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Electromagnetic Waves question

2018 · Shift 0 · Q47
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Electromagnetic Waves question

2018 · Shift 0 · Q47

JEE MainPhysicsElectromagnetic WavesMCQ+4 / −1
An EM wave from air enters a medium. The electric fields are E1→\overrightarrow {{E_1}}E1​​=E01x^cos⁡[2πv(zc−t)]{E_{01}}\widehat x\cos \left[ {2\pi v\left( {{z \over c} - t} \right)} \right]E01​xcos[2πv(cz​−t)] in air and E2→\overrightarrow {{E_2}}E2​​ =E02x^cos⁡[k(2z−ct)]{E_{02}}\widehat x\cos \left[ {k\left( {2z - ct} \right)} \right]E02​xcos[k(2z−ct)] in medium, where the wave number k and frequency uuu refer to their values in air. The medium is non-magnetic. If εr1{\varepsilon _{{r_1}}}εr1​​ and εr2{\varepsilon _{{r_2}}}εr2​​ refer to relative permittivities of air and medium respectively, which of the following options is correct ?
  1. A
    εr1εr2=4{{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = 4εr2​​εr1​​​=4
  2. B
    εr1εr2=2{{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = 2εr2​​εr1​​​=2
  3. C
    εr1εr2=14{{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = {1 \over 4}εr2​​εr1​​​=41​
  4. D
    εr1εr2=12{{{\varepsilon _{{r_1}}}} \over {{\varepsilon _{{r_2}}}}} = {1 \over 2}εr2​​εr1​​​=21​
View written solutionFree

Correct answer: C

  1. Identify the wave in air

The electric field in air is

E⃗1=E01x^cos⁡[2πν(zc−t)].\vec E_1 = E_{01}\hat x \cos\left[2\pi \nu\left(\frac{z}{c}-t\right)\right].E1​=E01​x^cos[2πν(cz​−t)].

Compare with the standard form:

E⃗=E0cos⁡(kz−ωt).\vec E = E_0 \cos(kz-\omega t).E=E0​cos(kz−ωt).

So for air,

k1=2πνc,ω=2πν.k_1 = \frac{2\pi \nu}{c}, \qquad \omega = 2\pi \nu.k1​=c2πν​,ω=2πν.

Thus the wave speed in air is

v1=ωk1=c.v_1 = \frac{\omega}{k_1} = c.v1​=k1​ω​=c.
  1. Identify the wave in the medium

The electric field in the medium is

E⃗2=E02x^cos⁡[k(2z−ct)].\vec E_2 = E_{02}\hat x \cos\left[k(2z-ct)\right].E2​=E02​x^cos[k(2z−ct)].

Here kkk is stated to be the wave number in air. Rewrite the phase:

k(2z−ct)=(2k)z−(kc)t.k(2z-ct)= (2k)z-(kc)t.k(2z−ct)=(2k)z−(kc)t.

So in the medium,

k2=2k,ω2=kc.k_2 = 2k, \qquad \omega_2 = kc.k2​=2k,ω2​=kc.

But since in air,

k=ωc,k = \frac{\omega}{c},k=cω​,

we get

omega2=kc=ω.omega_2 = kc = \omega.omega2​=kc=ω.

So frequency remains unchanged, as expected.

Hence the wave speed in the medium is

v2=ω2k2=kc2k=c2.v_2 = \frac{\omega_2}{k_2} = \frac{kc}{2k} = \frac{c}{2}.v2​=k2​ω2​​=2kkc​=2c​.
  1. Relate wave speed to permittivity

For a non-magnetic medium,

μr=1.\mu_r = 1.μr​=1.

Wave speed in a medium is

v=1με.v = \frac{1}{\sqrt{\mu \varepsilon}}.v=με​1​.

Therefore,

v∝1εrv \propto \frac{1}{\sqrt{\varepsilon_r}}v∝εr​​1​

when μr=1\mu_r=1μr​=1.

So,

v1v2=εr2εr1.\frac{v_1}{v_2} = \sqrt{\frac{\varepsilon_{r_2}}{\varepsilon_{r_1}}}.v2​v1​​=εr1​​εr2​​​​.

Now,

v1v2=cc/2=2.\frac{v_1}{v_2} = \frac{c}{c/2} = 2.v2​v1​​=c/2c​=2.

Thus,

2=εr2εr12 = \sqrt{\frac{\varepsilon_{r_2}}{\varepsilon_{r_1}}}2=εr1​​εr2​​​​

Squaring,

εr2εr1=4.\frac{\varepsilon_{r_2}}{\varepsilon_{r_1}} = 4.εr1​​εr2​​​=4.

Hence,

εr1εr2=14.\frac{\varepsilon_{r_1}}{\varepsilon_{r_2}} = \frac{1}{4}.εr2​​εr1​​​=41​.
  1. Match with options

The correct option is

εr1εr2=14\boxed{\frac{\varepsilon_{r_1}}{\varepsilon_{r_2}}=\frac14}εr2​​εr1​​​=41​​

which is Option C.

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