course PHY 202 ~PDߋyassignment #015
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15:45:55 **** Query Snell's Law Experiment For the first ray describe your measurements, how you use these measurements to obtain your angles, and the angle of the incoming ray and the angle of the transmitted ray
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RESPONSE --> ???
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19:27:52 STUDENT RESPONSE: The first ray that we were asked to do was 45 degrees in relation to the front of the tape case. This was a fairly simple angle to do, I simply took a square of paper and folded in half from corner to opposite corner. This gave me a piece of paper that was now a triangle that had a 90 degree angle and two 45 degree angles. So having the tape case squarely on the paper, I placed one of the base sides of my triangle against the front side of the tape case and used the hypotenuse side to draw a 45 degree angle on the page. Every angle was a multiple of 15 degrees so I then took the triangle and made two folds into one of the 45 degree corners so that I had a 15 degree angled corner in the paper now. I used this piece of paper and the 45 degree line on the page to make all the subsequent angles of (15, 30, 60, and 75) degrees. Whenever I made an incoming ray they all entered the tape case at the same designated point. So then I just made a mark on the page where the transmitted ray came back out of the tape case. From these marks I was able to draw the paths of the transmitted rays. I was sure to label them so that I did not confuse them. To measure how much each ray was defracted I took a ruler and extended each ray on down the page and used these lines as a standard to measure the transmitted rays.
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RESPONSE -->
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19:28:40 **** give the angle of the incoming ray and the angle of the transmitted ray for each ray you measured, all angles measured with respect to a line which is normal (i.e., perpendicular) to the side of the case
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RESPONSE --> 15-45 30-50 45-60 60-70 75-80
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19:28:45 STUDENT RESPONSE: 15 degrees ----- about 45 degrees 30 degrees ----- about 50 degrees 45 degrees ----- about 60 degrees 60 degrees ----- about 68 degrees 75 degrees ----- about 80 degrees These angles were not exact in the way that I measured them, but the data was still plenty able to illustrate the relationship between initial angle and transmitted angle. The less that the initial angle is, the greater the difference will be between it and the transmitted angle. The limit here will be an angle of 0 degrees because it will be completely parallel to the refracting surface (front edge of tape case) and therefore never be refracted at all. At 90 degrees a ray appeared to have no change in transmittance either which is what the graph suggested would happen. Of course if you go past 90 degrees you would be moving back down the graph because a 91 degree initial ray would be the same as an 89 degree ray, a 100 degree ray would be the same as an 80 degree ray and so on.
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RESPONSE --> right
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22:19:00 query gen phy problem 23.11 radius of curvature of 4.5 x lens held 2.2 cm from tooth
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RESPONSE --> 1/2.2cm + 1/image d = -1.17cm 1.7cm*image distance + 2.2cm*1.7cm = -2.2cm * image distance
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22:19:04 ** if the lens was convex then its focal length would be negative, equal to half the radius. Thus we would have 1 / 2.2 cm + 1 / image distance = -1 / 1.7 cm. Multiplying by the common denominator 1.7 cm * image distance * 1.7 cm we would get 1.7 cm * image distance + 2.2 cm * 1.7 cm = - 2.2 cm * image distance. Thus -3.9 cm * image distance = - 2.2 cm * 1.7 cm. Solving would give us an image distance of about 1 cm. Since magnification is equal to image distance / object distance the magnitude of the magnification would be less than .5 and we would not have a 4.5 x magnification. We have the two equations 1 / image dist + 1 / obj dist = 1 / focal length and | image dist / obj dist | = magnification = 4.5, so the image distance would have to be either 4.5 * object distance = 4.5 * 2.2 cm = 9.9 cm or -9.9 cm. If image dist is 9.9 cm then we have 1 / 9.9 cm + 1 / 2.2 cm = 1/f. Mult by common denominator to get 2.2 cm * f + 9.9 cm * f = 2.2 cm * 9.9 cm so 12.1 cm * f = 21.8 cm^2 (approx) and f = 1.8 cm. This solution would give us a radius of curvature of 2 * 1.8 cm = 3.6 cm, since the focal distance is half the radius of curvature. This positive focal distance implies a concave lens, and the image distance being greater than the object distance the tooth will be more than the focal distance from the lens. For this solution we can see from a ray diagram that the image will be real and inverted. The positive image distance also implies the real image. The magnification is - image dist / obj dist = (-9.9 cm) / (2.2 cm) = - 4.5, with the negative implying the inverted image whereas we are looking for a +4.5 magnification. There is also a solution for the -9.9 m image distance. We eventually get 2.2 cm * f - 9.9 cm * f = 2.2 cm * (-9.9) cm so -7.7 cm * f = -21.8 cm^2 (approx) and f = 2.9 cm, approx. This solution would give us a radius of curvature of 2 * 2.0 cm = 5.8 cm, since the focal distance is half the radius of curvature. This positive focal distance also implies a concave lens, but this time the object is closer to the lens than the focal length. For this solution we can see from a ray diagram that the image will be virtual and upright. The negative image distance also implies the virtual image. The magnification is - image dist / obj dist = -(-9.9 cm) / (2.2 cm) = + 4.5 as required; note that the positive image distance implies an upright image. **
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RESPONSE --> ok
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22:19:09 **** query univ phy problem 33.38 (34.28 10th edition) 3 mm plate, n = 1.5, in 3 cm separation between 450 nm source and screen. How many wavelengths are there between the source and the screen?
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RESPONSE --> ok
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22:19:12 ** The separation consists of 1.55 cm = 1.55 * 10^7 nm of air, index of refraction very close to 1, and 2.5 mm = 2.5 * 10^-6 nm of glass, index of refraction 1.4. The wavelength in the glass is 540 nm / 1.4 = 385 nm, approx.. So there are 1.55 * 10^7 nm / (540 nm/wavelength) = 2.27 * 10^4 wavelengths in the air and 2.5 * 10^-6 nm / (385 nm/wavelength) = 6.5 * 10^3 wavelengths in the glass. **
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RESPONSE --> ok
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