tag:blogger.com,1999:blog-59107693351736121442024-02-18T20:33:03.438-06:00PH 102 / Summer 2012General Physics II,<br>
University of Alabama<br>
Dr. LeClairpleclairhttp://www.blogger.com/profile/07263994015241270268noreply@blogger.comBlogger516125tag:blogger.com,1999:blog-5910769335173612144.post-1996256617947584492012-08-07T01:32:00.002-05:002012-08-07T08:00:52.551-05:00Grades are in<div dir="ltr" style="text-align: left;" trbidi="on">
Your exam 3 scores and final averages are now posted to eLearning. Your exam 2 scores have also been updated to reflect any gains from the extra credit problem. Exam 3 went very well, with a class average of nearly 90%, meaning there was no scaling.<br />
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Your final averages have taken into account dropping your lowest quiz, HW, and lab scores. Recall that exams are worth 20% each, HW is 20%, and labs and quizzes are worth 10% each.<br />
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Let me know if you see anything still missing or incorrect, or have questions about how your grades worked out. Just to head off a common question: I know some of you will find yourselves very close to a grade boundary, but at this point I can't really offer any additional extra credit. I will very carefully check borderline cases though.<br />
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Your final letter grades are also posted on myBama now.<br />
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UPDATE - your exam 2 score will only change if your extra credit score is higher than your lowest non-dropped question on exam 2, since the extra credit question replaces the lowest-non dropped question if it is higher. Thus, if your lowest non-dropped question on exam 2 was 7/10 and you got 7/10 on the extra credit, your exam score does not change.</div>pleclairhttp://www.blogger.com/profile/07263994015241270268noreply@blogger.com5tag:blogger.com,1999:blog-5910769335173612144.post-15306988444151843602012-08-03T11:35:00.002-05:002012-08-04T01:52:57.213-05:00More grade updates<div dir="ltr" style="text-align: left;" trbidi="on">
UPDATE: I have added HW5 grades, and made a few corrections as you've notified me of missing grades (some of which I am still tracking down). The only thing missing now <i>should</i> be exam 3 and the exam 2 bonus ... all labs, quizzes, and HW should be there. If you notice anything missing let me know as soon as you can, and I'll track it down one way or another. Should have exam 3 grades and a final overall average posted by Sunday night I expect.<br />
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Grades for everything except HW5, the bonus for Exam 2, and Exam 3 should now be on eLearning. Once Exam 3 is graded, I will post a final average as well. Recall that I will drop 1 quiz, 1 HW, and 1 lab.</div>pleclairhttp://www.blogger.com/profile/07263994015241270268noreply@blogger.com3tag:blogger.com,1999:blog-5910769335173612144.post-41373067077022391222012-08-02T23:25:00.000-05:002012-08-02T23:25:01.737-05:00Grade updates<div dir="ltr" style="text-align: left;" trbidi="on">
A few more things on eLearning. The grades that are <b>not</b> ready yet (and hence do not show up on eLearning) are:<br />
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HW4, HW5<br />
Q5, Q7<br />
L6, L9<br />
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I expect most of those things to show up some time tomorrow. Final grades will be ready by Sun-Mon I hope.</div>pleclairhttp://www.blogger.com/profile/07263994015241270268noreply@blogger.com0tag:blogger.com,1999:blog-5910769335173612144.post-4254261120900501952012-08-02T22:03:00.001-05:002012-08-02T22:03:22.027-05:00All quizzes now have solutions<div dir="ltr" style="text-align: left;" trbidi="on">
at least the ones from this summer, that is.</div>pleclairhttp://www.blogger.com/profile/07263994015241270268noreply@blogger.com4tag:blogger.com,1999:blog-5910769335173612144.post-86866136399596938092012-08-02T21:42:00.002-05:002012-08-02T21:42:12.223-05:00Summer 2009 final<div dir="ltr" style="text-align: left;" trbidi="on">
the relevant problems (and a couple more) have solutions.</div>pleclairhttp://www.blogger.com/profile/07263994015241270268noreply@blogger.com0tag:blogger.com,1999:blog-5910769335173612144.post-87766898941738525522012-08-02T20:50:00.003-05:002012-08-02T20:50:47.914-05:00Final time & location<div dir="ltr" style="text-align: left;" trbidi="on">
