Answer:
Following are the solution to the given question:
Explanation:
The input linear polarisation was shown at an angle of
. It's a very popular use of a half-wave plate. In particular, consider the case
, at which the angle of rotation is
. HWP thereby provides a great way to turn, for instance, a linear polarised light that swings horizontally to polarise vertically. Illustration of action on event circularly polarized light of the half-wave platform. Customarily it is the slow axis of HWP that corresponds to either the rotation. Note that perhaps the vector of polarization is "double-headed," i.e., the electromagnetic current swinging back and forward in time. Therefore the turning angle could be referred to as the rapid axis to reach the same result. Please find the attached file.
Answer:
a) the oscillation of this field is in phase, when the magnetic field goes in the negative direction of y, the elective field goes in the positive direction of the z axis
b) the direction of the magnetic field perpendicular to this electric field and the speed in the negative x the magnetic field goes in the x direction and in the direction (1, - 1.1)
Explanation:
a) the polarization the determined wave oscillates the electric field, which is the z axis
As the wave travels on the negative x-axis and the magnetic field is perpendicular, this field goes on the positive y-axis
the oscillation of this field is in phase, when the magnetic field goes in the negative direction of y, the elective field goes in the positive direction of the z axis
be) in the case of a polarization in the xi plane the magnetic field must go in the direction of the magnetic field perpendicular to this electric field and the speed in the negative x the magnetic field goes in the x direction and in the direction (1, - 1.1)
The area of the Earth (Ae) that is irradiated by is given by:
Ae = 4πRe^2, where Re = Distance from Sun to Earth
Substituting;
Ae = 4π*(1.5*10^8*1000)^2 = 2.827*10^23 m^2
On the Earth, insolation (We) = Psun/Ae
Therefore,
Psun (Rate at which sun emits energy) = We*Ae = 1.4*2.827*10^23 = 3.958*10^23 kW = 3.958*10^26 W