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Light is incident formals on as oil at an incident angle of $30^{\circ}$ After moving through the oil -1, oil-2, and glass it exters water . If the refraction index of glass and water are 1.5 and 1.3 respectively . Then the angle which the ray makes with normal in water .

$(a)\;\sin ^{-1}1 \\ (b)\;\sin ^{-1} \frac{1}{2} \\ (c)\;\sin ^{-1} \frac{1}{1.3}\\ (d)\;\sin ^{-1} \frac{1}{2.6} $

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As we know that
$y \sin 1$= constant
=> $ y +{air} -\sin _{i(air)}= y _{glass} \sin r _{glass)}$
$\sin i (glass) =\large\frac{y_{air}}{y_{glass}}$$\sin i_{air}$
Again $y_{glass} \sin i_{(glass)}=y _{water} \sin r_{water}$
From (i) and (ii)
$\qquad= \sin 30 =1.3 \sin r$
$\sin r =\large\frac{1}{2 \times 1.3} =\frac{1}{2.6};$
$r=\sin ^{-1} \large\frac{1}{2.6}$
Hence d is the correct answer.
answered Feb 11, 2014 by meena.p

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