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In biprismexperiment, 10th dark band is observed at 2.09 mm from the centralbright point on the screenwith red light of wavelength6400 Å. By how muchwill fringe width change if blue lightof wavelength 4800Å is used with thesame setting ? |
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Answer» Solution :In the biprism experiment, the `10^(th)`dark band isobserved.The distancebetween the`m^(th)`dark band with the central bright band is `x_(m) = (2m-1) (lambda D)/(2d)` Therefore, the DISTANCEFOR the `10^(th)`dark band is `x_(10) = ((2 xx 10)- 1) (lambdaD)/(2d) = (19lambdaD)/(2d)` Now, when red light is used, we have `(x_(10))_r = (19 lambda_(r)D)/(2d) "....."(i)` Similarly, for blue light, we have `(x_(10))_(b) = (19 lambdaD)/(2d) "....."(ii) ` Now, the fringewidth is `x = (lambdaD)/(d)` `:.` For red light, `x_(r) = (lambda_(r)D)/(d) "....."(iii)` `:.` For blue light, `x_(b) = (lambda_(b)D)/(d) "...."(IV)` From equations (i)and (iii), we get `(x_(10)) = (19x_(r))/(2) = 2.09mm` `:. x_(r) = (2 xx 2.09)/(19) = 0.22` mm Dividing equations (i) and (ii), we get `((x_(10))_(r))/((x_(10))_(b)) = ((19 lambda_(r)D)/(2d))/((19 lambda_(b)D)/(2d)) = (lambda_(r))/(lambda_(b))` `:. (x_(10))_(b) = (lambda_(b) xx (x_(10))_(r))/(lambda_(r))` `= (4800 xx 10^(-10) xx 2.09)/(6400 xx 10^(-10))` Now, from equations (ii) and (iv), we get `(x_(10))_(b) = (19x_(b))/(2) = 1.57 mm` `:. x_(b) = (2 xx 1.57)/(19) = 0.165 mm` Therefore, thechange in the fringe width when blue lightis used INSTEAD of red is `x_(r) - x_(b) = 0.22 - 0.165= 0.055 mm`. |
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