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Homework Help: Index of Refraction and actual location of fish in water

  1. Mar 15, 2010 #1
    1. The problem statement, all variables and given/known data
    After finishing your physics classes with the firm conviction that the questions
    were strange and out of touch with reality, you decide to take up spear fishing. While
    wading through water (of index of refraction=n and depth h) you spot a fish that appears
    to be a distance away touching the rocky soil at the bottom of the water. Under the
    water the light seems to travel a horizontal distance d. It exits the water at point p and
    then travels a horizontal distance D to your eye. Please calculate where the fish
    actually is in terms of vertical distance under the water and horizontal distance from
    you. It will be helpful to calculate the diagonal distance the light the light travels in and
    out of the water. (Take the index of refraction of air to be 1.)

    2. Relevant equations
    n1sintheta1 = n2sintheta2

    3. The attempt at a solution
    First I wanted to find the theta2 (refraction in water)
    nair sin theta air = nwater sin theta water
    1 * sin theta air = n sin theta water
    sin theta water = sin theta air / n

    Next, I tried to calculate the diagonal distance the light traveled while in the water, assuming there was no refraction:
    sin theta x (no refraction) = d / sqrt (d^2 + h^2)

    Then I tried to calculate the diagonal distance the light traveled while in the air:
    sin theta air = D / Air diagonal

    Then I combined the two to solve for Air diagonal:
    sin theta air = sin theta x
    D / Air diagonal = d / sqrt (d^2 + h^2)
    Air diagonal = D*sqrt(h^2 + d^2) / d

    Then I wanted to solve for the actual distance the light travels in the water (accounting for refraction):
    I made x be the distance between the actual fish and the virtual fish
    sin theta water = (d-x) / sqrt ( h^2 + (d-x)^2)
    Which I think simplifies to 1/h

    Now I'm not sure what to do, because it asks the vertical distance the fish is down in the water (I thought he would still be on the bottom, because the illusion is horizontal, not vertical) and I guess I could just add up the D + d-x to get the horizontal distance away from the person, but I can't have x in my answer.

    Can someone help? The diagram is attached. Thanks!

    Then I tried to find the diagonal distance

    Attached Files:

  2. jcsd
  3. Mar 17, 2010 #2
    OK - I managed to find a formula to solve for X, but now i'm having trouble with the algebra:

    [ x / sqrt(x^2 + h^2) ] = [ d / sqrt(d^2 + h^2) ]

    Can anyone help me remember how to get rid of the sqrt of x on the bottom of the left hand side? If I multiply by the sq rt of the ( ) I will have to do it on the other side, so I will still have the sq rt.

    If I square both sides of the equation, I think that I would get: x^2 / (x^2 + h^2) = d^2 / (d^2 + h^2) Is that right?

    If that is right, then my x's cancel, so I'm guessing that is wrong. Can someone help?
  4. Mar 17, 2010 #3


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    You did not take refraction of light into account. It does not go along a straight line from the fish to your eyes. Apply Snell's law.

  5. Mar 17, 2010 #4
    I did apply Snell's law, and I did tak the refraction of light into account. that is how I got the equation (x is the distance of the actual fish, where d is the distance of the virtual fish). I just don't know how to solve the algebra, can you help me with that?
  6. Mar 18, 2010 #5


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    Show your work. How did you take the refraction into account if you do not use the refractive index of water?

    As for solving the following equation,

    x^2 / (x^2 + h^2) = d^2 / (d^2 + h^2)

    Take the reciprocal of both sides:


    [tex]1+h^2/x^2=1+h^2/d^2\rightarrow h/x =h/d \rightarrow x=d[/tex]

  7. Mar 18, 2010 #6
    I must have done something wrong, because x can't equal d. x is less than d because of the index of refraction (in the diagram).

    Here is how I got the formula:
    the theta for the apparent angle of where the fish appears is I called theta apparent
    the theta for the angle of where the fish ACTUALLY is I called theta water

    sin theta air = sin theta apparent
    tan theta air = tan theta apparent
    d/h = d/h

    tan theta air = sin theta air / cos theta air
    sin theta air = n(water)sin theta water (because n of air is 1)
    cos theta air = h / sqrt(d^2 + h^2)

    n sin theta water / [h / (sqrt(d^2 + h^2)] = n sin theta water = d / (sqrt(d^2 + h^2)
    sin theta water = x / sqrt(x^2 + h^2) where x = distance of ACTUAL fish, not apparent fish (d is apparent fish, so x is less than d)

    n * x / sqrt(x^2 + h^2) = d / (sqrt(d^2 + h^2)

    But If solving for that gives x = d, then something is wrong. Do you know what I'm doing wrong?
  8. Mar 18, 2010 #7


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    Your equation did not contain n before. This new one is correct.

    n * x / sqrt(x^2 + h^2) = d / (sqrt(d^2 + h^2)

    Do the same as before. Square both sides, take the reciprocal, simplify...

  9. Mar 18, 2010 #8
    sorry, I see what I did - I forgot the n in the top equation before.

    so now I'm still confused as to how to solve for x...
  10. Mar 18, 2010 #9
    would solving that result in x = d/n?

    That is what I got. Here is how I got it:

    (x^2 + h^2) / x^2n^2 = (d^2 + h^2) / d^2
    1/n^2 + h^2/x^2 = 1 + h^2/d^2
    1/n^2x^2 = 1/d^2
    nx = d
    x = d/n

    Is that right? Thanks for your help on this I appreciate it by the way
  11. Mar 18, 2010 #10
    shoot, now when I did it again, I got x/n = d, so x = nd

    I think x = nd is correct...
  12. Mar 18, 2010 #11


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    1/n^2 + h^2/x^2 = 1 + h^2/d^2 - this is wrong.

    You have to divide both terms in the parentheses by n^2, or it is better if you multiply the equation

    (x^2 + h^2) /( x^2n^2 )= (d^2 + h^2) / d^2

    with n^2, obtaining

    1 + (h/x)^2= (1+( h/d^2) )n^2

  13. Mar 18, 2010 #12
    ok just tried to do it again and i'm completely lost.
  14. Mar 18, 2010 #13
    ok, i see how you got to here:

    1 + (h/x)^2= (1+( h/d^2) )n^2

    but not i'm still not sure how to solve for x?
  15. Mar 18, 2010 #14
    when I try to solve that, I get:

    x = (h^2 + d^2 + h^2n^2) / n^2
    x = (h + d + hn) / n
  16. Mar 18, 2010 #15


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    It is not correct.

    So we start from here:

    [tex]1 + (h/x)^2= (1+( h/d)^2 )n^2[/tex]

    (sorry, I put a parenthesis to the wrong place in my last post.)

    subtract 1.

    [tex](h/x)^2=(1+( h/d)^2 )n^2-1[/tex]

    Take the reciprocal:

    [tex](x/h)^2=\frac{1}{(1+( h/d)^2 )n^2-1}[/tex]

    take the square root and multiply by h:

    [tex]x=h\sqrt{\frac{1}{(1+( h/d)^2 )n^2-1}}[/tex]

    Check. I might make mistakes, too.

  17. Mar 18, 2010 #16
    I went through it a couple times and got the same thing. Thanks so much for your help!!!
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