Object Beneath the Water Problem

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The problem involves determining the apparent depth of a logo embedded in a transparent block submerged in water. The logo is located 4.03 cm beneath the block's surface, which is 2.15 cm underwater. The correct approach involves using the formula for apparent depth, applying the refractive indices of both the block and water. The calculations yield an apparent depth of 4.58 cm from the water's surface. This indicates how the logo appears to an observer looking from above the water.
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Homework Statement


A small logo is embedded in a thick block of transparent material (n = 1.81), 4.03 cm beneath the top surface of the block. The block is put under water (n = 1.333), so there is 2.15 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?


Homework Equations


apparent depth = d ((n_2)/(n_1))


The Attempt at a Solution


First was 4.03(1.33/215) it was wrong.
 
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For the block in water I used n2/n1 = real/apparent depth.
n2 is the block, n1 is the water. I got 2.96cm as the point in the block from which the light appears to come from into the water.
You can now take this point to be at a depth of 2.15 + 2.96 in WATER
then do n2/n1 = real/ apparent again ( I got 4.58cm from water surface)
 
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