In a YDSE experiment, d = 1mm, `lambda`= 6000Å and D= 1m. The minimum distance between two points on screen having 75% intensity of the maximum intensity will be

Options

A. `0.45 mm`
B. `0.40 mm`
C. `0.30 mm`
D. `0.20 mm`

Answer

D

Explanation

Lets look at the screen.

as we know that `75%` intensity will correspond to a point where intensity is `3 I_(0)`.
`{therefore I_(max)=4 I_(0)}`
`I=I_(0)+I_(0)+2sqrt(I_(0)) sqrt(I_(0)) cos (Deltaphi)`
`3I_(0)=2I_(0)(1+cos Deltaphi)`
`cos (Deltaphi)=(1)/(2)`
`Deltaphi=(pi)/(3),2pi-(pi)/(3), 2pi\+(pi)/(3),..........`
`Deltap=(lambda)/(6),lambda-(lambda)/(6),lambda+(lambda)/(6),.........`
`Deltap=(yd)/(D)`
`(yd)/(D)=(lambda)/(6)rArry=(D)/(d)xx(lambda)/(6),.........`
`y=(beta)/(6),beta-(beta)/(6),beta+(beta)/(6)`
`y=(lambda)/(6)xx(D)/(d)=(6000xx10^(-10)xx1)/(3xx10^(-3))=0.2 mm`