A plane electromagnetic wave of frequency 40 MHz travels in free space in the X-direction. At some point and at some instant, the electric field $\vec{E}$ has its maximum value of 750 N/C in Y-direction.
The propagation constant of the wave will be:
Answer & explanation
Correct answer: option 2
$\text{Propogation constant is }K = \frac{2\pi}{\lambda} = \frac{2\times 3.14}{7.5} = 0.838m^{-1}$