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A camera with a 100 $\mathrm{mm}$ focal length lens is used to photograph the sun and moon. What is the height of the image of the sun on the film, given the sun is $1.40 \times 10^{6} \mathrm{km}$in diameter and is $1.50 \times 10^{8} \mathrm{km}$ away?
$-0.933 \mathrm{mm}$
Physics 103
Chapter 25
Geometric Optics
Reflection and Refraction of Light
Rutgers, The State University of New Jersey
University of Michigan - Ann Arbor
University of Sheffield
Lectures
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we can find the height of the image by using the lenses formal of us. That is one where f is a clue. Nobody. I plus one over. D'oh! So we're fine, d I hear so d I will be one were f minus one word. D'oh minus one. Well, we know the definition of magnification, which is a dry, dry bitch. Oh, hi. Did the image you over? I don't They object. That is a cool minus D I right. Bye, dear. Then we can find a d r h. I hear height off the image, which is a dynasty I divided variety old times Mitchell. So, um, putting the values for a D. I hear we get minus a joe divided by the old times one minus f minus d. So now our job is to put the values from the given problem. His minus four minus 1.4 times 10 to the power nine readers is the height of the object in the distanced object. Me him is 1.5 times 10 to the power 11. So the writing this applying one where F, which is 10110 meter, minus one of the odors again. 1.5 times 10 to the power 11. So we find this. We get here. The height of the image we get here is minus mind points 33 times 10 to the power minus four meter, which is equal to zero point 933 middle meat. This is the height of the image.
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