course phys 202 Here is most of my back work for the second test. I was and am still unable to preform the labs for my lack of a lazer pointer. I will look at walmart tonight and if they have one I will do my lab this weekend.
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19:31:06 **** Query experiment 26 **** How nearly did your four rays come to converging? Did each ray reflect at the same angle from the normal as the angle of the incoming ray?
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RESPONSE --> I did not get to complete this experiment because my lack of a lazer pointer, and my in ablility to find one anywhere in my county.
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19:31:41 10-29-2005 19:31:41 GOOD STUDENT RESPONSE: I had a trouble on this experiment, although the procedure seems quite simple. I used a diet coke can as my 'mirror'. I used a ruler to measure the radius of the half-circle. Then I used a compass to draw the circle of the can on my paper. The radius fo the can was 2.8 cm.According to the video clips and the class notes, if I aim the laser beam to the left or to the right of the straight line drawn from the center of the can keeping the beam perpendicular to this center line, all reflected angles would cross that line at approximately R/2 or 1.4 cm. Before taking data for this experiment, I tested this idea. I was able to see that all the reflected angles appeared to cross the central line at the focal point of 1.4cm. I used trig to calculate the angle values. I measured the approriate sides of the triangle (all distances in cm). By dividing I think that it is adj/opp and taking the inverse tangent of this value, I calculated the angles. When I aimed the beam 1/8 of the container's radius((.35) to the right, I determined that the angle was approximately 28.5 degrees. When I aimed the third beam parallel to the central beam and approximately 2.8 of the container's radius to the left of the central beam, the angle was 33.7 degrees. When I aimed the fourth beam parallel to the central beam and approximately 3/8 of the container's radius to the right of the central beam the angle was 38.9 degrees. I am not sure how all of these angles are related. I know that there is probably great experimental error because I had a hard time estimating the reflected angle.
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NOTES -------> seem like good notes
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19:31:57 **** how did the radius of the 'mirror' cut from the can and the distance of the focal point from thie mirror compare?
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RESPONSE --> I do not know, explained earlier
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19:32:03 10-29-2005 19:32:03 STUDENT RESPONSE: Once again, I am not totally confident with this experiment. But, the radius of the can was 2.8 cm and the focal point of the mirror was R/2 or 1.4cm. INSTRUCTOR COMMENT: Sounds like you did this part right and that you got a result that agrees with the theory
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NOTES -------> ok
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19:32:13 **** For the circular lens were the rays entering the lens the diverted toward the normal?
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RESPONSE --> ?
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19:32:16 STUDENT RESPONSE: In this experiment I used a 2-liter Mountain Dew softdrink bottle. I used my compass and placed the pointer in the center of the bottom of the container, which was already marked. This enabled me to draw my bottle on my paper. The radius was 3.7cm. I was able to observe that as I would move the laser in one direction the laser beam would move in the opposite direction. I used the cassette with the ruler taped to it as my screen. I believe that because the bottle was so large, I was unable to accurately measure the angles as I moved to the left and to the right of the center line. I had no other container at my apartment, so I was unable to finish this experiment. You should understand that as the 'screen' is moved closer the movement of the point on the screen stops, then as you get the screen closer it starts moving in the same direction as the laser, rather than in the opposite direction. This shows that at the 'stopped' point the beams passing through the bottle will converge. **
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19:43:54 query gen phy problem 23.32 incident at 45 deg to equilateral prism, n = 1.52; and what angle does light emerge?
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RESPONSE --> I solved this problem through snell's law. I started with n=1 for air and n=1.58 for the glass. this gives me a feta 2 as 26.6 degrees. I am not sure if I did this part right, but I again used snell's law and n = 1.58 for both n's and got the second interior angle as 26.6 degrees. Using snell's law a third time I get the exiting angle as 45 degrees.
