course Phy 202
......!!!!!!!!...................................
10:13:56 **** Query experiment 29 **** how far from the V of the thread did you measure the separation of bright spots, how far apart were the spots, and what the you estimate was the average separation of threads at the point where you shined the light through
......!!!!!!!!...................................
RESPONSE --> The separation of the bright spots from the V of the thread was quite close to the discussed example in the class notes set, approximately 600cm. The spots were found to be approximately 1cm apart. I'd estimate the average spearation of the threads at the point where I shone the light through to be less than 0.01cm.
.................................................
......!!!!!!!!...................................
10:16:38 ** STUDENT RESPONSE: The distance from my V of thread to the wall that was my 'screen' was 6 meters, I tried to move it farther back but If I go much farther the dots are harder to see. I played with the cat too much with my laser pointer and have nearly used up the battery. ** It's hard not to do that with a cat. One battery per cat is about right. Beyond that they tend to get psychotic, though how you separate pyschosis from normal behavior in a cat I'm not sure. ** I measured the spots on the wall to be about 0.29 centimeters apart. The threads on my plastic were about 13 threads per centimeter where I tried to keep the lazer pointer which would make them have 7.7 * 10^-4 meters in between them.
......!!!!!!!!...................................
RESPONSE --> Haha, I play with my three cats and my two dogs with the laser pointer! OK.
.................................................
......!!!!!!!!...................................
10:19:10 **** explain how you determined the approximate wavelength of light from your data
......!!!!!!!!...................................
RESPONSE --> The wavelength is equal to (the separation between the spots / distance between the V of the thread and the screen) * distance between threads at the point where you shone the light through.
.................................................
......!!!!!!!!...................................
10:24:47 STUDENT RESPONSE: Well the first thing I did was to measure the distance between the dots which came out to be about 0.29 cm. Then since I knew the distance to my screen was right at 600 cm I used the trigonomic function for tangent to solve for the angle theta which in this case was my 1st order angle for my dot. tan of theta = opposite / adjacent opposite = 0.29cm adjacent = 600 cm so angle theta = 0.02768 Then I used the formula... sin of theta = m * wavelength / d to solve for wavelength m = 1 d = 7.7 * 10^-4 For the wavelength I got 3.72 * 10^-7 m or 372 nm which can not be right because it is not even on the visible spectrum. From this experiment I would expect to get around 700 nm because that is about what red should be and the lazer pointer has red light. I honestly expected to get something that was not even close due to crudeness of the apparati that I have rigged up. None the less, providing that I did everything correctly calculation wise and didn't miss anything, this technique should work if I tweeked it a bit more and refined my tools some more. INSTRUCTOR COMMENT: ** It's hard to be much more accurate than that with the crude apparatus, but it shows you what's going on here. **
......!!!!!!!!...................................
RESPONSE --> Use tangent to solve for angle theta, the first-order angle for the dot. Tan of theta = opposite / adjacent, so theta = 0.02768 (using this data). Then use the equation sin(theta) = (m * wavelength) / d, and get wavelength = 3.72 * 10^-4m = 372nm. OK.
.................................................
......!!!!!!!!...................................
10:26:16 gen phy problem 24.44 foil separates one end of two stacked glass plates; 28 lines observed for normal 650 nm light gen phy what is the thickness of the foil?
......!!!!!!!!...................................
RESPONSE --> I'm not sure what to do here. I'm not sure I really understand the question, and that may be part of the problem.
.................................................
......!!!!!!!!...................................
10:30:44 STUDENT SOLUTION: To solve this problem, I refer to fig. 24-31 in the text book as the problem stated. To determine the thickness of the foil, I considered the foil to be an air gap. I am not sure that this is correct. Therefore, I used the equation 2t=m'lambda, m=(0,1,2,...). THis is where the dark bands occur . lambda is given in the problem as 670nm and m=27, because between 28 dark lines, there are 27 intervals. Solve for t(thickness): t=1/2(2)(670nm) =9.05 *10^3nm=9.05 um INSTRUCTOR RESPONSE WITH DIRECT-REASONING SOLUTION:** Your solution looks good. Direct reasoning: ** each half-wavelength of separation causes a dark band so there are 27 such intervals, therefore 27 half-wavelengths and the thickness is 27 * 1/2 * 670 nm = 9000 nm (approx) **
......!!!!!!!!...................................
