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Four times In Fraunhoffer diffraction by single slit having width d. The perpendicular light of wavelength lambda incident on it. The distance between slit and screen is D. If the linear width of central maximum is halve to the width of slit then=.....

Answer»

`SQRT((lambdaD)/(4))`
`sqrt(lambdaD)`
`sqrt(4lambdaD)`
`(2lambdaD)`

Solution :The linear width of central maximum means linear DISTANCE between two first minimum. Condition for first order minimum.
`x_(1)=(lambdaD)/(d)`
`:.`Linear width of central maximum `=x_(1)+x_(1)`
`=2x_(1)`
`=(2lambdaD)/(d)`
but from GIVEN DATA `(2lambdaD)/(d)=(d)/(2)`
`:. d^(2)=4 lambdaD "" :. d=sqrt(4lambdaD)`


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