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1. 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.)
2. Homework 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 = (dx) / sqrt ( h^2 + (dx)^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 + dx 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
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. Homework 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 = (dx) / sqrt ( h^2 + (dx)^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 + dx 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
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