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19:52:45 10-29-2005 19:52:45 STUDENT SOLUTION: To solve this problem, I used figure 23-51 in the text book to help me visualize the problem. The problem states that light is incident on an equilateral crown glass prism at a 45 degree angle at which light emerges from the oppposite face. Assume that n=1.52. First, I calculated the angle of refraction at the first surface. To do this , I used Snell's Law: n1sin'thea1=n2sin'thea2 I assumed that the incident ray is in air, so n1=1.00 and the problem stated that n2=1.52. Thus, 1.00sin45 degrees=1.52sin'theta2 'thea 2=27.7 degrees. Now I have determined the angle of incidence of the second surface('thea3). This was the toughest portion of the problem. To do this I had to use some simple rules from geometry. I noticed that the normal dashed lines onthe figure are perpendicular to the surface(right angle). Also, the sum of all three angles in an equilateral triangle is 180degrees and that all three angles in the equilateral triangle are the same. Using this information, I was able to calculate the angle of incidence at the second surface. I use the equation (90-'thea2)+(90-'thea3)+angle at top of triangle=180degrees. (90-27.7degrees)=62.3 degrees. Since this angle is around 60 degrees then the top angle would be approx. 60 degrees. ###this is the part of the problem I am a little hesitant about. Thus, 62.3 degrees+(90-'thea3)+60 degrees=180 degrees-'thea)=57.7degrees 'thea=32.3 degrees This is reasonable because all three angles add up to be 180 degrees.62.3+60.0+57.7=180degrees Now, I have determined that the angle of incidence at the second surface is 32.3 degrees, I can calculate the refraction at the second surface by using Snell's Law. Because the angles are parallel, nsin'thea3=n(air)sin''thea4 1.52sin32.3=1.00sin (thea4) 'thea 4=54.3 degrees INSTRUCTOR COMMENT: Looks great. Here's my explanation (I did everything in my head so your results should be more accurate than mine): Light incident at 45 deg from n=1 to n=1.52 will have refracted angle `thetaRef such that sin(`thetaRef) / sin(45 deg) = 1 / 1.52 so sin(`thetaRef) = .707 * 1 / 1.52 = .47 (approx), so that `thetaRef = sin^-1(.47) = 28 deg (approx). We then have to consider the refraction at the second face. This might be hard to understand from the accompanying explanation, but patient construction of the triangles should either verify or refute the following results: This light will then be incident on the opposing face of the prism at approx 32 deg (approx 28 deg from normal at the first face, the normals make an angle of 120 deg, so the triangle defined by the normal lines and the refracted ray has angles of 28 deg and 120 deg, so the remaining angle is 32 deg). Using Snell's Law again shows that this ray will refract at about 53 deg from the second face. Constructing appropriate triangles we see that the angle between the direction of the first ray and the normal to the second face is 15 deg, and that the angle between the final ray and the first ray is therefore part of a triangle with angles 15 deg, 127 deg (the complement of the 53 deg angle) so the remaining angle of 28 deg is the angle between the incident and refracted ray. **
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NOTES -------> I understand the studetn answer, but in your explination I do not understand where the 120 degree angle came from. I can solve this problem the way the student did. I did not see the angles of the triangle when I solved this problem.
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19:52:49 **** query univ phy problem 34.86 (35.52 10th edition) f when s'=infinity, f' when s = infinity; spherical surface. How did you prove that the ratio of indices of refraction na / nb is equal to the ratio f / f' of focal lengths?
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19:52:51 ** The symbols s and s' are used in the diagrams in the chapter, including the one to which problem 62 refers. s is the object distance (I used o in my notes) and s' the image distance (i in my notes). My notation is more common that used in the text, but both are common notations. Using i and o instead of s' and s we translate the problem to say that f is the object distance that makes i infinite and f ' is the image distance that makes o infinite. For a spherical reflector we know that na / s + nb / s' = (nb - na ) / R (eqn 35-11 in your text, obtained by geometrical methods similar to those used for the cylindrical lens in Class Notes). If s is infinite then na / s is zero and image distance is s ' = f ' so nb / i = nb / f ' = (nb - na) / R. Similarly if s' is infinite then the object distance is s = f so na / s = na / f = (nb - na) / R. It follows that nb / f ' = na / f, which is easily rearranged to get na / nb = f / f'. THIS STUDENT SOLUTION WORKS TOO: All I did was solve the formula: na/s+nb/sprime=(nb-na)/R once for s and another time for sprime I took the limits of these two expressions as s and s' approached infinity. I ended up with f=-na*r/(na-nb) and fprime=-nb*r/(na-nb) when you take the ratio f/fprime and do a little algebra, you end up with f/fprime=na/nb **
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19:52:53 **** univ phy How did you prove that f / s + f' / s' = 1?
