course PHY 202 ??????f?x?yG??assignment #017?????€???????Physics II
......!!!!!!!!...................................
09:46:37 **** query experiment 27 **** explain how you determined the focal length of the convex lens, and give its focal length
......!!!!!!!!...................................
RESPONSE --> the focal length is the distance of the focal point from the center of the lens
.................................................
......!!!!!!!!...................................
09:46:50 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 **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
09:46:55 21:14:58
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
09:48:12 **** describe the ray diagram for the convex lens
......!!!!!!!!...................................
RESPONSE --> rays go in from the bottom and all are perpendicular to the lens and parallel to the other.
.................................................
......!!!!!!!!...................................
09:48:58 ** 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. **
......!!!!!!!!...................................
RESPONSE --> yes the focal point is horizontal in the center of the lens and vertically 4-5 inches above the lens.
.................................................
......!!!!!!!!...................................
09:49:02 21:16:46
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
11:23:36 **** explain how you determine the focal length of the concave lenses, and give their focal lengths
......!!!!!!!!...................................
RESPONSE --> focal length is the image distance for an object at infinity
.................................................
......!!!!!!!!...................................
11:23:39 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
11:42:55 **** query gen phy problem 24.7 460 nm light gives 2d-order max on screen; what wavelength would give a minimum?
......!!!!!!!!...................................
RESPONSE --> 2*460nm = 920nm sin(theta) = (m+0.5)lambda 920nm = (0+0.5)lambda = 1840 920nm = (1+0.5)lambda = 613.3 920nm = (2+0.5)lambda = 368 the 613.3 is in visible light
.................................................
......!!!!!!!!...................................
11:43:01 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. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
11:43:08 **** 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?
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
11:43:10 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
11:43:12 **** 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?
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
11:43:14 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. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................