20160226

Presentation: diffraction

Look at this fire hose nozzle. Just look at it. Pinching the flow of water makes it spread out more; while opening up the nozzle narrows the spread of water.

This is not meant to be a technically correct explanation for what we will see with waves that move through a single opening, but this very crude analogy will serve our purposes well enough.

Previously we considered the interference of waves from separate in phase sources, as monochromatic (same wavelength λ light) through two slits. Here we look at diffraction, which is the spread of light from a single slit.

First, terminology.

Notice how these parallel water wavefronts spread out after passing through the central opening in this inlet. This is an example of how waves diffract as they pass through a "single-slit" opening.

The relevant parameters are the wavelength λ of the parallel wavefronts and the slit opening width W, which affect how these waves diffract and spread out, given by the "half-angle" θ, as measured from the center line.

Second, quantifying the spread of these waves after diffracting through the singe slit.

If you squint (in order to increase the contrast of the diffracted wavefronts), you can make out a faint destructive region on either side of the center line, which forms the boundary of most of the diffracted wave energy. This is the first minima angle θ.

For our purposes will not derive this equation, as this would demand a non-trivial amount of calculus, or a very non-trivial end-run around calculus using qualitative arguments.
The equation for this diffraction minima is given by Wsinθ = mλ, where θ is the "half-angle" of the spread of the diffracted waves, and m = 1 (which if this doesn't freak you out by the resemblance to the double-slit maxima equation, it should). Just work with this, and let's see what it can tell us.

Much like constricting the nozzle would spread out the flow of water--but remember that this is nothing more than an analogy, and has no explanative power than reproducing the same result.
Since the slit width W and spread half-angle θ appear on both sides of the first diffraction minima equation, then making the slit opening smaller would result in increasing the spread of the diffracted waves.

Also making the slit opening larger would result in decreasing the spread of the diffracted waves.

An example of this is the diffraction of light through the circular aperture of a telescope with a "width" W (although the more correct equation in this case would have a correction for a diameter of a circular opening: Wsinθ = (1.22)(1)λ). This is the Whirlpool Galaxy M51 as seen by the NASA Spitzer Space Telescope and the European Space Agency Herschel Space Telescope, as observed with the same infrared wavelength λ. As the Spitzer Space Telescope has a much smaller mirror diameter, light from each part of the galaxy will diffract more and spread out more, resulting in a much less resolved image than from the Herschel Space Telescope, with a much larger mirror diameter, such that light from each part of the galaxy will diffract less and spread out less, resulting in a much better resolved image with finer details left intact.

Note the fainter fringes on either side of the central maximum 'spread.'
In laboratory you will shine a laser on your hair. Since the lasers are relatively low-powered (but don't shine them in your eyes), you won't be able to burn through your hairs, but light will diffract around either side of the hair shaft. This turns out to be entirely equivalent to light shining through a single slit of the same width as your hair, and the resulting diffraction pattern on a distant screen shows the spread of light contained within the first minima θ angles on either side of the center line. Depending on how thick your hair is will determine how little (or much) light will diffract and spread out on the screen.

20160223

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.

20160218

Astronomy quiz archive: eclipses/history of astronomy

Astronomy 210 Quiz 2, spring semester 2016
Cuesta College, San Luis Obispo, CA

Section 30674, version 1
Exam code: quiz02NAwL


Section 30674
0- 8.0 :   *** [low = 4.5]
8.5-16.0 :   ********** [mean = 15.9 +/- 7.2]
16.5-24.0 :   *****
24.5-32.0 :   ** [high = 32.0]
32.5-40.0 :  


Section 30676, version 1
Exam code: quiz02sL4g


Section 30676
0- 8.0 :   **** [low = 4.0]
8.5-16.0 :   ***************
16.5-24.0 :   **************** [mean = 19.7 +/- 7.4]
24.5-32.0 :   **************
32.5-40.0 :   * [high = 32.5]

20160215

Video: "Keck in Motion" excerpts

"Keck in Motion"
Andrew Cooper
vimeo.com/36442707

Excerpts for whole-class discussion (ask students: "what do you notice is going on?")





