An electromagnetic wave going through vacuum is described by $E=E_0 \sin (k x-\omega t) ; B = B_0 \sin (k x-\omega t)$. Then,
Answer & explanation
Correct answer: option 1
$\frac{E_0}{B_0}=\frac{\omega}{k}$ or $E_0 k=B_0 \omega$
An electromagnetic wave going through vacuum is described by $E=E_0 \sin (k x-\omega t) ; B = B_0 \sin (k x-\omega t)$. Then,
Correct answer: option 1
$\frac{E_0}{B_0}=\frac{\omega}{k}$ or $E_0 k=B_0 \omega$