RESPONSE --> That makes sense, I wasn't thinking! There are 27 dark bands caused by a half-wavelength of separation. The thickness is then 27 * 1/2 * 670nm = 9000nm. OK.
.................................................
......!!!!!!!!...................................
10:32:19 **** gen phy how many wavelengths comprise the thickness of the foil?
......!!!!!!!!...................................
RESPONSE --> I think maybe you use the equation sin(theta) = (m * wavelength) / d here, but I'm not sure about that.
.................................................
......!!!!!!!!...................................
10:34:49 GOOD STUDENT SOLUTION: To calculate the number of wavelengths that comprise the thickness of the foil, I use the same equation as above 2t=m'lambda and solve for m. 2(9.05 um)=m(6.70 *10^-7m) Convert all units to meters. m=27 wavelengths.
......!!!!!!!!...................................
RESPONSE --> Just kidding about which equation to use, I should have tried 2t = m*lambda and solved for m: 2(9.05um) = m(6.70*10^-7m), and m = 27 wavelengths. OK.
.................................................
wP{D]ěw՟ assignment #018 ȏ|cާ Physics II 12-30-2005
......!!!!!!!!...................................
10:40:12 **** query gen phy problem 24.36 width of 1st-order spectrum of white light (400 nm-750nm) at 2.3 m from a 7500 line/cm grating **** gen phy what is the width of the spectrum?
......!!!!!!!!...................................
RESPONSE --> I'm really not sure ho to determine this width of the spectrum.
.................................................
......!!!!!!!!...................................
10:56:44 GOOD STUDENT SOLUTION We are given that the spectrum is from 400-750 nm. We are also given that the screen is 2.3 meters away and that the grating is 7500 lines/cm. To find this I will find where 400 nm wavelength falls on the screen and also where 750 nm wavelength falls onto the screen. Everything in between them will be the spectrum. I will use the formula... sin of theta = m * wavelength / d since these are first order angles m will be 1. since the grating is 7500 lines/cm, d will be 1/7500 cm or 1/750000 m. Sin of theta(400nm) = 1 * (4.0 * 10^-7)/1/750000 sin of theta (400nm) = 0.300 theta (400nm) = 17.46 degrees This is the angle that the 1st order 400nm ray will make. sin of theta (750nm) = 0.563 theta (750nm) = 34.24 degrees This is the angle that the 1st order 750 nm ray will make. We were given that the screen is 2.3 meters away. If we draw an imaginary ray from the grating to to the screen and this ray begins at the focal point for the rays of the spectrum and is perpendicular to the screen (I will call this point A), this ray will make two triangles, one with the screen and the 400nm angle ray and one with the screen and the 750 nm angle ray. Using the trigonomic function; tangent, we can solve for the sides of the triangles which the screen makes up. Tan of theta = opposite / adjacent tan of 34.24 degrees = opposite / 2.3 meters 0.6806 = opposite / 2.3 meters opposite = 1.57 meters tan of 17.46 degrees = opposite / 2.3 meters opposite = 0.72 meters So from point A to where the angle(400nm) hits the screen is 0.72 meters. And from point A to where the angle(750nm) hits the screen is 1.57 meters. If you subtract the one segment from the other one you will get the length of the spectrum on the screen. 1.57 m - 0.72 m = 0.85 meters is the width of the spectrum on the screen. CORRECTION ON LAST STEP: spectrum width = 2.3m * tan (31.33)) - 2.3m * tan (17.45) = 0.68m
......!!!!!!!!...................................
