# Index of Refraction and actual location of fish in water

• skibum143
In summary: If x is less than d, thenn * x / sqrt(x^2 + h^2) = d / (sqrt(d^2 + h^2))Since x is less than d, the equation becomesn * x = d / (sqrt(d^2 + h^2))

## Homework Statement

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.)

## Homework Equations

n1sintheta1 = n2sintheta2

## 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

#### Attachments

• fishing.pdf
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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?
Thanks!

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.

ehild

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?

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:

$$\frac{x^2+h^2}{x^2}=\frac{d^2+h^2}{d^2}$$

simplify:
$$1+h^2/x^2=1+h^2/d^2\rightarrow h/x =h/d \rightarrow x=d$$

ehild

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)
therefore:

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?

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...

ehild

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...

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

shoot, now when I did it again, I got x/n = d, so x = nd

I think x = nd is correct...

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

ehild

ok just tried to do it again and I'm completely lost.

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?

when I try to solve that, I get:

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

skibum143 said:
when I try to solve that, I get:

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

It is not correct.

So we start from here:

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

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

subtract 1.

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

Take the reciprocal:

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

take the square root and multiply by h:

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

Check. I might make mistakes, too.

ehild

I went through it a couple times and got the same thing. Thanks so much for your help!