course Phy 122 When I completed Introductory Problem Set 6: #22-24, there was no solution for number 22. Could you give me the general solution to that problem?
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23:21:27 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 --> Rays parallel to the principal axis of a concave spherical mirror come to a focus at F, the focal point, as long as the mirror is small in width as compared to its radius of curvature, r, so that the rays are paraxial and make small angles with the axis. Figure 23-12 explains this concept very well and makes the relationship between r and F easy to see. The distance between the center of the curvature of the mirror and a point where a ray strikes is r, the radius of the curvature. The focal point is found at a focal length, f, from the center of the curvature to a point directly linear to it. The distance f is half of the distance r. Therefore, f = r/2 = 23cm/2 = 11.5cm. The focal distance is 11.5cm. In order for the image to form at infinity, the object must be placed at the focal distance, 11.5cm.
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23:22:12 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 --> I understand this concept. The book provided some very useful information.
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23:22:30 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 --> I am not a general physics student and this problem was not assigned to me.
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23:22:54 ** 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.
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23:23:06 **** 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 --> I am not a university physics student.
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23:23:14 ** 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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