A small tropical fish is at the center of a water-filled spherical fish bowl 28.

Question

A small tropical fish is at the center of a water-filled spherical fish bowl 28.0cm in diameter.

Part A

Find the apparent position of the fish to an observer outside the bowl. The effect of the thin walls of the bowl may be ignored.

Part B

Find the magnification of the fish to an observer outside the bowl.

Part C

A friend advised the owner of the bowl to keep it out of direct sunlight to avoid blinding the fish, which might swim into the focal point of the parallel rays from the sun. Is the focal point actually within the bowl?

Explanation / Answer

For part a) consider the fish to be the light source, and calculate the image position for light rays exiting the bowl.
We will be using this formula:

na/S + nb/S' = nb-na / R

na= 1.33 (water), nb=1;S = +14cm, R = -14cm (This radius is negative because the center of the bowl is on the same side as the light source (the fish))

1.33/14 + 1/S' = 1-1.33/-14

So, S' = -14cm
A negative value for S’ means the image is on the same side of the interface as the object (i.e. inside the bowl in this case). So the observer will see a virtual image of the fish at the center of the bowl

Part b)
Magnification can be found from this formula:
m = -naS'/nbS

m = -1.33(-14)/1*14

= 1.33
The fish appears larger by a factor of 1.33

Part c)
the light source is the sun, which is really far away (i.e. object distance is infinity). This is what they mean by “parallel rays from the sun”.
The focal point will be where the sun’s rays converge, so we need to find the image distance S’.
So,
na = 1; nb=1.33, S = infinity, R = +14cm (This radius is positive because the center of the bowl is on the opposite side as the light source (the sun))

1/infinity + 1.33/S' = 1.33-1/+14cm

So, S' = +56cm
This image is beyond the other side of the bowl (28cm away), so the fish will be safe.

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