I think the correct answer from the choices listed above is the second option. The relationship between the direction of energy and wave motion in a transverse wave would be the <span>energy direction is perpendicular to the motion of the wave. Hope this answers the question. Have a nice day.</span>
Answer:
Explanation:
Given
Initial Intensity of light is S
when an un-polarized light is Passed through a Polarizer then its intensity reduced to half.
When it is passed through a second Polarizer with its transmission axis 

here 


When it is passed through third Polarizer with its axis
to first but
to second thus 



When middle sheet is absent then Final Intensity will be zero