Answer:
The correct answer to the following question will be Option E (Leaning back the camera such that the plane seems to be in the yz-axis now.).
Explanation:
The given field is:
⇒ [tex]E=EO \ j[/tex]
The above-given field does have an E-field perpendicular to something like the plane.
So,
⇒ [tex]ECos\theta=ECos 90^{\circ}[/tex]
We know that the value of "Cos 90°" is zero.
⇒ [tex]Cos 90^{\circ}=0[/tex]
The other given choices are not related to the given circumstances. So that Option E seems to be the right answer.