RESPONSE --> I need to find where the 400nm wavelegnth and where the 759nm wavelength fall on the screen; everything in between will be the spectrum. Sin(theta) = (m*wavelength) / d; since these are first-order angles, m = 1. Sin(theta) 400nm = 0.300, so theta = 17.46deg. Sin(theta) 750nm = 0.563, so theta = 34.24deg. These are the two angles that first-order 400nm and 750nm rays will make. Tan(theta) = opposite / adjacent, and adjacent = 2.3m. Solving for the opposite distance, you get distances 0.72m and 1.57m. The spectrum width is then (2.3 * tan(31.33)) - (2.3 * tan(17.45)) = 0.68m. That makes sense. This may be a bad question, but how do you know that m=1, that these are first-order angles?
.................................................
......!!!!!!!!...................................
10:56:51 **** query univ phy 36.59 phasor for 8 slits
......!!!!!!!!...................................
RESPONSE --> Not in this class.
.................................................
......!!!!!!!!...................................
10:56:55 ** If you look at the phasor diagram for phi = 3 pi / 4 you will see that starting at any vector the fourth following vector is in the opposite direction. So every slit will interfere destructively with the fourth following slit. This is because 4 * 3 pi / 4 is an odd multiple of pi. The same spacing will give the same result for 5 pi / 4 and for 7 pi / 4; note how starting from any vector it takes 4 vectors to get to the antiparallel direction. For 6 pi / 4, where the phasor diagram is a square, every slit will interfere destructively with the second following slit. For phi = pi/4 you get an octagon. For phi = 3 pi / 4 the first vector will be at 135 deg, the second at 270 deg (straight down), the third at 415 deg (same as 45 deg, up and to the right). These vectors will not close to form a triangle. The fourth vector will be at 45 deg + 135 deg = 180 deg; i.e., horizontal to the left. The next two will be at 315 deg (down and toward the right) then 90 deg (straight up). The last two will be at 225 deg (down and to left) and 360 deg (horiz to the right). The resulting endpoint coordinates of the vectors, in order, will be -0.7071067811, .7071067811 -0.7071067811, -0.2928932188 0, 0.4142135623 -1, 0.4142135623 -0.2928932188, -0.2928932188 -0.2928932188, 0.7071067811 -1, 0 0, 0 For phi = 5 pi / 4 each vector will 'rotate' relative to the last at angle 5 pi / 4, or 225 deg. To check yourself the first few endpoints will be -0.7070747217, -0.7071290944; -0.7070747217, 0.2928709055; 0, -0.4142040038 and the final endpoint will again be (0, 0). For 6 pi / 4 you will get a square that repeats twice. For 7 pi / 4 you get an octagon. NEW PROBLEM: The longest wavelength is 700 nm and slit spacing is about 1250 nm. The path difference can't exceed the slit spacing, which is less than double the 700 nm spacine. So there are at most central max (path difference zero) and the first-order max (path difference one wavelength). Note that there will be a second-order max for wavelengths less than about 417 nm. **
......!!!!!!!!...................................
RESPONSE --> OK.
.................................................
~آZ̻ˬ։ͱ assignment #020 ȏ|cާ Physics II 12-30-2005
......!!!!!!!!...................................
13:19:42 Query introductory set #1, 1-8 Explain how we calculate the magnitude and direction of the electrostatic force on a given charge at a given point of the x-y plane point due to a given point charge at the origin.
......!!!!!!!!...................................
RESPONSE --> Magnitude is found by using the equation: (9*10^9nm^2/C^2 * charge one * charge two) / separation between the charges^2. This gives you the magnitude of the force, in Newtons. The direction can bd found by dividing the x and y components by their absolue values and taking the square root of thre bottom part (the absolute value portion).
.................................................
......!!!!!!!!...................................
13:24:38 ** The magnitude of the force on a test charge Q is F = k q1 * Q / r^2, where q1 is the charge at the origin. The force is one of repulsion if q1 and Q are of like sign, attraction if the signs are unlike. The force is therefore directly away from the origin (if q1 and Q are of like sign) or directly toward the origin (if q1 and Q are of unlike sign). To find the direction of this displacement vector we find arctan(y / x), adding 180 deg if x is negative. If the q1 and Q are like charges then this is the direction of the field. If q1 and Q are unlike then the direction of the field is opposite this direction, 180 deg more or less than this direction. **
......!!!!!!!!...................................
