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In the adjoining diagram, a wavefront $AB$, moving in air is incident on a plane glass surface $XY$. Its position $CD$ after refraction through a glass slab is shown also along with the normals drawn at $A$ and $D$. The refractive index of glass with respect to air ($\mu = 1$) will be equal to
$\frac{{\sin \theta }}{{\sin \theta '}}$
$\frac{{\sin \theta }}{{\sin \phi '}}$
$\frac{{\sin \phi '}}{{\sin \theta }}$
$\frac{{AB}}{{CD}}$
Solution
(b) In the case of refraction if $CD$ is the refracted wave front and $v_1$ and $v_2$ are the speed of light in the two media, then in the time the wavelets from $B$ reaches $C$, the wavelet from $A$ will reach $D$, such that
$t = \frac{{BC}}{{{v_a}}} = \frac{{AD}}{{{v_g}}}$$ \Rightarrow \frac{{BC}}{{AD}} = \frac{{{v_a}}}{{{v_g}}}$ …..$(i)$
But in $\Delta ACB,$ $BC = AC\sin \theta $ …..$(ii)$
while in $\Delta ACD,$ $AD = AC\sin \phi '$ …..$(iii)$
From equations $(i), (ii)$ and $(iii)$ $\frac{{{v_a}}}{{{v_g}}} = \frac{{\sin \theta }}{{\sin \phi '}}$
Also $\mu \propto \frac{1}{v} \Rightarrow \frac{{{v_a}}}{{{v_g}}} = \frac{{{\mu _g}}}{{{\mu _a}}} = \frac{{\sin \theta }}{{\sin \phi '}}$$ \Rightarrow {\mu _g} = \frac{{\sin \theta }}{{\sin \phi '}}$
