1
JEE Main 2018 (Offline)
+4
-1
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 ?
A
$${{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = 4$$
B
$${{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = 2$$
C
$${{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = {1 \over 4}$$
D
$${{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = {1 \over 2}$$
2
JEE Main 2018 (Online) 15th April Evening Slot
+4
-1
A plane polarized monochromatic EM wave is traveling in vacuum along z direction such that at t = t1 it is found that the electric field is zero at a spatial point z1. The next zero that occurs in its neighbourhood is at z2. The frequency of the electroagnetic wave is :
A
$${{3 \times {{10}^8}} \over {\left| {{z_2} - {z_1}} \right|}}$$
B
$${{1.5 \times {{10}^8}} \over {\left| {{z_2} - {z_1}} \right|}}$$
C
$${{6 \times {{10}^8}} \over {\left| {{z_2} - {z_1}} \right|}}$$
D
$${1 \over {{t_1} + {{\left| {{z_2} - {z_1}} \right|} \over {3 \times {{10}^8}}}}}$$
3
JEE Main 2018 (Online) 15th April Morning Slot
+4
-1
A monochromatic beam of light has a frequency $$v = {3 \over {2\pi }} \times {10^{12}}Hz$$ and is propagating along the direction $${{\widehat i + \widehat j} \over {\sqrt 2 }}.$$
It is polarized along the $$\widehat k$$ direction. The acceptable form for the magnetic field is :
A
B
C
D
4
JEE Main 2017 (Online) 9th April Morning Slot
+4
-1
The electric field component of a monochromatic radiation is given by

$$\overrightarrow E$$ = 2 E0 $$\widehat i$$ cos kz cos $$\omega$$t

Its magnetic field $$\overrightarrow B$$ is then given by :
A
$${{2{E_0}} \over c}$$ $$\widehat j$$ sin kz cos $$\omega$$t
B
$$-$$ $${{2{E_0}} \over c}$$ $$\widehat j$$ sin kz sin $$\omega$$t
C
$${{2{E_0}} \over c}$$ $$\widehat j$$ sin kz sin $$\omega$$t
D
$${{2{E_0}} \over c}$$ $$\widehat j$$ cos kz cos $$\omega$$t
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