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Finding the displacement of a object as viewed through water

  1. Nov 4, 2011 #1

    Krishna0703

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    hi everybody, suppose we are viewing an object in a pool whereby the object is actually at the bottom of the pool, ...so it will appear as if it has been displaced. what formula can we use to calculate the displacement of the object...??
    are there any conditions applicable..help me with this detail please
     
    Krishna0703, Nov 4, 2011
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  3. Nov 4, 2011 #2

    sophiecentaur

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    Are you a frustrated harpoon fisherman? I have spent ages trying to spear fish from the side of a boat. The Maths is not hard -
    http://www.physicstutorials.org/home/optics/refraction-of-light/apparent-depth-real-depth" [Broken] gives you the formula but it's hard to work it out on the fly and amidst the excitement of a massive Grey Mullett going past.
     
    Last edited by a moderator: May 5, 2017
    sophiecentaur, Nov 4, 2011
  4. Nov 4, 2011 #3

    Krishna0703

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    nah man, there do have a formula related to the refractive index...
     
    Krishna0703, Nov 4, 2011
  5. Nov 4, 2011 #4

    sophiecentaur

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    It's on that link for you to use at your leisure. The formula gives both the new angle and the new height. But, if you know Snell's Law, you should be able to derive it for yourself. At least, their derivation makes easy sense.

    I'll get back to my fishing when my future son in law buys a new harpoon to replace the one he chucked over the side by mistake.
     
    sophiecentaur, Nov 4, 2011
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