RESPONSE --> The direction is found by finding the arctan(y/x), and adding 180deg if the x component is negative. If q1 and Q are opposite signs, this is the direction opposite that of the field. OK.
.................................................
......!!!!!!!!...................................
13:28:19 Explain how we calculate the magnitude and direction of the electric field at a given point of the x-y plane point due to a given point charge at the origin.
......!!!!!!!!...................................
RESPONSE --> The magnitude is found similarily to the previous question, by (9*10^9nm^2/C^2 * charge one * charge two) / separation between the charges^2. The direction is found again as arctan(y/x), and adding 180deg if the x component is negative.
.................................................
......!!!!!!!!...................................
13:32:42 ** The magnitude of the force on a test charge Q is F = k q1 * Q / r^2, where q1 is the charge at the origin. The electric field is therefore F / Q = k q1 / r^2. The direction is the direction of the force experienced by a positive test charge. The electric field is therefore directly away from the origin (if q1 is positive) or directly toward the origin (if q1 is negative). The direction of the electric field is in the direction of the displacement vector from the origin to the point if q1 is positive, and opposite to this direction if q1 is negative. To find the direction of this displacement vector we find arctan(y / x), adding 180 deg if x is negative. If q1 is positive then this is the direction of the field. If q1 is negative then the direction of the field is opposite this direction, 180 deg more or less than this direction. **
......!!!!!!!!...................................
RESPONSE --> OK.
.................................................
yϰW∻zĵ}أ assignment #021 ȏ|cާ Physics II 12-30-2005
......!!!!!!!!...................................
17:37:45 Query introductory set #1, 9-16 Explain how to find the potential difference in volts between two given points on the x axis, due to a given charge at the origin.
......!!!!!!!!...................................
RESPONSE --> A volt is a measure of work per unit charge, and is measured in J / C. The potential difference is the difference between the final potential and the beginning potential. Work is the product of the charge and the potential difference. If you know the amount of work required to move the charge between the two points, then all you have to do is divide the work by the charge to get the potential difference.
.................................................
......!!!!!!!!...................................
17:43:24 ** Potential difference is the work per Coulomb of charge moved between the two points. To find this work you can multiply the average force on a Coulomb of charge by the displacement from the first point to the second. You can find an approximate average force by finding the force on a 1 Coulomb test charge at the two points and averaging the two forces. Multiplying this ave force by the displacement gives an approximate potential difference. Since the force is not a linear function of distance from the given charge, if the ratio of the two distances from the test charge is not small the approximation won't be particularly good. The approximation can be improved to any desired level of accuracy by partitioning the displacement between charges into smaller intervals of displacement and calculating the work done over each. The total work required is found by adding up the contributions from all the subintervals. University Physics students should understand how this process yields the exact work, which is the integral of the force function F(x) = k Q / x^2 between the two x values, yielding total work W = k * Q * 1 Coulomb ( 1 / x1 - 1 / x2) and potential difference V = k * Q ( 1 / x1 - 1 / x2). **
......!!!!!!!!...................................
RESPONSE --> Potential difference found by multiplying the average force on a Coulomb of charge by the displacement from the first point to the second. OK.
.................................................
......!!!!!!!!...................................
17:47:23 ** The work per Coulomb done between the two points is equal to the product of the electric field E and the displacement `dr. Thus for constant field E we have V = E * `dr. **
......!!!!!!!!...................................
RESPONSE --> Potential difference found by multiplying the average force on a Coulomb of charge by the displacement from the first point to the second. The work done per Coulomb between the two points = electric field E * displacement dr. So for a constant field E, V = E * `dr. OK.
.................................................
......!!!!!!!!...................................
17:50:35 Explain how to find the average electric field between two points given a specific charge and the work done on the charge by the electric field as the charge moves between the points.
