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JEE Main 2018 · Speed of EM Waves in a Medium

Speed of EM Waves in a Medium question from JEE Main 2018

Q23JEE Main 2018Difficult

Speed of EM Waves in a Medium

An EM wave from air enters a medium. The electric fields are

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] in air and

E⃗2=E02x^cos⁡[k(2z−ct)]\vec{E}_2 = E_{02} \hat{x} \cos\left[k(2z - ct)\right] in medium,

where the wave number k and frequency ν\nu refer to their values in air. The medium is non-magnetic. If εr1\varepsilon_{r_1} and εr2\varepsilon_{r_2} refer to relative permittivities of air and medium respectively, which of the following options is correct?

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Answer: (c)

  1. In air: k=2πν/ck = 2\pi\nu/c, so E1=E01cos⁡(kz−ωt)E_1 = E_{01}\cos(kz - \omega t) with ω=kc\omega = kc.
  2. In medium: cos⁡[k(2z−ct)]=cos⁡(2kz−ωt)\cos[k(2z - ct)] = \cos(2kz - \omega t), so wave number is 2k2k at the same ω\omega.
  3. v2=ω/(2k)=c/2⇒n=2v_2 = \omega/(2k) = c/2 \Rightarrow n = 2
  4. Non-magnetic: εr2/εr1=n2=4⇒εr1/εr2=1/4\varepsilon_{r_2}/\varepsilon_{r_1} = n^2 = 4 \Rightarrow \varepsilon_{r_1}/\varepsilon_{r_2} = 1/4
  5. So the answer is (c).
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