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(B) In the case of refraction, if $CD$ is the refracted wavefront and $v_1$ and $v_2$ are the speeds of light in air and glass respectively, then in t

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$\frac{\sin 	heta}{\sin \phi '}$

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(B) In the case of refraction, if $CD$ is the refracted wavefront and $v_1$ and $v_2$ are the speeds of light in air and glass respectively, then in the time the wavelet from $B$ reaches $C$, the wavelet from $A$ reaches $D$.Thus, the time taken is $t = \frac{BC}{v_a} = \frac{AD}{v_g}$, which implies $\frac{BC}{AD} = \frac{v_a}{v_g}$. .....$(i)$From the geometry of the diagram, in $\Delta ACB$, $BC = AC \sin 	heta$. .....$(ii)$In $\Delta ACD$, $AD = AC \sin \phi '$. .....$(iii)$Substituting $(ii)$ and $(iii)$ into $(i)$, we get $\frac{v_a}{v_g} = \frac{\sin 	heta}{\sin \phi '}$.Since the refractive index $\mu = \frac{v_a}{v_g}$, we have $\mu_g = \frac{\sin 	heta}{\sin \phi '}$.

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 along with the normals drawn at $A$ and $D$. The refractive index of glass with respect to air $(\mu = 1)$ will be equal to

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