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Light of wavelength 6000 overset(@)A is used to obtain interference fringe of width 6 mm in a young's double slit experiment. Calculate the wavelength of light required to obtain fringe of width 4 mm if the distance between the screen and slits is reduced to half of its initial value. |
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Answer» SOLUTION :`beta=(lambdaD)/d` `beta_1=(lambda_1D_1)/d beta_2=(lambda_2D_2)/d` `beta_1/beta_2=(lambda_1D_1)/(lambda_2D_2)` Substitution and simplification : Arriving upto `lambda^2` = 8000 Å Given, `lambda=6800xx10^(-10)m = 6xx10^(-7)` m `lambda.`=? `beta=6xx10^(-3)m , beta.=4xx10^(-3)m` D.=D/2 `rArr` fringe width , `beta=(lambdaD)/d` Hence, `(beta.)/beta=(lambda.D.)/d xx d/(lambdaD)` i.e., `(beta.)/beta=(lambda.D.)/d xx d/(lambdaD)` `therefore beta.=(lambda.beta)/(2lambda)` or `lambda.=(2lambdabeta.)/beta` i.e., `lambda.=(2xx6xx10^(-7)xx4xx10^(-3))/(6xx10^(-3))` i.e., `lambda=8xx10^(-7)` m or `lambda.`=800 Å Thewavelength required to produce FRINGES of width 4 mm is 8000 Å. |
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