......!!!!!!!!...................................
RESPONSE --> Average strength of electric field is measured as average force / unit charge. The average force, given the work done and the displacement, is work done / displacement. The magnitude of the force, then, is the absolute value of the force / charge, in N / C.
.................................................
......!!!!!!!!...................................
17:51:52 ** You get ave force from work and distance: Fave = `dW / `ds. You get ave electric field from work and charge: Eave = F / q. An alternative: Find potential difference `dV = `dW / q. Ave electric field is Eave = `dV / `ds **
......!!!!!!!!...................................
RESPONSE --> Fave = `dW / `ds, and Eave = F / q. I said this, but in words. OK.
.................................................
......!!!!!!!!...................................
17:53:16 In your own words explain the meaning of the electric field.
......!!!!!!!!...................................
RESPONSE --> An electric field is where there is at least one charge found, which creates opportunities for repulsion and attraction between chrages of alike and dissimilar charges.
.................................................
......!!!!!!!!...................................
17:55:40 STUDENT RESPONSE AND INSTRUCTOR COMMENT: electric field is the concentration of the electrical force ** That's a pretty good way to put it. Very specifically electric field measures the magnitude and direction of the electrical force, per unit of charge, that would be experienced by a charge placed at a given point. **
......!!!!!!!!...................................
RESPONSE --> An electric field is the concentration of the electrical field, that specifically measures the magnitude and direction of the electrical force, per unit of charge, that would be experienced by a charge placed at a given point. OK.
.................................................
......!!!!!!!!...................................
17:56:23 In your own words explain the meaning of voltage.
......!!!!!!!!...................................
RESPONSE --> Voltage is a measure of electrical activity and strength.
.................................................
......!!!!!!!!...................................
17:57:22 ** Voltage is the work done per unit of charge in moving charge from one point to another. **
......!!!!!!!!...................................
RESPONSE --> Voltage is the work done per unit of charge in moving a charge from one point to another. OK.
.................................................
......!!!!!!!!...................................
17:59:51 Query experiment 16 current flow and energy You should have submitted your writeup of this experiment. However, answer the following questions. First question: Why is it that when the circuit formed has the least resistance the generator is hardest to crank?
......!!!!!!!!...................................
RESPONSE --> Greater resistance means that fewer electrons are able to flow and so it is easier to crank. Similarly, lower resistance means that more electrons are flowing and so it's harder to crank.
.................................................
......!!!!!!!!...................................
18:03:55 ** Less resistance implies more current, meaning more charge per unit of time and therefore more work per unit of time. To perform this work at a constant cranking rate (hence a constant voltage, and also a constant distance per unit of time) requires more force (since the distance in a unit of time is constant, in order to do work at a greater rate the force must be greater). **
......!!!!!!!!...................................
RESPONSE --> Less resistance implies mor current, or more work per unit of time. To perform this work at a constant rate means that more force is required. OK.
.................................................
......!!!!!!!!...................................
18:07:27 Second question: Why do you think it is that the generator is harder to crank with two bulbs in parallel then with the same to bulbs and series?
......!!!!!!!!...................................
RESPONSE --> The bulbs in the series only had to focus on going through one bulb; there weren't options on where to go. The parallel bulbs are then harder to crank because of the ""options"" available to them: they must provide a constant force to 2 different bulbs, not just to 1, and so since there's more wire and places to go, there's also less resistance encountered.
.................................................
......!!!!!!!!...................................
18:10:19 ** In a parallel circuit the full voltage of the generator is applied to both branches of the circuit, since both are directly connected to the generator. In a series circuit, the circuit doesn't split and the voltage is divided (not equally unless the bulbs have equal resistance) between the two bulbs. So in the parallel circuit both bulbs experience greater voltage, and hence greater current, than in the series circuit. **
......!!!!!!!!...................................
RESPONSE --> In parallel circuit, both bulbs experience greater voltage and hence greater current than in the series circuit. This makes if harder to crank. OK.
.................................................
"