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An EM wave passes from air into a medium. The electric fields are E1 = E01*x_hat*cos[2*pi*nu*(z/c - t)] in air and E2 = E02*x_hat*cos[k*(2z - ct)] in the medium, where k and nu are the wave number and frequency in air. The medium is non-magnetic. If er1 and er2 are the relative permittivities of air and the medium, which is correct?
- er1/er2 = 2
- er1/er2 = 1/4
- er1/er2 = 1/2
- er1/er2 = 4
Correct answer: er1/er2 = 1/4
Solution
In air: phase 2*pi*nu*(z/c - t) gives omega = 2*pi*nu and v1 = c. In medium: k*(2z - ct) = 2k*z - kc*t, so omega2 = kc and wave number = 2k, giving v2 = omega2/(2k) = kc/(2k) = c/2. Frequency must match: omega2 = omega1 -> kc = 2*pi*nu. n2 = c/v2 = 2, so er2 = n2² = 4 (non-magnetic). er1 = 1. Thus er1/er2 = 1/4.
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