A) \[\frac{{{\in }_{{{r}_{1}}}}}{{{\in }_{{{r}_{2}}}}}=\frac{1}{4}\]
B) \[\frac{{{\in }_{{{r}_{1}}}}}{{{\in }_{{{r}_{2}}}}}=\frac{1}{2}\]
C) \[\frac{{{\in }_{{{r}_{1}}}}}{{{\in }_{{{r}_{2}}}}}=4\]
D) \[\frac{{{\in }_{{{r}_{1}}}}}{{{\in }_{{{r}_{2}}}}}=2\]
Correct Answer: A
Solution :
| [a] \[\overrightarrow{\text{E}}={{E}_{01}}\cos \left( 2\pi v\left( \frac{z}{c}-t \right) \right)\widehat{x}\,\,in\,\,air\] |
| \[k=\frac{2\pi v}{c}\] \[\text{speed=c}\] |
| \[{{\overrightarrow{\text{E}}}_{2}}={{E}_{02}}\cos (k(2z-ct))\widehat{x}\] |
| \[{{\overrightarrow{\text{E}}}_{2}}={{E}_{02}}\cos \left( \frac{2\pi v}{c}(2z-ct) \right)\widehat{x}\] |
| \[\text{speed=}\frac{\text{c}}{\text{2}}\] |
| \[\text{c}\propto \frac{1}{\sqrt{{{\varepsilon }_{0}}{{\varepsilon }_{{{r}_{1}}}}}}\] |
| \[\frac{c}{2}\propto \frac{1}{\sqrt{{{\varepsilon }_{0}}{{\varepsilon }_{{{r}_{1}}}}}}\] |
| \[2=\sqrt{\frac{{{\varepsilon }_{{{r}_{2}}}}}{{{\varepsilon }_{{{r}_{1}}}}}}\Rightarrow \frac{{{\varepsilon }_{{{r}_{2}}}}}{{{\varepsilon }_{{{r}_{1}}}}}=4\] |
| \[\frac{{{\varepsilon }_{{{r}_{1}}}}}{{{\varepsilon }_{{{r}_{2}}}}}=\frac{1}{4}\] |
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