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**25. A small logo is embedded in a thick block of
crown glass (n = 1.52), 3.20 cm beneath the top surface of the glass. The
block is put under water, so there is 1.50 cm of
water above the top surface of the block. The logo is viewed from directly
above by an observer in air. How far beneath the top surface of the water
does the logo appear to be? |
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Sketch below lays out
situation |
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We will do this problem in
steps. First
we will determine the apparent depth looking from water into glass. Then that image will be the object for the
air water interface. From that we will get the final apparent depth. We will use the apparent depth relationship |
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So for water glass, we have |
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So if we were in water, we would think the
logo was 2.80 cm below top glass surface instead of the actual 3.20 cm it
is. Now we use that for the air-water
interface, so while this image is 2.80 cm below water –glass , since the
water is 1.50 cm thick, the distance from the air water is 4.30 cm (2.80 cm +1.5
cm) so the equation becomes |
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So the log appears 3.23 cm below water
surface which is still inside the glass! |
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