Complex apparent and real depth problemm

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SUMMARY

The discussion focuses on calculating the apparent depth of a logo embedded in a block of transparent material when viewed from above the water's surface. The logo is located 3.59 cm beneath the block's top surface, with the block submerged under 1.86 cm of water. The correct approach involves applying the formula for apparent depth, specifically using the indices of refraction for glass (n = 1.73) and water (n = 1.333). The final calculated apparent depth of the logo from the observer's perspective is 4.162 cm beneath the water's surface.

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  • Understanding of refraction and indices of refraction
  • Familiarity with the concept of apparent depth in optics
  • Basic knowledge of the formula for calculating apparent depth
  • Ability to perform unit conversions and arithmetic operations
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Homework Statement



A small logo is embedded in a thick block of transparent material (n = 1.73), 3.59 cm beneath the top surface of the block. The block is put under water (n = 1.333), so there is 1.86 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 = dn_2/(n_1)

The Attempt at a Solution



What i tried doing is applying this equation twice. First with glass and water 3.59(1.33/1.73). Then with water and air 1.86(1/1.33). After that I added the two values of the apparent distances which I got 4.162
 
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pistonsfan321 said:
What i tried doing is applying this equation twice. First with glass and water 3.59(1.33/1.73).
OK. That's the apparent location of the first image below the block surface. What's the location of that image below the water surface? That's the distance you want when using the equation the second time.
 

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