Physics 208, Spring 08, Week 12Apr. 7, 2008
Problem 1
The magnetic field of a plane-polarized EM wave moving in the +z direction in vacuum is given by , in SI units. a) What are the wavelength and the period of the wave?
a) What are the wavelength, the period and the amplitude of the wave?
b) What is the direction of propagation of the wave?
c) Write the expression of the electric field wave explicitly indicating its
direction and amplitude. Check the units for the E-field amplitude.
d) Find the intensity (average power per square meter) carried by the EM
wave.
e) How much of it is carried by the E-field?
Remember that ε0 = 8.85 x 10-12 C2/ (N m2). You can then derive μ0 if you do not remember it…
Solution:
a) The amplitude of the wave is Bmax = 1.2 x 10-6 T, the wave number is m-1 and the angular frequency is .Hence the wavelength is
and the period is T = 8 x 10-7 s.
b) The direction of propagation of the wave is the z-axis.
c) The direction of the E-field is found considering that the waves propagates along z and this must be the direction perpendicular to the B-field (that is along y) and to the E-field that is E x B. The situation is shown in the figure.
So the E-field is in the x-direction and its expression is:
The E-field amplitude can be calculated as .
Units are m/s x T = m/s x N/C x s/m = N/C.
d)I = Sav = Pav /A = uav c
e) 50%
Short problem on Polarization
Which is the intensity of a light bulb of intensity I0 = 5 W/m2 transmitted by a polarizer and an analyzer with their transmission angles at 90 degrees one from the other. What is the transmitted intensity when the angle between the polarizer and analyzer is 45 degrees?
Solution:
The intensity transmitted by the polarizer is I0/2 because the light bulb light is not polarized and the plane of polarization is along the transmission axes of the polarizer.
After the analyzer no light passes because the axis of the polarizer and analyzer are perpendicular.
If the angle is 45 degrees the intensity is I0/2*cos^2(45degrees) = 1.25 W/m2.