JEE MAIN - Physics (2018 (Offline) - No. 3)
An EM wave from air enters a medium. The electric fields are
$$\overrightarrow {{E_1}} $$ = $${E_{01}}\widehat x\cos \left[ {2\pi v\left( {{z \over c} - t} \right)} \right]$$ in air and
$$\overrightarrow {{E_2}} $$ = $${E_{02}}\widehat x\cos \left[ {k\left( {2z - ct} \right)} \right]$$ in medium,
where the wave number k and frequency $$\nu $$ 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 ?
$$\overrightarrow {{E_1}} $$ = $${E_{01}}\widehat x\cos \left[ {2\pi v\left( {{z \over c} - t} \right)} \right]$$ in air and
$$\overrightarrow {{E_2}} $$ = $${E_{02}}\widehat x\cos \left[ {k\left( {2z - ct} \right)} \right]$$ in medium,
where the wave number k and frequency $$\nu $$ 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 ?
$${{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = 4$$
$${{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = 2$$
$${{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = {1 \over 4}$$
$${{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = {1 \over 2}$$
Explanation
Electric field in air,
$$\overrightarrow {{E_1}} $$ = E01 $$\widehat x$$ cos ( $${{2\pi vz} \over c}$$ $$-$$ 2$$\pi $$vt )
$$\therefore\,\,\,$$ 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)
$$\overrightarrow {{E_2}} $$ = E02 $$\widehat x$$ cos(2kz $$-$$ kct)
$$\therefore\,\,\,$$ 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 = $$\sqrt {{{{\varepsilon _{r2}}} \over {{\varepsilon _{r1}}}}} $$
$$ \Rightarrow $$$$\,\,\,$$ $${{{{\varepsilon _{r1}}} \over {{\varepsilon _{r2}}}}}$$ = $${1 \over 4}$$
$$\overrightarrow {{E_1}} $$ = E01 $$\widehat x$$ cos ( $${{2\pi vz} \over c}$$ $$-$$ 2$$\pi $$vt )
$$\therefore\,\,\,$$ 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)
$$\overrightarrow {{E_2}} $$ = E02 $$\widehat x$$ cos(2kz $$-$$ kct)
$$\therefore\,\,\,$$ 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 = $$\sqrt {{{{\varepsilon _{r2}}} \over {{\varepsilon _{r1}}}}} $$
$$ \Rightarrow $$$$\,\,\,$$ $${{{{\varepsilon _{r1}}} \over {{\varepsilon _{r2}}}}}$$ = $${1 \over 4}$$
Comments (0)
