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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

Text Solution

Verified by Experts

The correct Answer is:
`0.2mm`
`I=I_(1)+I_(2)+2sqrt(I_(1)I_(2))cosphi`
`I=2I_(0)(1+cos phi)`
since,`I_(max)=4I_(0)`
`(3)/(4)xx4I_(0)=2I_(0)(1=cosphi)`
`1=cosphi=(3)/(2) rArr cosphi=(1)/(2)`
`phi=pm(pi)/(3)` rad `rArr` path difference=`(lambda)/(6)`
`(xd)/(D)=(lambda)/(6) rArr x=(lambdaD)/(6d)`
distance=`2x` rArr distance =`(lambdaD)/(3d)`
distance=`(600xx10^(-9)xx1)/(3xx10^(-3)) rArr distance=200xx10^(-6)m`
distance=`2xx10^(-4)m rArr distance =0.2mm`
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