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A plate of thickness x made of refractive index $\mu $ is placed in front of one of the slits in a double slit experiment. What should be minimum thickness x . so that the central maxima turns to fringe with minimum intensity

$(a)\;(\mu -1) \frac{\lambda}{2} \\ (b)\;(\mu-1) \lambda \\ (c)\;\frac{\lambda}{2 (\mu-1)} \\ (d)\;\frac{\lambda}{\mu-1} $

1 Answer

The central fringe will be a dark fringe with minimum path differentiate is $\large\frac{\lambda}{2}$
Path difference created by a film of refractive index $\mu$ and thickness x is
$( \mu -1) x$
$\therefore (\mu-1) x =\large\frac{\lambda}{2}$
$x =\large\frac{\lambda}{2 ( \mu-1)}$
Hence c is the correct answer.
answered Jan 24, 2014 by meena.p

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