The discussion centers on interpreting the polarization of light in relation to Fresnel's equations. It clarifies that the plane of incidence differs from the glass interface, with the E-field components E|| and Eperp defined in relation to these planes. The question of how to determine the polarization of the reflected beam is raised, emphasizing the need for knowledge of the incident beam's polarization. Additionally, it is noted that Fresnel's equations provide information on the transmission fractions of E|| and Eperp, which can affect the overall polarization angle. Understanding these concepts is crucial for correctly applying Fresnel's equations to problems involving polarized light.
#1
quietrain
648
2
Homework Statement
The Attempt at a Solution
how do i interpret such a question? is there something wrong with the question?
in particular, the light is polarized at 45degrees wrt the plane of incidence(e.g x-y plane),
wouldn't that mean that
1) my E-field is orientated in the z=x line? (when we say light is polarized in a direction, it really means the E-field right?) [so what has this got to do with the E(parallel) and E(perpendicular)? since these E are wrt to the interface of the impact right?]
2) so how do i find the polarization of the reflected beam (E-field of reflected beam right?) , if the question didn't give me the polarization of the incident beam?
3)lastly, does fresnel's equation give me the polarising angle? because it seems that they only give me the incident and reflected and transmitted angles? thanks!
how do i interpret such a question? is there something wrong with the question?
in particular, the light is polarized at 45degrees wrt the plane of incidence(e.g x-y plane),
wouldn't that mean that
1) my E-field is orientated in the z=x line? (when we say light is polarized in a direction, it really means the E-field right?) [so what has this got to do with the E(parallel) and E(perpendicular)? since these E are wrt to the interface of the impact right?]
2) so how do i find the polarization of the reflected beam (E-field of reflected beam right?) , if the question didn't give me the polarization of the incident beam?
3)lastly, does fresnel's equation give me the polarising angle? because it seems that they only give me the incident and reflected and transmitted angles? thanks!
The plane of incidence is not the same thing as the plane of the glass interface. The plane of incidence is the plane in which the incident and reflected rays lie. In other words, it is the plane of the propagation (k) vectors. So, for example, if the directions are like so:
Code:
x
^
|
| k vectors lie in this plane
|
------->z
y points out of the page
Then the glass interface is the x-y plane, whereas the plane of incidence is the "page" (the x-z plane)
Therefore E|| is the component of the E-field that lies in the page, and Eperp is the component that points out of the page. These two components are equal, meaning that the E-field makes a 45 degree angle with the "page." In other words, it makes a 45-degree angle in the plane of polarization which is the plane perpendicular to k (yet another plane!)
The Fresnel equations can tell you what fraction of E|| is transmitted, and what fraction of Eperp is transmitted. Obviously if these fractions are not the same, then the two components will no longer be equal, and hence the angle of total E-field in the polarization plane will have rotated.
EDIT: As an example, if the direction of the propagation k vector were just the z-direction (which would mean normal incidence), then E|| would be in the x-direction, since it must lie in the x-z plane (the plane of incidence) but still be perpendicular to k. Eperp would be in the y-direction, since it must be perpendicular to the page, and also perpendicular to k. In this example, the plane of polarization (the plane in which E-field lies) would just be the x-y plane.
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#3
quietrain
648
2
hmm... there's a plane of polarization too?
ok, from your example(k in z direction), since the question says 45degrees wrt plane of incidence, so it means my E-field is in the x-y plane right?
so, the rperp means the fraction of reflected light in the perpendicular direction wrt the plane of incidence right? (x-y plane again)
, similarly, rparallel is parallel to plane of incidence (x-z plane) ?
i think i am getting it, becos i got confused with the parallel and perpendicular components of E when i was deriving the fresnel's equations from the boundary conditions.
the boundary conditions' parallel and perpendicular E and B are wrt to the interface right? not the plane of incidence.
i watched prof walter lewin's lecture on waves and he is a great teacher :)
The answer is (B) but I don't really understand why.
Based on formula of Young Modulus:
$$x=\frac{FL}{AE}$$
The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A##
I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...