A) \[\frac{\sin \theta }{\sin \theta '}\]
B) \[\frac{\sin \theta }{\sin \varphi '}\]
C) \[\frac{\sin \varphi '}{\sin \theta }\]
D) \[\frac{AB}{CD}\]
Correct Answer: B
Solution :
In the case of refraction if CD is the refracted wave front and v1 and v2 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 \[\Delta ACD,\] \[AD=AC\sin {\varphi }'\] .....(iii) From equations (i), (ii) and (iii) \[\frac{{{v}_{a}}}{{{v}_{g}}}=\frac{\sin \theta }{\sin {\varphi }'}\] Also \[\mu \propto \frac{1}{v}\Rightarrow \frac{{{v}_{a}}}{{{v}_{g}}}=\frac{{{\mu }_{g}}}{{{\mu }_{a}}}=\frac{\sin \theta }{\sin {\varphi }'}\]\[\Rightarrow {{\mu }_{g}}=\frac{\sin \theta }{\sin {\varphi }'}\]You need to login to perform this action.
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