Problem 21 - University Physics with Modern Physics

Question

A small tropical fish is at the center of a water-filled, spherical fish bowl 28.0 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 closer than the real position, magnified, but the focal point is outside the bowl.

Step 1: Understand the Problem

A fish is located at the center of a spherical fishbowl. Its diameter is given as 28.0 cm. We need to find the apparent position and magnification of the fish, ignoring wall thickness effects.

Step 2: Identify the Refraction Model

We'll use the refraction at a spherical interface model, given by the formula: \( \frac{n_1}{s} + \frac{n_2}{s'} = \frac{n_2 - n_1}{R} \), where \( n_1 = 1.33 \) (refractive index of water), \( n_2 = 1.0 \) (refractive index of air), \( s = -14.0 \) cm (distance from fish to the bowl surface, negative since it's measured from the fish to the bowl surface inwards), and \( R = 14.0 \) cm (radius of the bowl).

Step 3: Calculate Apparent Position (s')

Using the refraction formula: \( \frac{1.33}{-14} + \frac{1}{s'} = \frac{1.0 - 1.33}{14} \). Solve for \( s' \) to find the apparent position of the fish from the surface on the outside.

Step 4: Calculate Magnification

Magnification (\( m \)) is given by \( m = \frac{s'}{s} \). Using the values of \( s' \) and \( s = -14.0 \) cm obtained, find the magnification.

Step 5: Determine the Focal Point

The focal length \( f \) for the spherical refractor is given by \( \frac{n_2 - n_1}{f} = \frac{n_2 - n_1}{R} \). Since \( n_2 > n_1 \) and \( R = 14.0 \) cm, we calculate \( f \). Compare \( f \) to the radius to see if it lies within the bowl.

Key Concepts

Refraction

Refraction is a key concept in optics, describing how light bends as it passes through different mediums. This bending of light is due to the change in speed when light travels from one medium to another. The direction and degree to which light bends depends on the refractive indices of the two media. The classic example is a straw appearing "bent" when placed in a glass of water, illustrating how light changes direction at the water-air boundary.
For the fish in the spherical fishbowl, refraction occurs at the boundary between water and air. We use a refraction formula, involving the refractive indices of water ( _1 = 1.33") and air ( _2 = 1.0"). This formula allows us to calculate how the position of the fish seems to change to an observer outside the bowl. Understanding refraction is crucial for explaining many optical phenomena.

• Refraction changes the apparent position of objects.
• It is significant in designing lenses and optical instruments.
• The refractive index determines how much light rays will bend.

Spherical Interface

A spherical interface refers to a boundary with a circular cross-section. In optics, this typically involves light passing through surfaces of lenses or spherical containers, like the fishbowl. The curvature of the interface affects how light is refracted.
The fishbowl exercise involves calculating the apparent position of the fish using a formula specific to spherical interfaces. This formula considers the refractive indices of the two media and the radius of curvature of the sphere.
The radius of the fishbowl, 14.0 cm, plays a critical role in determining how light rays are bent as they exit the water and enter the air.

• The curvature affects how and where light rays converge or diverge.
• Spherical surfaces are common in lenses, mirrors, and bowls.
• Understanding spherical interfaces is vital for optical engineering.

Magnification

Magnification in optics refers to the change in size of an object's appearance due to refraction or reflection processes. It is a dimensionless quantity that indicates how much larger or smaller an object appears, compared to its actual size.
In the case of the fish in the bowl, magnification is calculated using the formula \( m = \frac{s'}{s} \), where \( s \) is the original distance of the fish from the surface, and \( s' \) is the apparent distance.
The magnification gives an idea of how big or small the fish seems to an outside observer, relative to its real size.

• Magnification is positive if the image appears upright.
• It is negative when the image appears inverted.
• A magnification greater than 1 means the image is larger.
• Magnification less than 1 means the image is smaller.

Refractive Index

The refractive index is a core concept that quantifies how much a medium can bend light. It is defined as the ratio of the speed of light in a vacuum to its speed in the given medium.
Different materials have different refractive indices, which affect how strongly light is bent. In this exercise, water has a refractive index of 1.33, meaning light travels slower in water than in air, which has a refractive index of 1.0.
The refractive index is essential in the refraction formula, influencing the calculation of s', the apparent position of the fish.

• A higher refractive index means greater bending of light.
• The refractive index can vary with wavelength, causing dispersion.
• It's crucial for designing and using optical lenses and systems.