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In an Young's experiment, the distances between two slits and that between slits and the screen are 0.05 cm and I m respectively. Find the distance between 3^(rd)bright and 5^(th) dark fringes. Take the wavelength of light equal to 5000Å |
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Answer» Solution :Here, d = 0.05 CM, D = 100 cm, `lamda=5000Å=5xx10^(-5)cm`, `x._(5)-x_(3)=?` n = 5 (dark) and n = 3 (bright) For `n^(th)` bright fringe, `(x_(n)d)/(D)=nlamda` TAKING n=3 `(x_(3)d)/(D)=3lamda` `:.x_(3)=(3lamdaD)/(d)""....(1)` For `n^(th)` dark fringe, `(x._(n)d)/(D)=(2n-1)(lamda)/(2)` Taking n=5 `(x._(n)d)/(D)=(9lamda)/(2)` `:.x._(5)=(9lamdaD)/(2D)""....(2)` Now, `x._(5)-x_(3)=(9lamdaD)/(2d)-(3lamdaD)/(d)=(3lamdaD)/(2d)` `=(3xx5xx10^(-5)xx100)/(2xx0.05)` `:.x._(5)-x_(3)=15xx10^(-2)cm=1.5mm` |
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