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RESPONSE -->
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19:52:55 ** We can do an algebraic solution: From nb / f ' = (nb - na) / R, obtained in a previous note, we get f ' = nb * R / (nb - na). From na / f = (nb - na) / R we get f = na * R / (nb - na). Rearranging na/s+nb/s'=(nb-na)/R we can get R * na / ( s ( na - nb) ) + R * nb / (s ' ( na - nb) ) = 1. Combining this with the other two relationships we get f / s + f ' / s / = 1. An algebraic solution is nice but a geometric solution is more informative: To get the relationship between object distance s and image distance s' you construct a ray diagram. Place an object of height h at to the left of the spherical surface at distance s > f from the surface and sketch two principal rays. The first comes in parallel to the axis, strikes the surface at a point we'll call A and refracts through f ' on the right side of the surface. The other passes through position f on the object side of the surface, encounters the surface at a point we'll call B and is then refracted to form a ray parallel to the axis. The two rays meet at a point we'll call C, forming the tip of an image of height h'. From the tip of the object to point A to point B we construct a right triangle with legs s and h + h'. This triangle is similar to the triangle from the f point to point B and back to the central axis, having legs f and h'. Thus (h + h') / s = h / f. This can be rearranged to the form f / s = h / (h + h'). From point A to C we have the hypotenuse of a right triangle with legs s' and h + h'. This triangle is similar to the one from B down to the axis then to the f' position on the axis, with legs h and f'. Thus (h + h') / s' = h / f'. This can be rearranged to the form f' / s' = h' / (h + h'). If we now add our expressions for f/s and f'/s' we get f / s + f ' / s ' = h / (h + h') + h' / (h + h') = (h + h') / (h + h') = 1. This is the result we were looking for. **
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19:52:57
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Tٗݙε`Dx assignment #017 yO̳aÉGɜ Physics II 10-29-2005
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19:53:28 **** query experiment 27 **** explain how you determined the focal length of the convex lens, and give its focal length
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RESPONSE --> I did not get to do this because of my lack of a lazer pointer.
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19:54:22 STUDENT ANSWER WITH EMBEDDED INSRUCTOR RESPONSES: if you sketch the ray diagram carefully you will find that this is not the case. Turning the lens around has no significant effect on the behavior of the lens; the only effect is that the effective center of a half-concave or half-convex lens will change slightly The first focal length that I calculated was for the convex lens.( bulging out from the case). There are several ways to determine the focal length , I first placed the ruler case behind the lens. (I carefully filled the lens with water). Aiming from a distance, I tried to keep the laser beam steady. I had my roomate move the ruler case back and forth until the laser point remained staionary on the case. Then I measured this length from the case to the lens. The focal length for the small convex lens was 13 cm and for the large convex lens 17.0cm. ** these are within 1 cm of the distances usually found ** To determine the focal length for the two concave ( curving in towards the cassette case), I used a different method. I filled the small lens with water and placed the ruler cassette approx. 10 cm from the lens as instructed. I varied the distance of the ruler while aiming the laser and obseved that there was no distance at which I could keep the laser point still. I also observed that the further the ruler is placed behind the lens, the larger and more blurry the laser point appears. However, I made two marks each about 1/2 cm apart from each other on the small lens. I marked the rays when the laser pointer was aimed at each dot. I noticed that the rays converged in front of the lens at approx. 4.5 cm from the front of the case. I had a little trouble on the large concave lens. The calking on the lens was leaky and it was hard to keep water in the lens. I'm sure this affected the accuracy of my results. However, I observed once again that the rays from the two dots( I made the dots the same as the first lens 1/2 apart from each other) converged in front of the lens at approximately 6.9cm. ** good description and the results are reasonable **
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RESPONSE --> good notes
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19:54:39 21:14:58
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RESPONSE --> ??
