#$&*
Phy 122
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.
** Question Form_labelMessages **
Solution for Problem Set 6 22
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Problem set 6 #22 reads...
Sketch a diagram explaining how the image forms, find the size and position of the image formed, whether the image will be upright or inverted, and whether real or virtual, under the following conditions:
An object 1.72 meters high is positioned 3 meters in front of a converging lens whose focal length is 65 millimeters.
Solution
The object distance is dObj = 3 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 * 3 m / ( 10 m + 3 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 = 3 meters * 0 = 0 meters.
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The given solution doesn't seem to populate with the values given in the question.
Here is what I came up with for the solution.
The object distance is dObj = 3 meters, and the focal length is f = .065 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) = (.065 m * 3 m) / ( .065 m + 3 m) = .0636meters.
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 = dImg/dObj = .0636 m/3m =.0212
so
hImg = hObj * .0212
hImg = 1.72 m * .0212 = .0365 meters
???Is this correct???
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If you draw the ray diagrams you will see that the principle rays converge at a point futher from the lens than the focal point. So .0636 meters, being less than the focal distance .064 meters, can't be right.
The problem is the equation dImg = f * dObj / (dObj + f), which you probably got from the given solution. That isn't correct. It should be dImg = f * dObj / (dObj - f).
To see this, we'll use o and i for object and image distance, and the equation 1 / o + 1 / i = 1 / f.
Rearranging 1 / o + 1 / i = 1 / f we get
1 / i = 1 / f - 1 / o = (o - f) / (f * o)
so that
i = f * o / (o - f).
The result will be image distance which exceeds the focal distance by about as much as your result fell short of it.
The ratio of sizes will still be about .021.
The image will be inverted, as a ray diagram will reveal.
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