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A ray of light travelling in glass enters glass water interface at an angle $\theta$. It finally emerges into air parallel to the surface of water. $ \mu_w=\large\frac{4}{3}$. What is the refractive index of glass.

 

$(a)\;\frac{4}{3} \sin \theta \\ (b)\;\frac{1}{\sin \theta} \\ (c)\;\frac{4}{3} \\ (d)\;1 $

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Since the ray finally emerges parallel to the water surface.
r must be the critical angle.
$\therefore \sin r =\large\frac{1}{\mu _{\Large water}}$
$\sin r =\large\frac{3}{4}$
At the water glass inerface
$\mu _g \sin \theta=\mu \sin r$
$\mu _g =\large\frac{\mu _w \times \sin r}{\sin \theta}$
$\qquad= \large\frac{\Large\frac{4}{3} \times \frac{3}{4}}{\sin \theta}$
$\qquad= \large\frac{1}{\sin \theta}$
Hence b is the correct answer.
answered Jan 16, 2014 by meena.p
 

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