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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Wave Optics
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In a YDSE with $d= 1mm$ and $D= 1m$ slabs of $(t= 1 \mu m , \mu =3)$ and $(t=0.5\;\mu m , \mu =2) $ are introduced in front of upper and lower slit respectively . Then the shift in fringe pattern is :-

(a) 3 mm upwards (b) 1 mm upwards (c)2 mm upwards (d)1.5 mm upwards

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1 Answer

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Optical path for light coming from upper slit $S_1$ is
$s_1P + 1 \mu m (3-1) =s_1 P + 2 \mu m$
Similarly optical path for light coming from $s_2$ is
$s_2P + 0.5 \mu m (2-1) =s_2 P + 0.5 \mu m$
Path difference : $ \Delta P= (s_2 P +0.5 \mu m)-(s_1P +2 \mu m)$
$\qquad= (s_2P-S_1P)-1.5 \mu m$
for central bright fringe $\S_2P-S_1P =0$
$y= \large\frac{1.5 \mu m}{1 mm}$$ \times 1m = 1.5 mm$
The whole pattern is shifted by $1.5 mm$ upwards
Hence d is the correct answer.


answered Feb 13, 2014 by meena.p
edited Jul 22, 2014 by thagee.vedartham

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