Question

Region 1, z < 0 and region 2, z > 0, are both perfect dielectrics $(\mu ={\mu }_0)$. A uniform plane wave travelling in the positive z-direction has a radian frequency of $3\times {10}^{10}$rad/sec. Its wave lengths in the two regions are ${\lambda }_1=5cm\text{and}{\lambda }_2=3cm$. The fraction of incident energy that is reflected and transmitted respectively are

• A. 0.0625 and 0.938
• B. 0.5 and 0.5
• C. 0.938 and 0.0625
• D. 0.4 and 0.6

Answer 1

The correct option is A 0.0625 and 0.938

${\epsilon }_{r1}=(\frac{2\pi c}{{\lambda }_1\omega }{)}^2\text{and}{\epsilon }_{r2}=(\frac{2\pi c}{{\lambda }_2\omega }{)}^2$
$\frac{{\epsilon }_{r1}}{{\epsilon }_{r2}}=(\frac{{\lambda }_2}{{\lambda }_1}{)}^2$
Now,
$\Gamma =\frac{{\eta }_2-{\eta }_1}{{\eta }_2+{\eta }_1}=\frac{{\eta }_0\sqrt{\frac{1}{{\epsilon }_{r2}}}-{\eta }_0\sqrt{\frac{1}{{\epsilon }_{r2}}}}{{\eta }_0\sqrt{\frac{1}{{\epsilon }_{r2}}}+{\eta }_0\sqrt{\frac{1}{{\epsilon }_{r2}}}}=\frac{\frac{{\lambda }_2}{{\lambda }_1}-1}{\frac{{\lambda }_2}{{\lambda }_1}+1}=\frac{{\lambda }_2-{\lambda }_1}{{\lambda }_2+{\lambda }_1}=\frac{3-5}{3+5}=-\frac{1}{4}$
Reflected, $|\Gamma {|}^2=\frac{1}{16}=6.25\times {10}^{-2}=0.0625$
Transmitted, $1-|\Gamma {|}^2=1-0.0625=0.938$