20160209

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]

20160204

Astronomy quiz archive: stars/sun/seasons/moon phases

Astronomy 210 Quiz 1, spring semester 2016
Cuesta College, San Luis Obispo, CA

Section 30674, version 1
Exam code: quiz01n0iR


Section 30674
0- 8.0 :   * [low = 8]
8.5-16.0 :   *****
16.5-24.0 :   *****
24.5-32.0 :   *** [mean = 25.5 +/- 10.5]
32.5-40.0 :   ******* [high = 40]


Section 30676, version 1
Exam code: quiz01sC4r


Section 30676
0- 8.0 :   ** [low = 8]
8.5-16.0 :   *****
16.5-24.0 :   *********************
24.5-32.0 :   ************* [mean = 24.9 +/- 7.8]
32.5-40.0 :   ************ [high = 40]

20160112

Astronomy in-class activity: planet-hunting

Astronomy 210 In-class activity 6 v.16.01.12, spring semester 2016
Cuesta College, San Luis Obispo, CA

Students find their assigned groups of three to four students, and work cooperatively on an in-class activity worksheet to determine where in the sky each naked-eye planet will be observed on a given date (here, February 4, 2016).




Previous posts:

20151211

Physics quiz archive: temperature, thermal equilibrium, heat transfer

Physics 205A Quiz 7, fall semester 2015
Cuesta College, San Luis Obispo, CA
Sections 70854, 70855, 73320, version 1
Exam code: quiz07zSOL



Sections 70854, 70855, 73320 results
0- 6 :   * [low = 3]
7-12 :   *****
13-18 :   *****************
19-24 :   ****************************** [mean = 21.2 +/- 5.6]
25-30 :   **************** [high = 30]

Astronomy quiz archive: Milky Way, cosmology

Astronomy 210 Quiz 7, fall semester 2015
Cuesta College, San Luis Obispo, CA

Section 70158, version 1
Exam code: quiz07srrH


Section 70158
0- 8.0 :   *** [low = 4.0]
8.5-16.0 :   **********
16.5-24.0 :   *********** [mean = 21.5 +/- 9.3]
24.5-32.0 :   *********
32.5-40.0 :   ****** [high = 40.0]


Section 70160, version 1
Exam code: quiz07nM0r


Section 70160
0- 8.0 :   *** [low = 4.0]
8.5-16.0 :   ******
16.5-24.0 :   ******* [mean = 20.7 +/- 9.6]
24.5-32.0 :   *******
32.5-40.0 :   ** [high = 40.0]

20151208

FCI pre-test comparison: Cuesta College versus UC-Davis (fall semester 2015)

Students at both Cuesta College (San Luis Obispo, CA) and the University of California at Davis were administered the 30-question Force Concept Inventory (Doug Hestenes, et al.) during the first week of instruction.

Cuesta College
Physics 205A
fall semester 2015    
UC-Davis
Physics 7B
summer session II 2002
N85 students*76 students*
low 2 2
mean    10.0 +/- 5.5 9.1 +/- 4.3
high2427

*Excludes students with negative informed consent forms (*.pdf)

Student's t-test of the null hypothesis results in p = 0.19 (t = 1.31, sdev = 4.93, degrees of freedom = 159), thus there is no significant difference between Cuesta College and UC-Davis FCI pre-test scores.

Later this semester (fall 2015), a comparison will be made between Cuesta College and UC-Davis FCI post-tests, along with their pre- to post-test gains.

D. Hestenes, M. Wells, and G. Swackhamer, Arizona State University, "Force Concept Inventory," Phys. Teach. 30, 141-158 (1992).
Development of the FCI, a 30-question survey of basic Newtonian mechanics concepts.

Previous FCI results: