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In a double slit experiment ,the separation between the slits is `d=0.25cm` and the distance of the screen `D=100`cm from the slits .if the wavelength of light used in `lambda=6000Å`and `I_(0)`is the intensity of the central bright fringe.the intensity at a distance `x=4xx10^(-5)`in form the central maximum is-A. `I_(0)`B. `I_(0)//2`C. `3I_(0)//4`D. `I_(0)//3` |
Answer» Correct Answer - C path diff.`(xd)/(D) rArr` path diff.=`(4xx10^(-5)xx0.25xx10^(-2))/(1)` Path diff.=`1xx10^(-7)` phase diff.=`("path diff.")/(lambda)xx2pi rArr` phase.diff=`(1xx10^(-7))/(6xx10^(-7))xx2pi` phase diff.=`(2pi)/(6) rArr` Phase.diff,=`(pi)/(3)rArr phi=60^(@)` `I_(R)=I_(1)+I_(2)+2sqrt(I_(1)I_(2))cos60^(@) rArr I_(R)=I+I+2Ixx(1)/(2)` `I_(R)=3I rArr I_(R)=(3I_(0))/(4)` |
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