lamda = 1.08*10^-6 = dtan@ = y/D
also sin@ nearly = tan@. (by calc it is .99)
but you have not provided D. i think its required...........
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13 Dec 2007 10:56:22 IST
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where are you?
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13 Dec 2007 11:32:27 IST
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The line refers to bright line of 1st order.
So, for a bright line in single slit diffraction,
asin5.2 degrees = 3k/2 (k = wavelength)
So, k = 2 x 0.012 mm x 0.09 / 3
= 7250 Angstroms,
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15 Dec 2007 02:47:02 IST
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thanks
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