Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

When E 0 = 390 N/C Angle = 44° What is the amplitude of the transmitted electric

ID: 1604485 • Letter: W

Question

When

E0 = 390 N/C
Angle = 44°

What is the amplitude of the transmitted electric field, in N/C?

Mastering Physics: Assignme x G What is the momentum of a 8 x XC Physics question I chegg.com x D prelab po.pdf B Take Test: Prelab PO Pola X C Iowa State University of Science and Technology [US] https://bb.its.iastate,edu -112L-ALL/112.. A Polaroid or polarizing filter is a transparent material whose molecules are arranged in such a way that only one particular direction of the electric field in electromagnetic waves is allowed to go through. The perpendicular components are absorbed. The direction in which the electric field is transmitted is called the axis of polarizatian of the filter. The wave in the images below propagates in the +z direction and is initially linearly polarized in the x direction. The wave goes through a Polaroid whose axis of polarization is indicated by the middle line. Let Eo be the amplitude of the electric field of the incoming wave, and let be the angle between th direction (the polarization of the incoming wave) and the axis of the Polaroid Note how, after the Polaroid filter: The resulting wave is linearly polarized in the direction of the Polaroid The amplitude of the resulting wave is equal to the projection of the original vector onto the direction of the axis, E: cose 0-0 0- 45 Polarization a Polaroid E' cos (00) E cos(45 0.71E 0> 70° 6- 90 Ed E cos(90°) 0 Er Ea cos (70) 0.34E 90 mjskeez

Explanation / Answer

here,
Eo = 390 N/C
= 44 degree

use Malus law:
I = Io*cos^2()
since i is proportional to E^2
above expression becomes,
E^2 = Eo^2 * cos^2 ()
E^2 = (390)^2 * cos^2 (44)
E^2 = 78704.1
E = 280.5 N/C

Answer: 280.5 N/C

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote