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Feb 10, 2019 at 8:12 comment added user143805 I accept that. How then do we find the sign? According to the Eq. (31.17) from Lecture 31: $\text{Field from glass} = -\frac{\eta q_e}{2\epsilon_0 c}\biggl[ i\omega\,\frac{q_eE_0}{m(\omega_0^2-\omega^2)}\, e^{i\omega(t-z/c)}\biggr] = |b|E_0 e^{i\omega t }e^{-i\pi / 2}$. But we should get a relative phase π for b and 0 for b′.
Jan 23, 2019 at 12:43 history answered Farcher CC BY-SA 4.0