Question

A small tropical fish is at the center of a water-filled spherical fishbowl 28.0 cm in diameter.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.b. Find the magnification of the fish to an observer outside the bowl.

Answer 1

This physics question requires knowledge of optics and principles of refraction to determine the apparent position and magnification of an object when viewed through a spherical medium. Calculating these properties involves using the lensmaker's equation and concepts such as refraction at spherical surfaces and magnification formulas.

Step-by-step explanation:

The student's question involves optics, particularly the principles of refraction and lens magnification in the context of observing objects through spherical surfaces like fishbowls or through a microscope's objective lenses. The question requires an application of the lensmaker's equation to find apparent positions, image formation, and magnifications due to lenses, taking into consideration the index of refraction of different materials, like glass, water, and Plexiglas. The fact that the walls of the fishbowl can be ignored simplifies the calculation as a single refractive surface.

Part a of the question refers to locating the apparent position of an object (the fish) inside a spherical medium when viewed by an outside observer. This phenomenon is due to refraction and can be calculated using the formula for refraction at spherical surfaces. Part b asks for the magnification of the fish as seen by the observer. This too involves refraction and can be calculated using magnification formulas associated with refractive surfaces.

For simplicity, solutions would require knowledge of the refractive indices involved and a methodical approach utilizing the principles of geometric optics, including Snell's Law and the lensmaker's equation.

Answer 2

To find the apparent position of the fish to an observer outside the bowl, we can use the lens equation:

1/p + 1/q = 1/f

where p is the distance from the object (the fish) to the center of the lens (in this case, the center of the fishbowl), q is the distance from the image to the center of the lens, and f is the focal length of the lens.

The focal length of a lens is given by the formula:

f = (n-1)R/2(n-1)

where R is the radius of the lens (in this case, the radius of the fishbowl) and n is the refractive index of the lens material (in this case, water).

Since the diameter of the fishbowl is 28.0 cm, the radius is 14.0 cm. The refractive index of water is approximately 1.33, so the focal length is approximately 8.5 cm.

To find the apparent position of the fish, we need to find p and q. If we let q be the distance from the image to the center of the lens, then the distance from the fish to the center of the lens is p = R - q.

Substituting these values into the lens equation, we get:

1/p + 1/q = 1/f

1/(R-q) + 1/q = 1/f

Solving this equation for q, we find that q = R/2 = 14.0 cm/2 = 7.0 cm.

This means that the apparent position of the fish is 7.0 cm from the center of the fishbowl.

To find the magnification of the fish to an observer outside the bowl, we can use the formula:

m = -q/p

Substituting the values we found above, we get:

m = -7.0 cm / (14.0 cm - 7.0 cm) = -1/2 = -0.5

This means that the fish is magnified by a factor of -0.5, or half its size.