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In a Young's double slit experiment the intensity at a point where the path difference is (lambda)/(6)(lambda being the wavelength of the light used) is I. If I_(0) denotes the maximum intensity then (I)/(I_(0))=…

Answer»

`(1)/(sqrt(2))`
`(sqrt(3))/(2)`
`(1)/(2)`
`(3)/(4)`

Solution :Maximumintensity `I_(0)=I.+I+2 sqrt(I.I) cos0^(@)`
`=I.+2I. [ :. cos 0^(@)=1]`
`=4I.`
and intensity at any POINT,
`I=I.+I.+2sqrt(I.I) cos PHI`
where `phi=(2PI)/(lambda)xx` path DIFFERENCE
`=(2pi)/(lambda)xx(lambda)/(6)=(pi)/(3)` rad
`:.I=2I.+2I"cos"(pi)/(3)`
`=2I.+I.=3I.`
`:.(I)/(I_(0))=(3I.)/(4I.)=(3)/(4)`


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