In a medium, the electric field in an EM wave is E=E0 sin 12 × 106(z-2 × 108 t) N / C The refractive index of the medium is

Question

In a medium, the electric field in an EM wave is E=E0 sin 12 × 106(z-2 × 108 t) N / C The refractive index of the medium is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

E=E0 sin [12 × 106(z-2 × 108 t)] =E0 sin (12 × 106 z-24 × 1014 t) ∴ k=12 × 106, ω=24 × 1014 ν=(ω/k)=(24 × 1014/12 × 106)=2 × 108 ; μ=(c/v)=(3 × 108/2 × 108)=(3/2)

Q. In a medium, the electric field in an EM wave is E=E0​sin12×106(z−2×108t)N/C The refractive index of the medium is

E=E0​sin[12×106(z−2×108t)] =E0​sin(12×106z−24×1014t) ∴k=12×106,ω=24×1014 ν=kω​=12×10624×1014​=2×108 ; μ=vc​=2×1083×108​=23​

Q. In a medium, the electric field in an EM wave is
$E = E_{0} sin 12 \times 1 0^{6} (z - 2 \times 1 0^{8} t) N / C$
The refractive index of the medium is