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Homework Help: Perceived/actual distance (block of glass with logo in water)

  1. May 7, 2014 #1
    1. The problem statement, all variables and given/known data

    A logo is embedded in a block of glass (n=1.52), 3.2cm beneath the top surface of the glass. The block is put under water so that there is 1.5cm of water above the top surface of the glass. 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 the observer?

    2. Relevant equations

    d' = d(n2)/n1
    Where n2 is the refraction index of the medium in which the observer is and n1 is the RI of the medium in which the object is observed.

    3. The attempt at a solution

    Because the observer is observing the logo in materials with greater refractory indexes than air, the logo will appear closer than its actual distance.

    The actual distance from the logo to air is 3.2 + 1.5 = 4.7cm

    There are two perceived distances, one is the perceived distance of the logo in the block and the other is the perceived distance of the logo in water. Added together, this would be the total perceived distance.

    Perceived distance of logo in block, as observed from air, is d(n2/n1) where n2 is air (≈1) and n1 is the block (given n = 1.52), so d' = 3.2 (1/1.52) = 2.1

    Perceived distance of logo in water, as observed from air, is d(n2/n1), where n1 is water and the amount of water given is 1.5, so d' = 1.5 (1/1.33) = 1.12

    Total distance = 1.12 + 2.1 = 3.23

    I have the answer right, but only because I just kept plugging numbers and guessing.

    Originally, I kept getting this wrong because I was using the total distance from air of the logo in water (3.2+1.5 = 4.7) as the actual distance, so I had d' = 4.7 (1/(1.33 +1.52), which was wrong.

    What I don't get is why you can't use total distance of the logo from air (4.7) somehow to get the answer. Why is the distance of the logo in water only 1.5 and not 4.7? I'm probably missing something totally obvious or just overthinking it, but any insight would be welcome. Thanks!
  2. jcsd
  3. May 7, 2014 #2


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    Science Advisor
    Homework Helper
    Gold Member

    The two distances are subject to different weighting factors, so adding them together with no weighting factor cannot tell you anything useful.
    The problem is analogous to driving some distance at one speed and another distance at a different speed, and asking how long the journey takes. Adding the two distances is not a useful operation.
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