course PHY 202 vz׆lassignment #026
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09:30:25 Principles of Physics and General College Physics Problem 23.08. How far from a concave mirror of radius 23.0 cm must an object be placed to form an image at infinity?
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RESPONSE --> GIven the formula: Focal length=radius of mirror/2, thus f=23/2=11.5 cm.
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09:30:35 07-22-2007 09:30:35 Recall that the focal distance of this mirror is the distance at which the reflections of rays parallel to the axis of the mirror will converge, and that the focal distance is half the radius of curvature. In this case the focal distance is therefore 1/2 * 23.0 cm = 11.5 cm. The image will be at infinity if rays emerging from the object are reflected parallel to the mirror. These rays would follow the same path, but in reverse direction, of parallel rays striking the mirror and being reflected to the focal point. So the object would have to be placed at the focal point, 11.5 cm from the mirror.
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NOTES ------->
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09:30:39 Recall that the focal distance of this mirror is the distance at which the reflections of rays parallel to the axis of the mirror will converge, and that the focal distance is half the radius of curvature. In this case the focal distance is therefore 1/2 * 23.0 cm = 11.5 cm. The image will be at infinity if rays emerging from the object are reflected parallel to the mirror. These rays would follow the same path, but in reverse direction, of parallel rays striking the mirror and being reflected to the focal point. So the object would have to be placed at the focal point, 11.5 cm from the mirror.
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RESPONSE --> ok
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09:49:01 query gen phy problem 23.14 radius of curvature of 4.5 x lens held 2.2 cm from tooth
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RESPONSE --> d1/do=1/f where d0= distance of object from the center of the mirror d1=distance of image from center of mirror. f=focal length of the object 4.5/2.2=1/f; Thus f= .48
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10:20:54 ** 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, I don't really understand the first part of the solution, but I start to understand when the two equations are introduced: 1 / image dist + 1 / obj dist = 1 / focal length and | image dist / obj dist | = magnification = 4.5, It's alot easier for me to start from the magnification equation to solve for the focal length and then the radius. In my first solution, I didn't understand that 4.5 was my magnification but now I do. -
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course PHY 202 vz׆lassignment #026
......!!!!!!!!...................................
09:30:25 Principles of Physics and General College Physics Problem 23.08. How far from a concave mirror of radius 23.0 cm must an object be placed to form an image at infinity?
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RESPONSE --> GIven the formula: Focal length=radius of mirror/2, thus f=23/2=11.5 cm.
.................................................
......!!!!!!!!...................................
09:30:35 07-22-2007 09:30:35 Recall that the focal distance of this mirror is the distance at which the reflections of rays parallel to the axis of the mirror will converge, and that the focal distance is half the radius of curvature. In this case the focal distance is therefore 1/2 * 23.0 cm = 11.5 cm. The image will be at infinity if rays emerging from the object are reflected parallel to the mirror. These rays would follow the same path, but in reverse direction, of parallel rays striking the mirror and being reflected to the focal point. So the object would have to be placed at the focal point, 11.5 cm from the mirror.
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NOTES ------->
.......................................................!!!!!!!!...................................
09:30:39 Recall that the focal distance of this mirror is the distance at which the reflections of rays parallel to the axis of the mirror will converge, and that the focal distance is half the radius of curvature. In this case the focal distance is therefore 1/2 * 23.0 cm = 11.5 cm. The image will be at infinity if rays emerging from the object are reflected parallel to the mirror. These rays would follow the same path, but in reverse direction, of parallel rays striking the mirror and being reflected to the focal point. So the object would have to be placed at the focal point, 11.5 cm from the mirror.
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RESPONSE --> ok
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09:49:01 query gen phy problem 23.14 radius of curvature of 4.5 x lens held 2.2 cm from tooth
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RESPONSE --> d1/do=1/f where d0= distance of object from the center of the mirror d1=distance of image from center of mirror. f=focal length of the object 4.5/2.2=1/f; Thus f= .48
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10:20:54 ** 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, I don't really understand the first part of the solution, but I start to understand when the two equations are introduced: 1 / image dist + 1 / obj dist = 1 / focal length and | image dist / obj dist | = magnification = 4.5, It's alot easier for me to start from the magnification equation to solve for the focal length and then the radius. In my first solution, I didn't understand that 4.5 was my magnification but now I do. -
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