A) \[{{\mu }_{1}}={{\mu }_{2}}\]
B) \[{{\mu }_{2}}={{\mu }_{3}}\]
C) \[{{\mu }_{3}}={{\mu }_{4}}\]
D) \[{{\mu }_{1}}={{\mu }_{4}}\]
Correct Answer: D
Solution :
From Snells law \[_{1}{{\mu }_{2}}=\frac{\sin \,i}{\sin \,r}=\frac{{{\mu }_{2}}}{{{\mu }_{1}}}\] ?. (i) \[_{2}{{\mu }_{3}}=\frac{\sin \,{{r}_{1}}}{\sin \,{{r}_{2}}}=\frac{{{\mu }_{3}}}{{{\mu }_{2}}}\] ?. (ii) \[_{3}{{\mu }_{4}}=\frac{\sin \,{{r}_{2}}}{\sin \,{{r}_{3}}}=\frac{{{\mu }_{4}}}{{{\mu }_{3}}}\] ?. (iv) Multiplying Eqs. (i), (ii) and (iii), we get \[\frac{{{\mu }_{2}}}{{{\mu }_{1}}}\times \frac{{{\mu }_{3}}}{{{\mu }_{2}}}\times \frac{{{\mu }_{4}}}{{{\mu }_{3}}}=\frac{{{\mu }_{4}}}{{{\mu }_{1}}}=\frac{\sin \,i}{\sin \,{{r}_{3}}}\] \[\Rightarrow \] \[{{\mu }_{1}}\sin \,i={{\mu }_{4}}\sin \,{{r}_{3}}\] Since, AB is parallel to CD, so \[i={{r}_{3}}\] Therefore, \[{{\mu }_{1}}={{\mu }_{4}}\]
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