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In Young's double-slit experiment , the slits are 0.5 mm apart and interference is observed on a screen placed at 1.0 m from the slits. It is found that the 9th bright fringe is at 8.835 mm from the 2nd dark fringe on the same side of the interference pattern. Find the wavelength of light used. |
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Answer» Solution :Data: `d=0.5mm= 5 xx 10^(-4)m, D=1.0m`, `x_(9)-x_(2)= 8.835 MM = 8.835 xx 10^(-3)`m (on the same side of the CENTER of the interference pattern) `x_(n) = (nlambdaD)/(d)`…………….(bright FRINGE) `x_(m)^(') = ((2m-1)lambdaD)/(2d)`………………..(dark fringe) `therefore x_(9) - x_(2)^(') = (9lambdaD)/d - ((2 xx 2-1)lambdaD)/(2d)` `=(9lambdaD)/d - (3lambdad)/(2d) = ((18-3)lambdaD)/(2d) = (15lambdaD)/(2d)` `therefore` The wavelength of light used, `lambda= (2d(x_(9)-x_(2)^(')))/(15D)` `=(2 xx 5 xx 10^(-4) xx 8.835 xx 10^(-3))/(15 xx 1.0)` `= 5.89 xx 10^(-7)m = 5890 Å` |
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