00:01
Hello students, in this question given as a electromagnetic field which is moving in the y direction and incident on a dielectric slab.
00:10
The dielectric slab is from the z direction from z is equal to 0 to up to z greater than z.
00:18
So, in the first part of the question we have to calculate the polarization.
00:27
Since electric field is in the y direction and which is perpendicular to the plane of incidence, therefore the wave will be the perpendicular polarized.
00:39
From the given incident electric field we can see that the propagation vector is will be equal to the k 4 in x direction plus 3 in z direction.
00:52
From ray diagram we can write tan theta i will be equal to the k propagation vector along the x direction and propagation vector of incident electric field along the z direction will be equal to the 4 upon 3.
01:17
Now incident angle will be equal to the 53 .15 degree.
01:23
In the second part of the question we have to calculate the reflected electric field.
01:30
The reflected electric field will also be in the free space.
01:34
Therefore from the ray diagram we can write the propagation vector is equal to k r x x direction this will be equal to the minus k r z in z direction.
01:49
So, this will be 4 x cap minus 3 z cap.
01:58
Now we can write the magnitude of propagation vector of reflected wave will be equal to the 5.
02:07
Hence we can say that the incident angle the reflected angle will be equal to the incident angle which is 53 .15 degree.
02:21
Now we will apply the snell's law on transmitted and the incident boundary.
02:28
So, sin theta t transmitted angle will be equal to the n1 by n2 sin theta i.
02:38
Reflective index n1 n2 c will be equal to the c under root mu1 epsilon 1 permittivity and permeability of free space and permittivity and permeability of the dielectric medium.
02:56
And in this question given as a relative permittivity and relative permeability value which is given as epsilon 2 by epsilon 1 is equal to 2 .5 and mu 2 by mu 1 is equal to...