Object Beneath the Water Problem

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SUMMARY

The problem involves calculating the apparent depth of a logo embedded in a transparent block when viewed from above the water's surface. The block has a refractive index of 1.81 and is submerged under water with a refractive index of 1.333. The logo is located 4.03 cm beneath the block's top surface, and after applying the formula for apparent depth, the logo appears to be 4.58 cm beneath the water surface. This calculation utilizes the relationship between real and apparent depths based on the refractive indices of the materials involved.

PREREQUISITES
  • Understanding of refractive indices and their significance in optics
  • Familiarity with the formula for apparent depth: apparent depth = d (n2/n1)
  • Basic knowledge of light behavior when transitioning between different media
  • Ability to perform calculations involving real and apparent depths
NEXT STEPS
  • Study the principles of refraction and Snell's Law
  • Learn about optical phenomena in different media, focusing on refractive indices
  • Explore practical applications of apparent depth in optical devices
  • Investigate the effects of varying refractive indices on light behavior in layered materials
USEFUL FOR

Students in physics or optics courses, educators teaching light behavior, and anyone interested in understanding the principles of refraction and apparent depth in various media.

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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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