In young's double slit experiment, if the distance between the slits is halved and the distance between the slits and the screen is doubled, the fringe width becomes

Options

A. 1.0 mm
B. 2.0 mm
C. 0.5 mm
D. 1.5 mm

Answer

A

Explanation

d = 1.0 mm, D = 1m , `lambda` = 600 nm
`I_(P)= 75%` of maximum or `I_(P)= 3I_(0)`
where `I_(0)` is the intensity of a single wave
Now, `I_(P)= 3I_(0)= (sqrt(I_(0)))^(2)+ (sqrt(I_(0)))^(2)+ 2 sqrt( I_(0) xx I_(0)) cosphi`
`:. Cosphi = cos(pi)/(3),` also `Deltax = (yd)/(D)` Now, `Deltax= (lambda)/(2pi) xx (pi)/(3) = (lambda)/(6) :. y= (lambdaD)/(6d)= (600 xx 10^(-19))/(6 xx 0.1 xx 10^(-3))`
or y= 1mm