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A 4 cm thick layer of water covers 6 cm thick glass slab. A coin is placed at the bottom of slab and is being observed from the air side along the normal to surface. Then the apparent position of coin surface is

$(a)\;14 \;cm \\ (b)\;7\;cm \\ (c)\;21\;cm \\ (d)\;8\;cm $

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Using equation the total apparent shift is
$s= h_1 (1- \large\frac{1}{\mu_1}) +h_2 (1- \frac{1}{\mu_2})$
or $s= 4 \bigg( 1-\large\frac{1}{4/3}\bigg) $$+6 \bigg( 1- \large\frac{1}{3/2}\bigg)$
$\qquad= 3.0 cm$
Thus $h= h_1+h_2 -s= 4+6 -3$
$\qquad= 7\;cm$
Hence b is the correct answer.
answered Feb 13, 2014 by meena.p

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