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The figure shows total internal reflection taking place in three materials of refractive indices $n_1,n_2$ and $n_3$ where $\theta _{\large c_1},\theta _{\large c_2},\theta _{\large c_3}$ are critical angles in the respective mediums. Then



$(a)\;n_1 > n_2 > n_3 \\ (b)\;n_3 < n_1 < n_2 \\ (c)\;n_1 < n_2 < n_3 \\ (d)\;n_2 < n_1 < n_3 $

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$\sin \theta _c =\large\frac{1}{\mu}$
Since third material has maximum critical angle, 2nd material least
$ \theta _{\Large c_3} > \theta _{\Large c_1} > \theta _{\Large c_2}$
We have $ n_3 < n_1 < n_2$
Hence b is the correct answer.
answered Jan 17, 2014 by meena.p

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