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In the ideal YDSE , when a glass plate $(refractive index\;1.5)$ of thickness t is introduced in path of one of interfering beams (wave - length $\lambda$) the intensity at position where central maximum occurred preciously remains unchanged . The minimum thickness of glass plate is :

$(a)\;2 \;\lambda \\ (b)\;\frac{\lambda}{3} \\ (c)\;\frac{2 \lambda}{3} \\ (d)\;\lambda $

1 Answer

Shift $=\large\frac{D}{d} $$(\mu-1) t$
$\qquad= \large\frac{D \lambda}{d}$
=> $t= \large\frac{\lambda}{\mu-1}$$=2 \lambda$
Hence a is the correct answer.
answered Feb 18, 2014 by meena.p

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