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19:54:55 **** describe the ray diagram for the convex lens
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RESPONSE --> wish I could
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19:55:01 10-29-2005 19:55:01 ** The ray diagram of the convex lense is that the rays appear to enter the lense from the bottom of the screen. All of the rays are perpendicular to the lense and parallel to each other. The focal point is horizontally in the center of the lense and vertically about 4-5 inches above the lense. All of the rays pass through this point. So rays that are toward the right side of the lense break toward the left in varying amounts so that they pass through the focal point. The rays that are toward the left side of the lense break toward the right in varying amounts so that they pass through the focal point. The ray that is in the center of the lense does not change, it just goes straight and still goes through the focal point. **
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NOTES -------> good notes
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19:55:08 21:16:46
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RESPONSE --> ok
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19:56:30 **** explain how you determine the focal length of the concave lenses, and give their focal lengths
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RESPONSE --> I could use the law that states that 1/object dist +1/image dist = 1/f
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19:56:57 ** GOOD STUDENT EXPLANATION: To determine the focal length for the two concave ( curving in towards the cassette case), I used a different method. I filled the small lens with water and placed the ruler cassette approx. 10 cm from the lens as instructed. I varied the distance of the ruler while aiming the laser and obseved that there was no distance at which I could keep the laser point still. I also observed that the further the ruler is placed behind the lens, the larger and more blurry the laser point appears. However, I made two marks each about 1/2 cm apart from each other on the small lens. I marked the rays when the laser pointer was aimed at each dot. I noticed that the rays converged in front of the lens at approx. 4.5 cm from the front of the case. I had a little trouble on the large concave lens. The calking on the lens was leaky and it was hard to keep water in the lens. I'm sure this affected the accuracy of my results. However, I observed once again that the rays from the two dots( I made the dots the same as the first lens 1/2 apart from each other) converged in front of the lens at approximately 6.9cm. **
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RESPONSE --> looks good to me
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20:16:36 **** query gen phy problem 24.7 460 nm light gives 2d-order max on screen; what wavelength would give a minimum?
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RESPONSE --> I have looked and thought, but have not found out how to solve this problem.
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20:22:41 10-29-2005 20:22:41 STUDENT SOLUTION FOLLOWED BY INSTRUCTOR COMMENT AND SOLUTION: The problem states that in a double-slit experiment, it is found that bule light of wavelength 460 nm gives a second-order maximun at a certain location on the screen. I have to determine what wavelength of visible light would have a minimum at the same location. To solve this problem I fist have to calculate the constructive interference of the second order for the blue light. I use the equation dsin'thea=m'lambda. m=2 (second order) dsin'thea=(2)(460nm) =920nm Now, I can determine the destructive interference of the other light, using the equation dsin'thea=(m+1/2)'lambda=(m+1/2)'lambda m+(0,1,2...) Now that I have calculated dsin'thea=920nm, I used this value and plugged it in for dsin'thea in the destructive interference equation.(I assumed that the two angles are equal) because the problem asks for the wavelength at the same location. Thus, 920nm=(m+1/2)'lambda. m=(0,1,2,...) I calculated the first few values for 'lambda. For m=0 920nm=(0+1/2)'lambda =1.84*10^nm For m=1 920nm=(1+1/2)'lambda =613nm For m=2 920nm=(2+1/2)'lambda=368 nm From these first few values, the only one of thes wavelengths that falls in the visible light range is 613nm. Therefore, this would be the wavelength of visible light that would give a minimum. INSTRUCTOR COMMENT AND SOLUTION: good. More direct reasoning, and the fact that things like sines are never needed: ** The key ideas are that the second-order max occurs when the path difference is 2 wavelengths, and a minimum occurs when path difference is a whole number of wavelengths plus a half-wavelength (i.e., for path difference equal to 1/2, 3/2, 5/2, 7/2, ... of a wavelength). We first conclude that the path difference here is 2 * 460 nm = 920 nm. A first-order minimum (m=0) would occur for a path difference of 1/2 wavelength. If we had a first-order minimum then 1/2 of the wavelength would be 920 nm and the wavelength would be 1860 nm. This isn't in the visible range. A minimum would also occur If 3/2 of the wavelength is 920 nm, in which case the wavelength would be 2/3 * 920 nm = 613 nm, approx.. This is in the visible range. A niminum could also occur if 5/2 of the wavelength is 920 nm, but this would give us a wavelength of about 370 nm, which is outside the visible range. The same would be the case for any other possible wavelength. **
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NOTES -------> I knew the equation, but I did not think of solving for one and setting the other equal. I understand now how to use the wave length and the knowledge that the the wavelengt times the order gives me a value to set equal to the distructive interfierence, and solve for each m+1/2 wavelength until I find one that falls within the visible spectrum
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20:22:47 **** query univ phy problem 35.52 (37.46 10th edition) normal 477.0 nm light reflects from glass plate (n=1.52) and interferes constructively; next such wavelength is 540.6 nm. How thick is the plate?
