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Phase difference =(2pi)/lambda xx...

Phase difference `=(2pi)/lambda xx`_____

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In Young's doubIe sIit experiment using monochromatic Iight of waveIength lambda the intensity of Iight at point on the screen where path difference is lambda is k units What the intensity of Iight at a point where the path difference is (lambda)/(3) ?

If the path difference is lambda , what will be the phase difference?

In Young's double slit experiment the intensity of light on the screen where the where the path difference is lambda is k ( lambda being the wavelength of light used). The intensity at a point where the path difference is (lambda)/(4) will be

The intensity at the central maxima in Young' s double slit experiment is I_(0) . Find out the intensity at a point where the path difference is (lambda)/(6) , (lambda)/(4) and (lambda)/(3) .

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The unit of both x and y is m in the expression y=A sin[(2pi)/(lambda)(ct-x)]. Then

Statement I : If the phase difference between the light waves passing through the slits in the Young's experiment is pi radian the central fringe will be dark. Statement II. Phase differene is equal to (2 pi)/(lambda) times the path difference .

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