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In a YDSE experiment, d = 1mm, lambda= 6...

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

A

`0.45 mm`

B

`0.40 mm`

C

`0.30 mm`

D

`0.20 mm`

Text Solution

Verified by Experts

The correct Answer is:
D
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`
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