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A concave mirror produces a real image that is three times as large as the object. (a) If the object is 22 $\mathrm{cm}$ in front of the mirror, what isthe image distance? (b) What is the focal length of this mirror?
(a) 66 $\mathrm{cm}$(b) 17 $\mathrm{cm}$
Physics 103
Chapter 26
Geometrical Optics
Wave Optics
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Okay, so for this question, port A. We will be looking for to the image distance with the given information. And we know at the con Cave mirror would produce a real image that is three times as tall as the object. So we know that the height of the image equals three times the height of the object, then magnificent. Therefore, we can also say that the magnification would equal height of the image divided by height of the object. And then we would substitute the negative three for em since, well, that's be the magnification here. So we would have negative three equals negative distance of the image about by distance of the object. Therefore, distance of the image one equal three times the distance of the object. And here we can substitute. Since we know the distance of the object, we can substitute in the 22 centimeters which is given so that the distance of the image equals three times a 22 centimeters and therefore we get the distance of the image is equivalent to 66 centimeters. Because of this sense of the image from the mirror and now for poor be, we will once again calculates the focal length of the mirror. Using the Formula One over the focal length equals one over the distance of the object, this one divided by the distance of the image. And since we know that the distance of the object this twe two centimeters from the distance of the image, it's 66 centimeters. We just plugged that in, so we have that one over. The focal length was one divided by 22 centimeters, plus one over 66 centimeters, and this is also equivalent to 4/66 centimeters, and therefore local ING is equivalent to 16.62 centimeters, which is approximately 17 centimeters for the focal length.
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