Solveeit Logo
Login

Question

Physics Question on Electromagnetic waves

An EMEM wave from air enters a medium. The electric fields are E1⃗=E01x^cos⁡[2πv(zc−t)]\vec{E_1} = E_{01} \hat{x} \cos \left[ 2 \pi v\left( \frac{z}{c} - t \right) \right] in air and E⃗2=E02x^cos⁡[k(2z−ct)]\vec{E}_2 = E_{02} \hat{x} \cos [k (2 z - ct)] in medium, where the wave number kk and frequency ?? refer to their values in air. The medium is non-magnetic . If ∈r1\in_{r_1} and ∈r2\in_{r_2} refer to relative permittivities of air and medium respectively, which of the following options is correct ?

A

∈r1∈r2=4\frac{\in_{r_1}}{\in_{r_2}} = 4

B

∈r1∈r2=2\frac{\in_{r_1}}{\in_{r_2}} = 2

C

∈r1∈r2=14\frac{\in_{r_1}}{\in_{r_2}} = \frac{1}{4}

D

∈r1∈r2=12\frac{\in_{r_1}}{\in_{r_2}} = \frac{1}{2}

Answer

∈r1∈r2=14\frac{\in_{r_1}}{\in_{r_2}} = \frac{1}{4}

Explanation

Solution

E⃗1=E01x^cos⁡[2πv(zc−t)]     \vec{E}_{1}=E_{01} \hat{x} \cos \left[2 \pi v\left(\frac{z}{c}-t\right)\right] \,\,\,\,\, air
E⃗2=E02x^cos⁡[k(2z−ct)]     \vec{E}_{2}=E_{02} \hat{x} \cos [k(2 z-c t)] \,\,\,\,\, medium
During refraction, frequency remains unchanged, whereas wavelength gets changed.
∴k′=2k    \therefore k'=2 k \,\,\,\, (From equations)
⇒2πλ′=2(2πλ0)\Rightarrow \frac{2 \pi}{\lambda'}=2\left(\frac{2 \pi}{\lambda_{0}}\right)
⇒λ′=λ02\Rightarrow \lambda'=\frac{\lambda_{0}}{2}
⇒v=c2\Rightarrow v=\frac{c}{2}
⇒1μ0ε2=12×1μ0ε1\Rightarrow \frac{1}{\sqrt{\mu_{0} \varepsilon_{2}}}=\frac{1}{2} \times \frac{1}{\sqrt{\mu_{0} \varepsilon_{1}}}
⇒ε1ε2=14\Rightarrow \frac{\varepsilon_{1}}{\varepsilon_{2}}=\frac{1}{4}