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In Young's double-slit experiment, the fringe pattern shifts by a distance x_(0) when a mica sheet, 1.964 mu m thich and of refractive index 1.6, covers one of the slits. If the mica sheet is removed and the slite-to-screen distance doubled, the new fringe width is equal to x_(0). Find the wavelength of light used. |
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Answer» Solution :Data: `n_(m) = 1.6, B=1.964, mu m= 1.964 xx 10^(-6)m, D_(2) = 2D_(1), X_(2) = x_(0)` The fringe shift with the mica sheet, `x_(0) = D_(1)/d (n_(m)-1)b` Subsequent to the removal of the mica sheet and doubling the slits-to-screen DISTANCE, the new fringe width is `X_(2) = (lambdaD_(2))/d = (lambda(2D_(1)))/(d)` Since, `X_(2) = x_(0)`, `(2lambdaD_(1))/(d) = D_(1)/d(n_(m)-1)b` `THEREFORE` The wavelength of the LIGHT used, `lambda= (n_(m)-1)/(2).b-(1.6-1)/2.(1.964 xx 10^(-6))` `=0.3 xx 1.964 xx 10^(-6) = 5.892 xx 10^(-7)m = 5892 Å` |
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