Direction of oscillations of a polarised light wave

[songoku]

Homework Statement

Please see below

Relevant Equations

not sure

My answer is (C) because light can only travel in one direction after being polarised so the oscillation is also in the same direction as the direction of travel. Am I correct?

Thanks

 

[Orodruin]

Homework Statement: Please see below
Relevant Equations: not sure

Am I correct?

No. Light is a transversal wave. It has no longitudinal component.

 

[BvU]

Comes from solving the Maxwell equations :

E and B are perpendicular to each other and to the direction of wave propagation, and are in phase with each other. A sinusoidal plane wave is one special solution of these equations. Maxwell's equations explain how these waves can physically propagate through space. The changing magnetic field creates a changing electric field through Faraday's law. In turn, that electric field creates a changing magnetic field through Maxwell's modification of Ampère's circuital law. This perpetual cycle allows these waves, now known as electromagnetic radiation, to move through space at velocity c.

$\$

 

[jbriggs444]

My answer is (C) because light can only travel in one direction

There is a tendency to imagine light waves like other transverse waves. Like water waves or transverse waves in a stretched string. In those cases there is physical motion perpendicular to the propagation direction.

Light is not like that. There is no physical motion from side to side. It is (as @BvU points out) the directions of the changing fields that are perpendicular to the direction of travel.

 

[songoku]

No. Light is a transversal wave. It has no longitudinal component.

Comes from solving the Maxwell equations :


$\$

There is a tendency to imagine light waves like other transverse waves. Like water waves or transverse waves in a stretched string. In those cases there is physical motion perpendicular to the propagation direction.

Light is not like that. There is no physical motion from side to side. It is (as @BvU points out) the directions of the changing fields that are perpendicular to the direction of travel.

I see, the answer must be (D).

Thank you very much for the help Orodruin, BvU, jbriggs444

 

[Steve4Physics]

My answer is (C) because light can only travel in one direction after being polarised so the oscillation is also in the same direction as the direction of travel. Am I correct?

No. To add to what's already been said,...

EM waves are difficult to visualise but this may help (at the risk of being accused of oversimplification).

Imagine a plane polarised EM wave passing by a small test-charge (red dot in diagram). The wave’s oscillating E-field produces an oscillating force on the test-charge. The test-charge oscillates along a line; the line matches the directions of the wave’s E-field.

For example, if the charge oscillates along the red line in diagram, then the E-field directions match the red line. There in no test-charge motion (therefore no component of the E-field) in the direction of propagation (z) because we’re dealing with a transverse wave.

Edit. The line lies in a plane (blue) perpendicular to the direction of propagation - and this is the plane referred to in answer option D.

 

[kuruman]

There is no physical motion from side to side.

Perhaps there is if you put an antenna in the path of the EM wave.

 

[Orodruin]

No. To add to what's already been said,...

EM waves are difficult to visualise but this may help (at the risk of being accused of oversimplification).

Imagine a plane polarised EM wave passing by a small test-charge (red dot in diagram). The wave’s oscillating E-field produces an oscillating force on the test-charge. The test-charge oscillates along a line; the line matches the directions of the wave’s E-field.

View attachment 356600

For example, if the charge oscillates along the red line in diagram, then the E-field directions match the red line. There in no test-charge motion (therefore no component of the E-field) in the direction of propagation (z) because we’re dealing with a transverse wave.

Edit. The line lies in a plane (blue) perpendicular to the direction of propagation - and this is the plane referred to in answer option D.

… and for this reason the sky at an angle of 90 degrees from the Sun is highly polarised!

This is the reason I look like a crazy person tilting my head from side to side when wearing polarized sunglasses on a sunny day.

(other fun exercises when wearing polarized sunglasses include, but are not limited to: looking at the glass panes of windows at different angles, looking at the LCD screens at bus stops - most are ”correctly” oriented so show really well with your head straight and completely black if you turn it by 90°)

 

[kuruman]

This is the reason I look like a crazy person tilting my head from side to side when wearing polarized sunglasses on a sunny day.

You can eliminate the appearance of craziness and get more of an angle of rotation if you kept your head fixed, took your glasses off and turn them. That's what I do to verify the axis of polarization relative to the glasses. I turn the glasses and look for a minimum at obliquely reflected light off a linoleum floor which I know is (at least partially) polarized in the direction parallel to the floor.

 

[Steve4Physics]

Q: How can you spot a physicist on a sunny day?
A: See @Orodruin's and @kuruman's Post ##8 and #9.

 

[Orodruin]

You can eliminate the appearance of craziness and get more of an angle of rotation if you kept your head fixed, took your glasses off and turn them. That's what I do to verify the axis of polarization relative to the glasses. I turn the glasses and look for a minimum at obliquely reflected light off a linoleum floor which I know is (at least partially) polarized in the direction parallel to the floor.

Turning your smartphone from portrait to landscape mode is also fun

 

[berkeman]

This is the reason I look like a crazy person tilting my head from side to side when wearing polarized sunglasses on a sunny day.

Well, it wouldn't be so bad if your dog would stop imitating you as you walk him...

https://www.gettyimages.com/photos/dog-head-tilted

 

[DaveE]

https://www.khanacademy.org/science...s/v/polarization-of-light-linear-and-circular