Physics quiz archive: electromagnetic waves, polarization, reflection/refraction

Physics 205B Quiz 1, spring semester 2020
Cuesta College, San Luis Obispo, CA
Sections 30882, 30883, version 1
Exam code: quiz01PxP7



Sections 30882, 30883 results

0- 6 :	* [low = 6]
7-12 :	**
13-18 :	*********
19-24 :	*********** [mean = 21.9 +/- 5.9]
25-30 :	************ [high = 30]

Physics quiz archive: electromagnetic waves, polarization, reflection/refraction

Physics 205B Quiz 1, spring semester 2019
Cuesta College, San Luis Obispo, CA
Sections 30882, 30883, version 1
Exam code: quiz01aN7u




Sections 30882, 30883 results

0- 6 :	* [low = 3]
7-12 :	****
13-18 :	*******
19-24 :	***************** [mean = 20.8 +/- 5.7]
25-30 :	******** [high = 27]

Physics quiz archive: electromagnetic waves, polarization, reflection/refraction

Physics 205B Quiz 1, spring semester 2018
Cuesta College, San Luis Obispo, CA
Sections 30882, 30883, version 1
Exam code: quiz01AM0l




Sections 30882, 30883 results

0- 6 :
7-12 :	* [low = 9]
13-18 :	****
19-24 :	**********
25-30 :	************** [mean = 24.7 +/- 4.9] [high = 30]

Physics quiz archive: electromagnetic waves, polarization, reflection/refraction

Physics 205B Quiz 1, spring semester 2017
Cuesta College, San Luis Obispo, CA
Sections 30882, 30883, version 1
Exam code: quiz01Om6A


Sections 30882, 30883 results

0- 6 :
7-12 :	**** [low = 9]
13-18 :	**********
19-24 :	******* [mean = 19.7 +/- 6.0]
25-30 :	******* [high = 30]

Presentation: double-slit interference

Here we have microwaves (as discussed previously, a long wavelength form of electromagnetic radiation) from two side-by-side in-phase sources, interfering at a detector that can be moved at various locations to detect their interference, whether constructive or destructive (as translated into an audio signal). (Video link: "MIT Physics Demo--Microwave Interference.")

In the previous presentation we discussed the conditions for constructive or destructive interference for waves (of the same wavelength) due to phase and/or path differences. In this presentation we discuss the very specific case of waves (again, of the same wavelength) from two side-by-side in-phase sources, which we will see has been classically called "double-slit" interference.

First, path-length differences.

The waves we are considering will come from two sources that are in phase, so we do not need to concern ourselves with the out of phase sources. Since source phase differences don't matter here--only path differences--then we must pay careful attention to the difference in path length: how much longer the wave from one source travels than the wave from the other source, as they reach and interfere at the position of the detector, as it moves from side-to-side.

We are going to make the assumption that the detector is sufficiently (approaching infinitely?) far away from the two sources (spaced apart by a distance d) that the two waves will travel along a parallel path 1 and path 2. Then the location of the detector can be specified merely by the angle θ (where θ = 0° would be on the center line).

In this case, for the angle θ shown, waves travel longer along path 2. How much longer the waves travel along this longer path can be given by the relation ∆l = dsinθ. (There is a trigonometry derivation using the right triangle for this relation, but the focus here is on relating ∆l with the resulting constructive interference (maxima) or destructive interference (minima), and later on during problem-solving we'll use the ∆l = dsinθ relation to find these maxima and minima θ angles, without worrying too much about how to derive this ∆l = dsinθ relationship.)

Here's the simple case (θ = 0°) where both waves leave their slits to travel equal distances to the distant detector towards the right. Do the waves start from their slits in phase or out-of-phase? Is the path difference a whole wavelength or a half-wavelength? Does constructive interference (maxima) or destructive interference (minima) occur at the detector?

In this case, both waves leave their slits to travel unequal distances to the distant detector towards the right, located at an angle of θ = +23° off the (dashed) center line, such that path 1 is shorter than path 2. Do the waves start from their slits in phase or out-of-phase? Is the path difference a whole wavelength or a half-wavelength? Does constructive interference (maxima) or destructive interference (minima) occur at the detector? What would happen if the detector were instead placed at θ = –23° (same angle but on the other side of the center line)?

Now in this case, both waves leave their slits again to travel unequal distances to the distant detector towards the right, located at an angle of θ = –51° off the (dashed) center line, such that path 1 is longer than path 2. Do the waves start from their slits in phase or out-of-phase? Is the path difference a whole wavelength or a half-wavelength? Does constructive interference (maxima) or destructive interference (minima) occur at the detector? What would happen if the detector were instead placed at θ = +51° (same angle, but on the other side of the center line)?

Second, where we are going with this path-length difference relation: locating where these two sources interfere constructive (maxima) or destructively (minima).

As discussed before, for in phase sources, the difference in path length will be some integer m multiple of a wavelength for constructive interference, or will be some integer and a half (m + 1/2) multiple of wavelength for destructive interference.

