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Two lights of wavelenght 560 nm and are used in Young's double slit experiment. Find the least distance from the centeral fringe where the bright fringe of the two wavelenght coindes. Give D = 1 m and d = 3 mm. |
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Answer» SOLUTION :`lambda_(1) = 560 NM = 560 xx 10^(-9) m:` `lambda_(2) = 420 xx 10^(-9) m,` `D = 1m, d = 3 mm = 3 xx 10^(-3) m` For a GIVE y,n and `lambda`are inversely proportional Let `n^(th)` order bright fringe of `lambda_(1)` conicides with `(n + 1)^(th)` order bright fringe of `lambd_(2)`. EQUATION for `n^(th)` bright fringe is, `y_(n) = n (lambdaD)/(d)` `Here, n(lambda_(1)D)/(d)=(n+1)(lambda_(2)D)/(d)"" (aslambda_(1)gtlambda_(2))` `nlambda_(1)=(n+1)lambda_(2)(or)(lambda_(1))/(lambda_(2))=(n+1)/(n)` `1+(1)/(n)=(560xx10^(-9))/(420xx10^(-9))(or)1+(1)/(n)=(4)/(3)` `(1)/(n)=(1)/(3)(or)n=3` Thus the `3^(rd)` hright fringe of `lambda_(1) and 4^(th)` bright fringe of `lambda_(2)` coincide at the least distancy The least distance from the central fringe where the bright fringes of the TWO wavelenght coincides is, `y_(n)=n(lambdaD)/(d)` `y_(n)=3xx(560xx10^(-9)xx1)/(3xx10^(-3))=560xx10^(-6)m` `y_(n)=0.560xx10^(-3)m=0.560mm` |
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