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In a Young's double slit experiment , th...

In a Young's double slit experiment , the path difference at a certain point on the screen ,between two interfering waves is `1/8`th of wavelenght .The reatio of the intensity at this point to that at the center of abright fringe is close to :

Text Solution

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Given path difference `Deltax=(lamda)/8`
Phase difference `(Delta phi)` is given by
`Delta phi =(2pi)/(lamda)(Deltax)`
`Delta phi=((2pi)lamda)/(lamda)(lamda)/8=(pi)/4`
For two sources in different phases
`I=I_(0)cos^(2)((pi)/8)`
`I/(I_(0))=cos^(2)((pi)/8)`
`=(1+"cos"(pi)/4)/2=(1+1/(sqrt(2)))/2=0.85`
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