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Phy 242
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Question_Set6_introset
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On set 6 intro set # 22, I am confused by the solution.
The question is:
Where will the image form if an object 1.5 meters high is positioned 4 meters in front of a converging lens whose focal length is 50 millimeters, and how large will the image be? Will the image be upright or inverted? Will it be real or virtual? Sketch a diagram explaining how the image forms.
The solution says:
The object distance is dObj = 4 meters, and the focal length is f = 10 m. If the image distance is dImg, we have
1 / dObj + 1/ dImg = 1 / f.
We solve this equation for dImg, obtaining
dImg = f * dObj / (dObj + f) = 10 m * 4 m / ( 10 m + 4 m) = 0 meters.
The ratio of image size hImg to object size hObj is the same as the ratio of image distance to object distance, or
hImg / hObj = g m / (b m) = 0,
so
hImg = hObj * 0 = 4 meters * 0 = 0 meters.
None of the values match up and 0 doesn't make sense for the image distance and height. Can you please show me how to find the distance that the image will form, height of image? I don't understand this solution from the picture that shows below the problem.
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The random number generator sometimes doesn't get the right numbers.
You have all the information you need to do these calculations.
You should verify that dImg comes out somewhere around 3 meters, hImg / hObj around .7 and hImg around 1 meter.
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Remember that the explanations in the given solutions are correct, as far as I'm aware, but that the numbers sometimes go haywire.
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