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When a thin flake of mica (n_(m) = 1.58) covers one slit of a double - slit interfernce setup, the fringe pattern shifts byseven fringe widths. If the wavelength of the light used is5500Å, what is the thichness of the mica flake ? |
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Answer» Solution :Data : `n _(m)= 1. 58, lambda = 5.5 xx 10^(-7) m, x_(0) = 7 X` `x_(0) = D/d (n_(m) - 1) b, "" X = (lambda D)/d` where `x_(0) -=` the lateral SHIFT of the central bright fringe, `b -=` the THICKNESS of themica flake of refractive index `n_(m), d -=` theseparation between the slits, `D -=` the slit - to - SCREEN DISTANCE, `X -=` the fringe width and ` lambda -=` the wavelength of light . Since `x_(0) = 7 X`, ` D/d (n_(m) -1) b = (7 lambdaD)/d` ` :. b = (7 lambda)/(n_(m) -1) = (7(5.5 xx 10^(-7)))/(1.58 -1)` ` = (3.85 xx 10^(-6))/(0.58)` `= (38.5 xx 10^(-6))/(5.8) ` ` = 6.639 xx 10^(-6) m` ` = 6.639 mu m` `{:(log 38.5,," "1.5855),(log 5.8,,ul(-0.7634)),(,," "0.8221):}` AL ` 0.8221 = 6.639` |
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