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An em wave going through vacuum is described by `E=E_0sin(kx-omegat)` `B=B_0sin(kx-omegat)`A. `E_(0)k = B_(0)omega`B. `E_(0) omega = B_(0)k`C. `E_(0) B_(0) = omegak`D. None of these

Answer» Correct Answer - A
From `k = (2pi)/(lambda)` and `omega = 2piv`
we get, `k/(omega) = ((2pi)/(lambda))/(2piv) = v/(lambda) = 1/c`
As, `(E_(0))/(B_(0)) = c rArr (k)/(omega) = (B_(0))/(E_(0))`
`E_(0)k = B_(0)omega`


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