Problem 23 - College Physics

Question

A spherical fishbowl. A small tropical fish is at the center of a water-filled spherical fishbowl \(28.0 \mathrm{~cm}\) in diameter. (a) Find the apparent position and magnification of the fish to an observer outside the bowl. The effect of the thin walls of the bowl may be ignored. (b) 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?

Step-by-Step Solution

Answer

The fish appears 56.0 cm away, magnified 3 times. The focal point is inside the bowl.

Step 1: Understanding the Problem

We are tasked with finding two things: (a) the apparent position and magnification of the fish as perceived by an observer, and (b) the location of the focal point of parallel rays entering a spherical fishbowl compared to the fishbowl's size.

Step 2: Identify Known Values

Given the fishbowl is a sphere with a diameter of \(28.0\ \text{cm}\), the radius \(R = 14.0\ \text{cm}\). The fish is positioned at the center, so the object distance \(s = 0.0\). The refractive index of water \(n = 1.33\).

Step 3: Calculating Apparent Position of the Fish

Use the spherical refraction formula: \(\frac{n_1}{s} + \frac{n_2}{s'} = \frac{n_2 - n_1}{R}\), where \(n_1 = 1\) (air), \(n_2 = 1.33\) (water), \(s\) is the object distance, and \(s'\) is the image distance in water. Since the fish is at the center, \(s = 14.0\ \text{cm}\). Rearranging for \(s'\) gives \(s' = \frac{n_2R}{n_2 - n_1}\).

Step 4: Solving for the Apparent Position

Substitute the known values into the equation: \[s' = \frac{1.33 \times 14.0}{1.33 - 1} \approx 56.0\ \text{cm}\]. Thus, the apparent position is \(56.0\) cm from the center, behind the spherical surface from the observer's perspective.

Step 5: Calculating Magnification

Magnification is given by \(m = \frac{n_1 s'}{n_2 s}\). Substituting known values: \(m = \frac{1 \times 56.0}{1.33 \times 14.0} \approx 3.0\). Thus, the magnification is \(3.0\), meaning the fish appears three times larger.

Step 6: Determining the Focal Point Location

The focal length \(f\) can be found using \(f = \frac{R}{2(n_2 - n_1)}\). Substituting known values, \(f = \frac{14.0}{2 \times (1.33 - 1)} \approx 21.2\ \text{cm}\). Since this distance is less than the radius, the focal point is within the bowl.

Key Concepts

Apparent Position

When an object is viewed through a curved surface like a fishbowl, the light rays bend, causing the object's position to appear different from where it really is. This is known as the apparent position. The formula used to find this position is based on the principles of spherical refraction, which consider the bending of light as it moves from one medium to another. In this exercise, the fish is placed inside a spherical water-filled fishbowl.
The spherical refraction formula applied here is:

• \[ \frac{n_1}{s} + \frac{n_2}{s'} = \frac{n_2 - n_1}{R} \]
• Where \( n_1 \) is the refractive index of air (1.00), \( n_2 \) is the refractive index of water (1.33), \( s \) is the object distance (the actual position of the fish), and \( s' \) is the image distance or apparent position.

The calculation results in the apparent position of the fish being 56 cm behind the surface of the sphere. This means to an observer outside the bowl, the fish appears to be further away than its actual position inside the bowl center.

Magnification

Magnification refers to how much larger or smaller an object appears compared to its actual size. It's especially useful in optics where objects viewed through lenses or curved surfaces often seem different in size. In the context of spherical refraction, magnification can be determined by the formula:

• \[ m = \frac{n_1 s'}{n_2 s} \]
• Where \( m \) is magnification, and the other parameters are the same as mentioned earlier.

According to this calculation, the fish seems three times larger than its real size due to the fishbowl's spherical shape. This happens because light rays bend more significantly when passing from water to air, making the object appear bigger to an observer outside the fishbowl.

Focal Point

The focal point is a critical concept in optics, describing the exact spot where parallel rays of light converge after passing through a lens or a curved surface. In our spherical fishbowl scenario, determining whether the focal point is inside or outside the bowl helps understand the light behavior through the bowl.
The relation for the focal length in spherical refraction is:

• \[ f = \frac{R}{2(n_2 - n_1)} \]
• Here, \( f \) represents the focal length, and again, the other symbols follow the previously noted meanings.

Through calculation, it is found that the focal point, located at 21.2 cm from the sphere's surface, falls within the fishbowl. This means any parallel rays like those from the sun could indeed focus inside the fishbowl, potentially blinding the fish if it swims into that region. It's a compelling example of why optical awareness is important where lenses and refraction are involved.