course Phy 202 6/29 7PM 017. `Query 15
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Given Solution: 3 `aRecall 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: ********************************************* Question: `qquery gen phy problem 23.11 radius of curvature of 4.5 x mirror held 2.2 cm from tooth YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We want to know the curvature of the mirror and the radius of this mirror. 2.20cm away from the tooth and 4.5x magnification. We can have either a convex or concave mirror in this example. First we should find the image distance which will be: +-2.20cm(4.5) which gives an image distance of +-9.9 We can use this to find freq which will then be used to find the radius. F=2.20cm+9.90cm=12.1 We then use F to solve for radius using: 2(freq) which gives 22.2cm.
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Given Solution: 1 `a** 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. There is also a solution for the -9.9 m image distance, which would correspond to a positive magnification (i.e., an upright image). The image in this case would be 'behind' the mirror and therefore virtual. For this case the equation is 1 / (-9.9 cm) + 1 / (2.2cm) = 1 / f, which when solved give us f= (-9.9 cm * 2.2 cm) / (-9.9 cm + 2.2 cm) = 2.9 cm, approx. This solution would give us a radius of curvature of 2 * 2.9 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): This problem is extremely confusing. What parts am I supposed to know I feel like I solved it very wrong and for only a small section.