The final is tomorrow morning at 8am in room 203, i.e., the lecture hall.</div>pleclairhttp://www.blogger.com/profile/07263994015241270268noreply@blogger.com0tag:blogger.com,1999:blog-5910769335173612144.post-4188164193779006032012-08-02T20:50:00.001-05:002012-08-02T20:50:22.572-05:00Spring 2008 final<div dir="ltr" style="text-align: left;" trbidi="on">
The questions relevant for tomorrow's exam now have answers, and some of them solutions. </div>pleclairhttp://www.blogger.com/profile/07263994015241270268noreply@blogger.com0tag:blogger.com,1999:blog-5910769335173612144.post-73553484615148598822012-08-02T20:11:00.001-05:002012-08-02T21:04:48.229-05:00Relax. Read your textbook. That is all.<div dir="ltr" style="text-align: left;" trbidi="on">
<a href="http://faculty.mint.ua.edu/~pleclair/ph102/Exams/cheet-sheet_ex3.pdf">If you're not going to that, you can check out the formula sheet though</a>.</div>pleclairhttp://www.blogger.com/profile/07263994015241270268noreply@blogger.com0tag:blogger.com,1999:blog-5910769335173612144.post-58681141447870929372012-08-02T17:37:00.002-05:002012-08-02T17:37:55.247-05:00Summer 08 final exam<div dir="ltr" style="text-align: left;" trbidi="on">
has a partial solution up. All the optics problems have answers at least. </div>pleclairhttp://www.blogger.com/profile/07263994015241270268noreply@blogger.com7tag:blogger.com,1999:blog-5910769335173612144.post-5697870056826714432012-08-02T17:18:00.003-05:002012-08-02T17:18:41.188-05:00Old finals<div dir="ltr" style="text-align: left;" trbidi="on">
Once I finally get home, I will start posting answers to as many of the old final questions as I can. I have many more requests than I can probably do tonight, but I'll do what I can. Start looking for them around 7pm.</div>pleclairhttp://www.blogger.com/profile/07263994015241270268noreply@blogger.com3tag:blogger.com,1999:blog-5910769335173612144.post-90800543277195151952012-08-02T01:59:00.002-05:002012-08-02T18:48:40.853-05:00A few more example problems<div dir="ltr" style="text-align: left;" trbidi="on">
<a href="http://faculty.mint.ua.edu/~pleclair/ph102/Exams/q-prac.pdf">on quantum/atomic physics.</a> I should be able to post some more tomorrow evening, but these are highly relevant for the exam.</div>pleclairhttp://www.blogger.com/profile/07263994015241270268noreply@blogger.com8tag:blogger.com,1999:blog-5910769335173612144.post-4974899845988817042012-08-02T01:34:00.002-05:002012-08-02T01:36:54.884-05:00Exam 2 extra credit hints<div dir="ltr" style="text-align: left;" trbidi="on">
So the problem is a tough-looking one, but I swear this is an incredibly useful circuit. The circuit I gave you is a filter, and one fair tactic is to google "RLC filters" and see if you can find a match. Inelegant, but effective. I will tell you it is a "bandpass" filter, but you'll probably spend just as much time trying to understand the pages you find as you would just working the thing out with a little help. I say that because you will run across a lot of electrical engineering pages which speak a different breed of math than you are used to, and the language barrier will be tough to overcome, not because this is a particularly tough circuit.<br />
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We know reactances add like resistances. The whole circuit has L and C in parallel, and that combination in series with a resistance. The total reactance is then $X_{tot} = X_R + X_{L||C}$, where $X_{L||C}$ is the reactance of the capacitor and inductor in parallel, which you'd find by adding the $L$ and $C$ reactances like you would parallel resistors. The resistor's reactance is just $X_R=R$. The current in the circuit is then the input voltage divided by the total reactance, $I=V_{in}/X_{tot}$.<br />
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What is the output voltage? You're measuring the output voltage across the $L$ and $C$ parallel combination, so it must be $V_{out}=IX_{L||C}$. Find the equivalent reactance of $L$ and $C$ in parallel, add the resistor's reactance, and you have the total reactance. That gets you the current, and the current times the equivalent reactance of $L$ and $C$ in parallel gives you the output. This will all be a function of frequency as well as $R$, $L$, and $C$.<br />
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Qualitatively, you can ignore the $L$ and $C$ one at a time. Without the $C$, it is a high-pass. Without the $L$, it is a low-pass. The whole circuit looks like a combination of the two - hates both high and low frequencies, but will like a middle range in between.<br />
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To sketch the plot, presume some values for $R$, $L$, and $C$. I would say perhaps $R\!=\!100\,\Omega$, $L\!=\!1\,$mH, and $C\!=\!1\,\mu$F. What is it good for? Any time you want to let a small range of frequencies through, but nothing else. Radio comes to mind.</div>pleclairhttp://www.blogger.com/profile/07263994015241270268noreply@blogger.com0tag:blogger.com,1999:blog-5910769335173612144.post-44503624328997884612012-08-02T00:54:00.000-05:002012-08-02T00:57:22.259-05:00HW5 solutions<div dir="ltr" style="text-align: left;" trbidi="on">
<a href="http://faculty.mint.ua.edu/~pleclair/ph102/Homework/HW5_31Jul12_optics_SOLN.pdf">can be found here</a>. The exam problems on this material will (promise) be easier than the HW questions - the exam problems will probably be somewhere between these problems and the optics questions on your quizzes.<br />
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These HW problems were hard (mostly) intentionally - given that we had to cover the material rather quickly, I wanted you to wrestle with some harder problems in the hopes it would help the material sink in a bit better. Whether that works or not is debatable, but I think it makes sense at this point that at least reading the solutions for these harder problems will help you understand what's happening in simpler situations, even if you didn't quite solve them correctly on the HW.</div>pleclairhttp://www.blogger.com/profile/07263994015241270268noreply@blogger.com0tag:blogger.com,1999:blog-5910769335173612144.post-7625413167260792232012-08-01T01:04:00.002-05:002012-08-01T01:04:31.663-05:00Possible extra credit on exam 2<div dir="ltr" style="text-align: left;" trbidi="on">
<a href="http://faculty.mint.ua.edu/~pleclair/ph102/Exams/Ex2-makeup_Sum12.pdf">can be had by solving this problem</a>. Partial credit will be generous, and there is <i>no way</i> that doing or not doing the problem can <i>hurt</i> your grade. I will happily provide hints if you ask relatively direct and specific questions.</div>pleclairhttp://www.blogger.com/profile/07263994015241270268noreply@blogger.com3tag:blogger.com,1999:blog-5910769335173612144.post-17670160915817156322012-07-31T23:50:00.001-05:002012-07-31T23:50:15.821-05:00Wednesday's lab<div dir="ltr" style="text-align: left;" trbidi="on">
<a href="http://physics.ua.edu/lab10x/ph102/PDF/Atomic_Spectra_LAB.pdf">will be on atomic spectra</a>. Specifically, hydrogen and mercury, with a chance of helium. </div>pleclairhttp://www.blogger.com/profile/07263994015241270268noreply@blogger.com7tag:blogger.com,1999:blog-5910769335173612144.post-79686894170189139562012-07-31T21:06:00.002-05:002012-08-01T22:45:05.083-05:00Homework 5 hints<div dir="ltr" style="text-align: left;" trbidi="on">
I'll start off with number 1 and keep adding to this throughout the evening.<br />
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1. The apparent depth is where the light appears to have come from under the water (according to the observer) if refraction were not present. Extrapolate the light back from the observer, ignore refraction, and when the extrapolated ray reaches the same <i>horizontal</i> position as the actual object, you're at the apparent depth. Draw a little picture, it will help.<br />
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a. The angle of incidence can be related to the angle the (real) ray in the water makes with the surface via Snell's law. Once you have this angle, since you know the real depth you can find the lateral (horizontal) position of the object with some trigonometry. Now worry about the extrapolated ray under the water. You know its angle with respect to the surface is the same as the ray above the water by construction. Using the lateral position of the object and this angle, you can find the apparent depth.<br />
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b. You can generalize your result from a if you are careful about setting the angle to zero, or you can realize this is the same as a spherical surface with an infinite radius. For a spherical surface you know $n_1/p+n_2/q=(n_2-n_1)/R$. If $R$ tends to infinity, this gives $n_1/p+n_2/q$. The apparent depth is $|q|$, the real depth $p$, and that's that.<br />
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2. You are given $p+q=L$, and the lens equation tells you $1/p + 1/q = 1/f$. What will determine the minimum value for L? If we want to focus an image, we need to have a <i>real</i> image, not a virtual one. For a convex lens, this will happen only when $p>f$. We also know that $q$ must be positive for the image to be real. Combining these two facts, we really only need to make sure that $q$ is <i>real and positive</i> in order to have a real image formed. If $q$ is negative, the image is virtual. If $q$ is imaginary, so is the image ...<br />
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What you need to do, then, is solve one equation for $q$ and put it in the other. The resulting expression for $q$ will be quadratic, the condition that the roots are real is that the discriminant is positive. Start by noting that $1/p + 1/q = 1/f$ can be rewritten as $1/f = (p+q)/pq$. Knowing $p+q=L$, change the numerator. Knowing that $p=q-L$ (since $p+q=L$), change the denominator. Solve for $q$ in terms of $p$ and $L$. Do that stuff I just mentioned above. A "4" should be involved.<br />
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3. The previous problem should have given you two possible solutions for $q$ that give the minimum $L$. Those solutions should be solutions to the quadratic equation, and look something like: $q\in\{q_1,q_2\}=\{x+y, x-y\}$ where $x$ and $y$ are quantities involving $L$ and $f$ (and likely a "4"). The "$y$" quantity is likely to look like $\sqrt{\ldots}$. Those are your two possible $q$ positions, which imply two possible corresponding $p$ positions $p\in\{p_1,p_2\}=\{L-q_1,L-q_2\}$.<br />
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You know the magnification factor in each case in terms of the q's and p's. In the first case, $M_1=-q_1/p_1=a/h$, and in the second $M_2=-q_2/p_2=b/h$. Solve both of those for $h$, and multiply them to get an equation for $h^2$. In the resulting equation, you will have to show that $p_1p_2/q_1q_2=1$. Use your equations for $q$ above and multiply your two solutions together, cancellations will occur and it will end up simply. Do the same for the $p$'s, noting that $p_1=L-q_1$ and $p_2=L-q_2$. If you are careful in your algebra, the desired result will follow.<br />
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4. a. There is only one sort of spherical mirror that <i>always</i> gives virtual images. For any spherical mirror, the focal point is at a distance $R/2$. However: if the (virtual) image forms on the opposite side of the mirror as the image, what does that imply about the <i>sign</i> of the focal length? Remember the sign conventions!<br />
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b. You know $M=-q/p$, or $q=-Mp$. You also know $1/p+1/q=1/f$. Taking careful notice of the sign of $f$, you can plug in $q=-Mp$ and $|f|=R/2$ and solve this thing for $p$ in terms of $M$ and $R$, both of which are known. Your value for $p$ should come out to be positive.</div>pleclairhttp://www.blogger.com/profile/07263994015241270268noreply@blogger.com2tag:blogger.com,1999:blog-5910769335173612144.post-44872838626173782612012-07-31T11:30:00.001-05:002012-07-31T11:30:41.502-05:00Notes for today<div dir="ltr" style="text-align: left;" trbidi="on">
I will follow <a href="http://faculty.mint.ua.edu/~pleclair/PH253/Notes/Spring_2010/wave-particle.pdf">this</a>, more or less.</div>pleclairhttp://www.blogger.com/profile/07263994015241270268noreply@blogger.com3tag:blogger.com,1999:blog-5910769335173612144.post-80635998654003640162012-07-30T22:13:00.003-05:002012-07-30T22:13:26.110-05:00Stuck in Memphis<div dir="ltr" style="text-align: left;" trbidi="on">
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So. I write this from a dingy hotel in Memphis, thanks to Delta airlines. My flight back was cancelled, too late to get a different flight out or start driving. This has a number of important consequences:<br />
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1) The extra credit assignment is not ready, and will not be until Tuesday afternoon some time. I will let you know.<br />
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2) <b>I am not going to make it right at 11am tomorrow</b>. My best-case scenario will get me to UA in time to start class at 12:00, so let's start then. Better an hour than nothing. Should my travel plans go further awry, I will have someone leave a note on the door so you don't wait around for me ...<br />
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3) I am unlikely to be very responsive to email before noon tomorrow - I have been on minimal internet connections all weekend, and tonight is not great either. I will reply to what emails I can.<br />
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4) If you are wondering about what is going to be on the final, keep wondering until Tuesday night.<br />
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5) If you are wondering about an alternate final exam time and have a good pre-arranged reason to do so, find me after class tomorrow.<br />
</div>pleclairhttp://www.blogger.com/profile/07263994015241270268noreply@blogger.com0tag:blogger.com,1999:blog-5910769335173612144.post-58256131279365554062012-07-30T09:36:00.003-05:002012-07-30T09:36:43.446-05:00Today's lab<div dir="ltr" style="text-align: left;" trbidi="on">
<a href="http://faculty.mint.ua.edu/~pleclair/ph102/Labs/LED_Planck/LED_Planck.pdf">you will measure Planck's constant</a>.</div>pleclairhttp://www.blogger.com/profile/07263994015241270268noreply@blogger.com3tag:blogger.com,1999:blog-5910769335173612144.post-26067440624260094042012-07-26T23:54:00.002-05:002012-07-26T23:54:29.289-05:00HW5<div dir="ltr" style="text-align: left;" trbidi="on">
<a href="http://faculty.mint.ua.edu/~pleclair/ph102/Homework/HW5_31Jul12_optics.pdf">is out</a>. Due Tuesday. We'll go over some of these in class on Tuesday if you get stuck.</div>pleclairhttp://www.blogger.com/profile/07263994015241270268noreply@blogger.com1tag:blogger.com,1999:blog-5910769335173612144.post-89919093522603558342012-07-26T12:52:00.003-05:002012-07-26T12:52:54.704-05:00Left my phone at home ...<div dir="ltr" style="text-align: left;" trbidi="on">
so if you need to reach me today (Thurs), email is basically it.</div>pleclairhttp://www.blogger.com/profile/07263994015241270268noreply@blogger.com0tag:blogger.com,1999:blog-5910769335173612144.post-28851318944261323832012-07-26T03:21:00.002-05:002012-07-26T03:21:38.984-05:00Updated grades<div dir="ltr" style="text-align: left;" trbidi="on">
Your exam 2 and homework 3 grades are now available on eLearning. Enjoy, and let me know if anything seems mistaken.<br />
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The average for exam 2 was 78.6 with a standard deviation of 14.3. The A/B/C/D/F breakdown was 15/11/10/10/6. I am happy with it overall. More intelligent comments after I've slept a bit.</div>pleclairhttp://www.blogger.com/profile/07263994015241270268noreply@blogger.com2tag:blogger.com,1999:blog-5910769335173612144.post-61866002718020985192012-07-26T01:57:00.000-05:002012-07-26T01:57:04.711-05:00Thursday's lab<div dir="ltr" style="text-align: left;" trbidi="on">
you will do 2 short experiments. First, <a href="http://faculty.mint.ua.edu/~pleclair/ph102/Labs/reflection_refraction/reflection_refraction.pdf">demonstrating the laws of reflection and refraction</a>, and second, <a href="http://faculty.mint.ua.edu/~pleclair/ph102/Labs/lenses/lenses.pdf">playing with lenses</a>. Print both procedures, you should finish relatively quickly.<br />
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There will be a quiz in the lab, on refraction.</div>pleclairhttp://www.blogger.com/profile/07263994015241270268noreply@blogger.com0tag:blogger.com,1999:blog-5910769335173612144.post-73196460846365249682012-07-25T00:51:00.001-05:002012-07-25T00:51:30.765-05:00HW4 solutions<div dir="ltr" style="text-align: left;" trbidi="on">
<a href="http://faculty.mint.ua.edu/~pleclair/ph102/Homework/HW4_24July12_magn-ind_SOLN.pdf">are here</a>. I don't have a solution typed up for the last problem right now (and it is exceedingly unlikely to happen before the exam), but nothing like it will show up on the exam. Also, I did it all in class today ... if you missed that, you can ask one of your classmates.</div>pleclairhttp://www.blogger.com/profile/07263994015241270268noreply@blogger.com2tag:blogger.com,1999:blog-5910769335173612144.post-87620383672908697552012-07-24T21:51:00.003-05:002012-07-24T21:51:39.688-05:00Summer 2010 Exam 2<div dir="ltr" style="text-align: left;" trbidi="on">
<a href="http://faculty.mint.ua.edu/~pleclair/ph102/Exams/Sum_2010/ph102_Sum10_Exam_2_SOLN.PDF">This handwritten solution is as good as it is going to get</a>. The last problem I didn't solve since it isn't relevant to the upcoming exam, and time is running short ...</div>pleclairhttp://www.blogger.com/profile/07263994015241270268noreply@blogger.com0