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A beam of light consisting of two wave lengths 4200 Å and 5600 Å is used to obtain interference fringes in Young's double slit experiment. The distance between the slits is 0.3 mm and the distance between the slits and the screen is 1.5 m. Compute the least distance of the point from the central maximum, where the bright fringes due to both the wavelengths coincide. |
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Answer» Solution :Here , `lambda_1`=4200Å=`4200xx10^(-10)` m `lambda_2`=5600Å = `5600xx10^(-10)` m d=0.3 mm = `0.3xx10^(-3)` m D=1.5 m Let `n^(th)` bright fringe due to `lambda_1`= 4200 Å COINCIDE with `(n-1)^(th)`bright fringe due to `lambda_2`= 5600 Å `nlambda_1=(n-1)lambda_2` i.e., `nxx4200xx10^(-10)=(n-1)5600xx10^(-10)` `n=(n-1)8/6` 6n=8n-8 n=4 `THEREFORE` The least distance REQUIRED `X=nlambdaD/d` `=(4xx4200xx10^(-10) xx1.5)/(0.3xx10^(-3))=(25200xx10^(-10))/(0.3xx10^(-3))=84000xx10^(-7)` X=840 nm |
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