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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Wave Optics
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In the ideal YDSE when a glass plate (RI= 1.5) of thickness t is introduced in path of one of interfering beams , the intensity at the position where central maximum occurred previously remains unchanged. The minimum thickness of glass plate is :

$(a)\;\lambda \\ (b)\;3 \lambda \\ (c)\;4 \lambda \\ (d)\;2 \lambda $

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Path difference due to slab should be integral multiple of $\lambda$ or $\Delta x = n \lambda$
or $(\mu-1)t=n \lambda$$ \qquad n=1,2,3...$
or $t= \large\frac{n \lambda}{\mu-1}$
For minimum value of t , we have n =1
$\therefore t= \large\frac{\lambda}{\mu-1}=\frac{\lambda}{1.5-1}$
$\qquad= 2 \lambda$
Hence d is the correct answer.


answered Feb 20, 2014 by meena.p
edited Jul 24, 2014 by thagee.vedartham

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