Question 145: EM waves

Question

An EM wave from air enters a medium. The electric fields are
$$\overrightarrow {{E_1}} $$ = $${E_{01}}\widehat x\cos \left[ {2\pi v\left( {{z \over c} - t} ight)} ight]$$  in air and 
$$\overrightarrow {{E_2}} $$ = $${E_{02}}\widehat x\cos \left[ {k\left( {2z - ct} ight)} ight]$$  in medium, where the wave number k and frequency $$
u $$ refer to their values
in air. The medium is non-magnetic. If $${\varepsilon _{{r_1}}}$$ and $${\varepsilon _{{r_2}}}$$ refer to relative permittivities of air and medium
respectively, which of the following options is correct ?

Accepted Answer

Electric field in air,
$$\overrightarrow {{E_1}} $$  =  E
01
$$\widehat x$$ cos ( $${{2\pi vz} \over c}$$ $$-$$ 2$$\pi $$vt )
$$	herefore\,\,\,$$ Velocity in air = $${{2\pi v} \over {{{2\pi v} \over c}}}$$ = c
Also,    c = $${1 \over {\sqrt {\mu \varepsilon {r_1}{\varepsilon _0}} }}$$ . . . . . . (1)
$$\overrightarrow {{E_2}} $$  =  E
02
$$\widehat x$$ cos(2kz $$-$$ kct)
$$	herefore\,\,\,$$ Velocity in medium = $${{kc} \over {2k}}$$ = $${c \over 2}$$
Also, $${c \over 2}$$ = $${1 \over {\sqrt {\mu {\varepsilon _{r2}}\,{\varepsilon _0}} }}$$ . . . . . (2)
As, medium is non magnetic,
So,   $$\mu $$
medium
= $$\mu $$
air
= $$\mu $$
Dividing (1) by (2),  we get
2 = $$\sqrt {{{{\varepsilon _{r2}}} \over {{\varepsilon _{r1}}}}} $$
$$ \Rightarrow $$$$\,\,\,$$ $${{{{\varepsilon _{r1}}} \over {{\varepsilon _{r2}}}}}$$ = $${1 \over 4}$$

Answer

$${{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = 4$$

Answer

$${{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = 2$$

Answer

$${{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = {1 \over 4}$$

Answer

$${{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = {1 \over 2}$$

EM waves

Choose the correct answer: