The practice final exam has been posted on Blackboard in the Course Documents section. Answers will be given in the Dec. 7 lecture slides.
In this blog thread, we will discuss any questions you have about that exam.
This thread is also the place for you to ask questions about any of the course material, old homework problems, etc.
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36 comments:
Im completely stuck on #10. I know this has something to do with snell's law, but i dont know where to start. any tips?
Start with Sec. 23.4, and especially the assigned homework problem #9 from that chapter (SV8, Ch. 23, Problem #22). Parallel light rays can be regarded as coming from a source at infinity. You are given the radius of curvature of the interface and the image distance.
Will CHAPTER 30 be on the final exam? (It was not lectured on, but it was in the homework). Thanks!!
The material in chapter 30 that I covered in lecture and in homework problems, namely fission and fusion, is one of the topics that may be on the exam.
Professors,
Lecture from 11/02/2010 slide #14 question 3. How did you know that m was = 1.5?
For single slit diffraction, bright fringes occur at m=+/- 1.5, +/- 2.5, ... half integers except for 0.5. So the first bright fringe (outside of the central max at m=0) has m=1.5.
It's important to distinguish this from two-slit interference, which has bright fringes - constructive interference maxima - at integer m, 0, +/-1, +/-2, etc.
if we did not answer a clicker question in lecture, should we still consider it fair game for the final?
Our statement is that an excellent way to prepare for the final exam is to review all clicker and homework problems. It would be overinterpreting for students to think that the ONLY material you need to know was the subject of clicker or homework problems. When clicker questions were dropped from lectures, it was usually because of time constraints, not because either of your instructors thought they weren't interesting, educational, or important. So I would say that an unanswered clicker question is just about as important as one that was answered. If there are ones you are specifically interested in, please post them and we'll reply with the solutions.
Question emailed to me after the review session:
"I had a question regarding problem 6 on the practice final. When I use this equation B=((uo)(I))/(2pi(r)) I keep getting 25 amps. I saw that in the review you arrived at E, 7.96 Amps which you rounded up to 8, but I was wondering why you did not use pi at all in the equation. Thank You."
The answer is that the equation with pi in the denominator is for a long wire, observed at a distance r. The equation I used, (no pi in denominator) is for the field at the center of a circular loop. These are SV8 equations 19.11 and 19.15. It illustrates the point that knowing the equation alone is not as important as knowing where to use it.
for quest 17 on the practice final, i get 8 fringe shifts from the equation you gave us from the slides, and also in the book, its says theres one fringe shift for every lambda/4. Since theres 2lambda in this question, shouldnt it go from D -C -D -C -D- C-D- C- D. Please explain professor!!
From SV8 p 840: "As mirror M1 is moved ... by lambda/4, the destructive interference that initially produced a dark fringe has changed to constructive, and we now observe a bright fringe. The term fringe shift is used to describe the change in a fringe from dark to light or from light to dark." It does seem like an odd definition, that it takes two fringe shifts to get from dark to dark again, but that's the convention. In which case, moving the mirror by lambda constitutes 4 fringe shifts, D-C-D-C-D. OK?
can you post the link to the lab manuel please
does the anti/nutrino have E when released i know we don't need for formula but i want to know for knowledge
In beta decay, the reaction energy is divided between the electron and the antineutrino (or positron and neutrino) in a statistical manned. The total energy given to the two of them comes from the difference of masses between the original and daughter nuclide. The curve of electron energies I showed in class illustrates this. That's why beta decay questions ask for the energy released in the reaction, not the energy of the electron.
Please clarify what you mean by lab manual. The writeups for all the labs? Errors and Uncertainties?
Jack,
The url for the PHY 124 lab and all its documentation is
http://www.ic.sunysb.edu/class/phy122ps/labs/dokuwiki/doku.php?id=phy124off:phy124_main_page
BTW, you'll find this url in every post I made in this blog notifying all the PHY 122/4 students that a new Lab Pretest was posted and available. Just look.
Prof. Koch
Professors,
Just to clarify, the sheet of notes we are permitted to bring to the final can be typed as well? Are there any limitations as to what I can put on the typed version?
The sheet of notes you bring can be prepared in any way you wish, as long as it can be read without mechanical or optical assistance. Reading glasses excepted.
