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=.....

Options

A. `sqrt((lambdaD)/(4))`
B. `sqrt(lambdaD)`
C. `sqrt(4lambdaD)`
D. `(2lambdaD)`

Answer

C

Explanation

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)`