Question
An EM wave from air enters a medium. The electric fields are $$\overrightarrow {{E_1}} = {E_{01}}\hat x\cos \left[ {2\pi v\left( {\frac{z}{c} - t} \right)} \right]$$ in air and $$\overrightarrow {{E_2}} = {E_{02}}\hat x\cos \left[ {k\left( {2z - ct} \right)} \right]$$ in medium, where the wave number $$k$$ and frequency $$v$$ refer to their values in air. The medium is nonmagnetic. If $${ \in _{{r_1}}}$$ and $${ \in _{{r_2}}}$$ refer to relative
permittivities of air and medium respectively, which of the
following options is correct?
A.
$$\frac{{{ \in _{{r_1}}}}}{{{ \in _{{r_2}}}}} = 4$$
B.
$$\frac{{{ \in _{{r_1}}}}}{{{ \in _{{r_2}}}}} = 2$$
C.
$$\frac{{{ \in _{{r_1}}}}}{{{ \in _{{r_2}}}}} = \frac{1}{4}$$
D.
$$\frac{{{ \in _{{r_1}}}}}{{{ \in _{{r_2}}}}} = \frac{1}{2}$$
Answer :
$$\frac{{{ \in _{{r_1}}}}}{{{ \in _{{r_2}}}}} = \frac{1}{4}$$
Solution :
Velocity of EM wave is given by $$v = \frac{1}{{\sqrt {\mu \in } }}$$
Velocity in air = $$\frac{\omega }{k} = C$$
Velocity in medium = $$\frac{C}{2}$$
Here, $${\mu _1} = {\mu _2} = 1$$ as medium is non-magnetic
$$\therefore \frac{{\frac{1}{{\sqrt {{ \in _{{r_1}}}} }}}}{{\frac{1}{{\sqrt {{ \in _{{r_2}}}} }}}} = \frac{C}{{\left( {\frac{C}{2}} \right)}} = 2 \Rightarrow \frac{{{ \in _{{r_1}}}}}{{{ \in _{{r_2}}}}} = \frac{1}{4}$$