So now let's put in our approximation for the path difference ∆l = dsinθ, for two waves from side-by-side sources reaching a (distant) detector located at an angle θ. What we will wind up with is a relation between the angle θ that a distant detector is located at, and the condition for either constructive (maxima) or destructive (minima) interference to occur. So given the wavelength λ of the two side-by-side sources, and the separation distance d between the two side-by-side sources, then plugging in different integer m values (0, ±1, ±2, ±3, etc.) allows us to solve for different θ angles where either constructive (maxima) or destructive (minima) interference occurs.

You will demonstrate this for yourselves in recreating a classic experiment in laboratory. Using laser light (of a given wavelength λ) that illuminates two very closely spaced together slits (two in-phase sources spaced a distance d apart), there will appear bright (maxima, or constructive interference) regions and dark (minima, or destructive interference) regions on a screen (the detector) at certain θ angles, as predicted by the double-slit interference maxima/minima equations.

Physics quiz archive: electromagnetic waves, reflection/refraction

Physics 205B Quiz 1, spring semester 2016
Cuesta College, San Luis Obispo, CA
Sections 30882, 30883, version 1
Exam code: quiz01sN0w



Sections 30882, 30883 results

0- 6 :	*** [low = 6]
7-12 :	**
13-18 :	************
19-24 :	****************** [mean = 20.7 +/- 6.1]
25-30 :	******** [high = 30]

Physics quiz archive: electromagnetic waves, reflection/refraction

Physics 205B Quiz 1, spring semester 2015
Cuesta College, San Luis Obispo, CA
Sections 30882, 30883, version 1
Exam code: quiz01c0Co



Sections 30882, 30883 results

0- 6 :	*** [low = 6]
7-12 :	***
13-18 :	***********
19-24 :	************************* [mean = 19.9 +/- 5.7]
25-30 :	****** [high = 30]

Physics quiz archive: electromagnetic waves, reflection/refraction

Physics 205B Quiz 1, spring semester 2014
Cuesta College, San Luis Obispo, CA
Sections 30882, 30883, version 1
Exam code: quiz01ksB4



Sections 30882, 30883 results

0- 6 :
7-12 :	* [low = 9]
13-18 :	***********
19-24 :	****************** [mean = 22.6 +/- 5.0]
25-30 :	*********** [high = 30]

Physics quiz archive: electromagnetic waves, reflection/refraction

Physics 205B Quiz 1, spring semester 2013
Cuesta College, San Luis Obispo, CA
Section 30882, version 1
Exam code: quiz01b3Es



Section 30882 results

0- 6 :	* [low = 6]
7-12 :	****
13-18 :	**************
19-24 :	************ [mean = 19.1 +/- 5.9]
25-30 :	**** [high = 30]

Presentation: polarization

Look at this iPad. Just look at it. Through this pair of polarizing sunglasses. Wait, what? How could you not see that? (Video link: "iPad display vs. polarized dark glasses: Portrait view problem.")

In this presentation we will develop a simple model of generating and receiving light with antennas, and transmission of light through filters.

First, developing a simple model of light, and sending and receiving light using antennas.

Our working model of light (considering that we are covering this topic before electromagnetism) is a transverse wave, much like along this length of rope. The periodic disturbances travel along the rope, while the disturbances themselves wiggle back-and-forth sideways to the rope, in this case vertically (due to the person shaking the rope up-and-down), but these disturbances could also be horizontal, if the person shakes the rope side-to-side. The direction of these disturbances (which are perpendicular to the direction along the rope) correspond to the polarization of the transverse wave.

The sources of light are vibrating charges, such as atoms and electrons. If these charges vibrate in random directions, such as this molten sample, then they will generate unpolarized light, meaning that all possible polarization directions (vertical, horizontal, diagonal, etc.) perpendicular to the direction of motion are included.

In contrast, a radio antenna is an example of a polarized light source, as electrons are forced to flow up-and-down the length this vertical antenna, such that the light (in the form of radio waves) will in this case have only vertical polarization.

Let's extend this concept of sources and receivers of polarized light.

This is a schematic model of what happens when electrons are force to oscillate up-and-down along a vertical antenna. This generates radio waves with transverse peaks and troughs that are aligned vertically. In order to best receive this radio wave, a receiving antenna would need to be aligned vertically as well.

Here we have a practical example of a vertical antenna array. This will produce vertically polarized radio waves in all directions (except for straight up above the airplane, and straight down below the airplane--can you figure out why?).

We can demonstrate the production and reception of vertically polarized radio waves. At left is a vertical antenna with electrons that oscillate up-and-down, generating vertically polarized radio waves. Holding a vertical antenna nearby will pick up the vertically polarized radio waves, causing electrons to oscillate up-and-down along the length of this receiver, producing a current that illuminates the bulb. Keep in mind that the energy to illuminate the bulb is transmitted from the vertical source antenna to the vertical receiver antenna.