Will ch. 29/30 homework count toward our grade. (It isn't posted on BB).
for chapter 28 question 9 EL = −(Z − x)^2*(13.6 eV)/(n^2) how exactly do you determine 'x' I understand the way the book did it which is the electrons between the initial and final state but when I plug in x=2 because the K shell which is the final state has 2 electrons on it my answer is wrong and its suppose to be x=3. my element is Co Z=27
HW on Ch 29,30 does count - it's in the Bb gradebook now.
Ch 28 #9, the x-ray energy is the difference between initial and final states. The final state has the electron in the K shell, for which x = 1. For Cr, Z=24, the initial state energy is -13.6*(24-1)^2=-7194 eV.
The initial has the electron in the L shell, which means that there are the two K electrons inside of it, and the one companion electron in the L shell. So Zeff = 24-3=21. Also, the L shell means n=2, so the energy of an electron in the L shell is -13.6*(21^2)/4=-1499 eV.
The energy difference is 5694 eV, and so the wavelength is c/f = c*h/(5694 * e) = 0.218 nm.
This business of figuring out Zeff is a bit ad hoc, and so I can promise you, no questions on the final that depend on you to determine that
Quick question on mass defect problems: Do you need to take into account 2 electrons for an alpha particle? If the answer is YES, why so? isn't an alpha particle 42He++
The two electrons are needed to balance charge. For example, if a atom U235 decays to Th231 + alpha, that balances the number of protons and neutrons, but we also have to worry about electrons. U235 was neutral (at least that's the mass we look up in Appendix B), and so is Th231. So the rest of the equation must also be neutral, i.e., we have to account for the two extra electrons that were riding around with the U, which the Th doesn't want. If we write the reaction as U235 -> Th231 + He++ + 2 e-, that's neutral. Then you can see that there has to be a helium ion and two electrons on the right side. Let's take the two electrons and combine them with the He++, and call it a He atom - then we can look up its mass in the table, a neutral He atom.
great explanation! thank you
Question on Q15 on the practice final. it asks for the angle between two stars that it can barely resolve over the entire range of visible light. I understand we have to use the resolvable angle equation but ..why wouldn't you take the wavelengths for the two extremes (i.e 400 nm and 700nm), find their respective angles and subtract them to find the angle between the two stars? Why just the 700? Thanks.
Heisenberg: Do we use delta E * delta t > or equal to h/(2pi) OR OR do we use E * delta t > or equal to h/(4pi) ??? I've seen it both ways and I want to know which we should use for the final. (I'm also asking about the one using momentum, too).
Asking same question as kid before me:
Question on Q15 on the practice final. it asks for the angle between two stars that it can barely resolve over the entire range of visible light. I understand we have to use the resolvable angle equation but ..why wouldn't you take the wavelengths for the two extremes (i.e 400 nm and 700nm), find their respective angles and subtract them to find the angle between the two stars? Why just the 700? Thanks.
Telescope: I'm looking through my telescope (which has an objective lens or mirror of diameter D) at two stars, and I can just barely resolve them if they are an angular distance 1.22*lambda/D (radians). If they are violet, that's a certain angle calculated by putting in 400 nm for the wavelength. If they are red, I do the same calculation with 700 nm. That's a larger number, so I would say that the stars have to be that far apart for me to be able to resolve them across the entire visible spectrum.
Uncertainty principle: We did not have anything to do with uncertainty relation between E and t, (more's the pity), so don't worry about that. (FYI, it would be DeltaE * Deltat >= h/(4*pi).) For uncertainty of position and momentum, it's DeltaX * DeltaP >= h/(4*pi). You'll see different values in place of 4*pi in different books because it comes down to exactly how the uncertainty is defined.
I'm a bit confused by number 9 on the chapter 29/30 homework. I got part A easily, and got part B by playing with numbers but I dont understand the concept behind it.
sorry/nevermind- i see it on the other blog
Wow....the final was a killer.
Yeah, no kidding the final was a killer. It just brutally murdered my GPA. I had an 80 average in this class until today. I got a 26 on this test, so thanks for the early Christmas gift. >:(
The final was cruel. It is really frustrating when the practice test is nothing like the real one.
geez that final was something else eh?
Glad our physics teachers taught us everything that was on there, or taught us at all for that matter. =DDD
(lies)
::sarcasm:: I just want to thank the physics department for their wonderfully crafted instruction. Once again you have gone above and beyond my expectations of what science professors at stony brook can accomplish if they put their minds to it! ::sarcasm::
I miss Professor Hemmick.
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