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The wave length of light in 1st liquid is $2500 \;A^{\circ}$, and that in 2nd liquid is $5000\; A^{\circ}$, What is the critical angle of 1st liquid with respect to 2nd liquid.

$(a)\;60^{\circ} \\ (b)\;45^{\circ} \\ (c)\;30^{\circ} \\ (d)\;90^{\circ} $

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$\sin c= \large\frac{\mu_2}{\mu_1} =\frac{\mu \;in \;2nd \;medium}{\mu \;in \;1st \;medium}$
$\qquad= \large\frac{\lambda _1}{\lambda_2}$
$\qquad= \large\frac{wave\;length\; in\; 1st\; medium}{wave\;length\;in\;2nd\;medium}$
$\qquad= \large\frac{2500 \;A^{\circ}}{5000 A^{\circ}}$
$\therefore$ Critical angle $C=30^{\circ}$
Hence c is the correct answer.
answered Jan 17, 2014 by meena.p

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