Here is a schematic model of what happens when electrons are force to oscillate side-to-side along a horizontal antenna. This generates radio waves with transverse peaks and troughs that are aligned horizontally. In order to best receive this radio wave, a receiving antenna would need to be aligned horizontally as well.

Here we have a practical example of a horizontal antenna array. This will produce horizontally polarized radio waves in all directions (except towards the upper right of this picture, and away towards the lower left of this picture--can you figure out why?).

We can also demonstrate the production and reception of horizontally polarized radio waves. At left is a horizontal antenna with electrons that oscillate side-to-side along its length, generating horizontally polarized radio waves. Holding a vertical receiver antenna nearby will not pick up these horizontally polarized radio waves, and no energy is received to illuminate the bulb. However, if the receiver antenna is aligned horizontally, then it will pick up the horizontally polarized radio waves, causing electrons to oscillate side-to-side along the length of this receiver and producing a current that illuminates the bulb. So for best reception, the relative orientations of the source antenna and receiving antenna should match.

Second, let's consider what we can do to the polarization of light (in this case, visible light) while it is en route from a source to a receiver.

In this movie, a T-1000 Terminator made of liquid metal is able to walk through the bars of this gate, but the gun it is holding sideways is caught between the bars, and must be held vertically in order to pass through the bars. This is the concept behind our "fencepost model" of the transmission (or blocking) of light with different polarizations: anything that is aligned parallel to the opening will pass through, while anything aligned perpendicular to the opening will be blocked.

So in this model, light is represented by oscillations that travel along a rope. If the polarization of this light--the sideways direction of the oscillations on the rope--is parallel the direction of the openings between the fenceposts, then it will be transmitted through. If the polarization of this light is perpendicular to the direction of the openings between the fenceposts, then it will be blocked. The set of fenceposts is said to be a polarizer, and the orientation of the openings between the fenceposts is the transmission axis of the polarizer.

Let's separately consider the effects of a polarizer on filtering either unpolarized or polarized light. For unpolarized light here, traveling out of the plane of this page--with every possible sideways oscillation--is drawn with just horizontal and vertical polarization directions, and we assume that every other possible diagonal polarization direction is merely the superposition of these two polarization directions. If unpolarized light (with an original intensity S̄_unpolarized = 1) is incident on a polarizer, then only the polarization direction parallel to the transmission axis (here, vertical) will pass through, and the polarization direction perpendicular to the transmission axis (here, horizontal) will be blocked. The result is that only one-half of the original intensity S̄_unpolarized of unpolarized light will pass through the polarizer, and the remaining light will be vertically polarized with an intensity of S̄_polarized = 0.5.

If instead polarized light (with an original intensity S̄_polarized = 1) is incident on a polarizer, then S̄_through, the intensity of light transmitted through will be given by Malus' law, which depends on the square of the cosine of the angle θ measured between the polarization of the incident light, and the transmission axis of the polarizer. As expected, if the polarization of the incident light is parallel to the transmission axis of the polarizer (such that θ = 0°), then all of this light will pass through the polarizer, and S̄_through = 1.

If the transmission axis of the polarizer was perpendicular to the polarization of the incident light (such that θ = 90°), then all of this light would be blocked (making S̄_through = 0), as the square of cos(90°) = 0.

Now most generally the transmission axis of the polarizer may be at some arbitrary angle θ with respect to the polarization of the incident light. The intensity of the light that will pass through the polarizer is merely calculated via Malus' law, while the (lessened intensity) of light that passed through will have a polarization direction that exactly matches the transmission axis of the polarizer.

Interesting things happen when polarizers are stacked such that light must pass through several of them. Here unpolarized light is projected through the first (smaller) polarizer. If the transmission axis of this first polarizer is horizontal, then one-half of the unpolarized light will pass through, and the light that passes through is now horizontally polarized.

When the light that passed through the first polarizer is now completely blocked by the second polarizer, the transmission axis of the second polarizer is:
(A) horizontal.
(B) vertical.
(C) diagonal, upper left to lower right.
(D) diagonal, lower left to upper right.
(E) (Unsure/guessing/lost/help!)

Back to the iPad (here, held in portrait mode) and polarizing sunglasses from the start of this presentation. Polarizing sunglasses are typically manufactured with a vertical transmission axis (to filter out horizontally polarized glare while fishing or driving).

Light from the screen of the iPad (in portrait mode) is:
(A) horizontally polarized.
(B) vertically polarized.
(C) diagonally (upper left to lower right) polarized.
(D) diagonally (lower left to upper right) polarized.
(E) unpolarized.
(F) (Unsure/guessing/lost/help!)

Now consider a Nexus 4 smartphone (again, held in portrait mode), and polarizing sunglasses (with a vertical transmission axis).

Light from the screen of the Nexus 4 is:
(A) horizontally polarized.
(B) vertically polarized.
(C) diagonally (upper left to lower right) polarized.
(D) diagonally (lower left to upper right) polarized.
(E) unpolarized.
(F) (Unsure/guessing/lost/help!)