Where will the image of the fish appear in relation to the observer?

[T7]

Hi,

This is undoubtedly an easy question, but I am having "one of those days"...

A fish in a pond is 2m below the water's surface. An observing eye is 0.75m above the water and a horizontal distance of 1m from the fish. Given that the water has a refractive index of 1.33, where is the image of the fish going to appear to the observer? (I am supposed to use my answer to this question to show whether or not the refractive index = real depth / apparent depth, and then state whether the position of the image changes as the observer's eye moves further away horizontally).

Without first knowing a priori where the incident ray hits the water, I'm not sure how I can diagram and calculate this one.

Any hints?

Cheers,

T7

 

[HallsofIvy]

Draw a picture. Put the fish where ever you want, your eye where ever you want.
Draw in the perpendiculars to the water from your eye and the fish. Label the appropriate sides of the "eye" right triangle 0.75 m and 1 m. From that you can figure out the angle the ray (from your eye to the water) makes with the water (tan theta= opposite side/near side). Knowing the index of refraction, you can find the angle the ray (from the water to the fish) makes with the water. Use THAT, together with the 2 meters depth to determine the position of the fish. Now, find a point on the straight line from your eye (without bending) that has the same distance along that line as the distance from the water to the fish.

 

[T7]

Draw a picture. Put the fish where ever you want, your eye where ever you want.
Draw in the perpendiculars to the water from your eye and the fish. Label the appropriate sides of the "eye" right triangle 0.75 m and 1 m. From that you can figure out the angle the ray (from your eye to the water) makes with the water (tan theta= opposite side/near side). Knowing the index of refraction, you can find the angle the ray (from the water to the fish) makes with the water. Use THAT, together with the 2 meters depth to determine the position of the fish. Now, find a point on the straight line from your eye (without bending) that has the same distance along that line as the distance from the water to the fish.

Thanks for the answer. I'll give it a go - though at first glance I am unsure how finding the angle of a line from the eye to a point on the water just above the fish will give me what I need to calculate the angle of refraction from, since the actual light ray from the fish to the eye will intersect the water's boundary at some point x<1m from the eye. But I'll look at this again when I have some more time.

Cheers!

 

[GCT]

$$n1/p + n2/q=(n1-n2)/R,~R=infinite~thus~-p(n2/n1)=q$$

Solve for q, this will be the respective location of the virtual image above the fish

Simply draw out a right triangle, the with the image and observer at two of the vertices.

All rays emerging from the fish at a point will refract in an outwardly fashion and all will form a virtual image at a particular point, this is the image point.