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RESPONSE --> ok
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20:22:48 ** The path difference for constructive interference is an integer multiple of the wavelength. The path difference here is twice the thickness. Wavelengths in glass are 477 nm / 1.52 and 540.6 nm / 1.52. So we know that double the thickness is an integer multiple of 477 nm / 1.52, and also an integer multiple of 540.6 nm / 1.52. We need to find the first integer multiple of 477 nm / 1.52 that is also an integer multiple of 540.6 nm / 1.52. We first find an integer multiply of 477 that is also an integer multiply of 540.6. Integer multiples of 540.6 are 540.6, 1081.2, 1621.8, etc. Dividing these numbers by 477 we obtain remainders 63.6, 127.2, etc. When the remainder is a multiple of 477 then we have an integer multiple of 477 which is also an integer multiple of 540.6. SInce 477 / 63.6 = 8.5, we see that 2 * 477 / 63.6 = 17. So 17 wavelengths of 477 cm light is the first multiple that is equivalent to an integer number of wavelengths of 540.6 cm light. 17 * 477 = 8109. Since 8109 / 540.6 = 15, we see that 17 wavelengths of 477 nm light span the same distance as 15 wavelengths of 540.6 nm light. It easily follows that that 17 wavelengths of (477 nm / 1.52) light span the same distance as 15 wavelengths of (540.6 nm / 1.52) light. This distance is 17 * 477 nm / 1.52 = 5335 nm. This is double the thickness of the pane. The thickness is therefore pane thickness = 5335 nm / 2 = 2667 nm. IF INTERFERENCE WAS DESTRUCTIVE: n * 477 nm / 1.52 = (n-1) * 540.6 nm / 1.52, which we solve: Multiplying by 1.52 / nm we get 477 n = 540.6 n - 540.6 n * (540.6 - 477 ) = 540.6 n * 63.6 = 540.6 n = 540.6 / 63.6 = 8.5. This is a integer plus a half integer of wavelengths, which would result in destructive interference for both waves. Multiplying 8.5 wavelengths by 477 nm / 1.52 we get round-trip distance 2667 nm, or thickness 1334 nm. **
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20:22:50 **** query univ phy prob 35.50 (10th edition 37.44): 700 nm red light thru 2 slits; monochromatic visible ligth unknown wavelength. Center of m = 3 fringe pure red. Possible wavelengths? Need to know slit spacing to answer?
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20:22:52 STUDENT SOLUTION: The pure red band at m = 3 suggests that there exists interference between the wavelength of the red light and that of the other light. Since only the red light is present at m = 3 it stands to reason that the wavelength of the other light is a half of a wavelength behind the red wavelength so that when the wavelength of the red light is at its peak, the wavelength of the other light is at its valley. In this way the amplitude of the red light is at its maximum and the amplitude of the other light is at it minimum - this explains why only the red light is exhibited in m = 3. INSTRUCTOR COMMENT At this point you've got it. At the position of the m=3 maximum for the red light the red light from the further slit travels 3 wavelengths further than the light from the closer. The light of the unknown color travels 3.5 wavelengths further. So the unknown wavelength is 3/3.5 times that of the red, or 600 nm. You don't need to know slit separation or distance (we're assuming that the distance is very large compared with the wavelength, a reasonable assumption for any distance we can actually see. **
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SH assignment #018 yO̳aÉGɜ Physics II 10-31-2005
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19:25:45 **** 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?
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RESPONSE --> The way I solved this problem, I used example 24-6 and the equation sinfeta=mlamda/d the two angles I got were .313 degrees and .573 degrees. I then solved for the right triangle using the tangent of the extremely small angle, and multiplied the 2.3 distance from the screen to get each distance x1, and x2. the difference between x1 and x2 is teh width of the spectrum.
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19:25:45 **** 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?
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RESPONSE -->
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19:30:14 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
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RESPONSE --> I used the right formula, but somehow I got the wrong answer, I think because I did not divide 1 by 7.5*10^5. Other than this I did the rest of the problem correctly
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19:30:21 **** query univ phy 36.59 phasor for 8 slits
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19:30:23 ** 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. **
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