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Ch. 32 - Light: Reflection and Refraction
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
All textbooksGiancoli Douglas 5th editionCh. 32 - Light: Reflection and RefractionProblem 64
Chapter 31, Problem 64

A fish is swimming in water inside a thin spherical glass bowl of uniform thickness. Assuming the radius of curvature of the bowl is 32.0 cm, locate the image of the fish if the fish is located: (a) at the center of the bowl; (b) 20.0 cm from the side of the bowl between the observer and the center of the bowl. Assume the fish is small.

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Understand the problem: The fish is inside a spherical glass bowl filled with water. The bowl acts as a spherical refracting surface. We need to locate the image of the fish for two cases: (a) when the fish is at the center of the bowl, and (b) when the fish is 20.0 cm from the side of the bowl. The radius of curvature of the bowl is given as 32.0 cm. We'll use the formula for refraction at a spherical surface.
Recall the formula for refraction at a spherical surface: \( \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \), where \( n_1 \) is the refractive index of the medium where the object is located (water in this case), \( n_2 \) is the refractive index of the medium where the image is formed (air), \( u \) is the object distance, \( v \) is the image distance, and \( R \) is the radius of curvature of the spherical surface.
For part (a): When the fish is at the center of the bowl, the object distance \( u \) is equal to the radius of curvature \( R \). Substitute \( u = R \), \( n_1 = 1.33 \) (refractive index of water), \( n_2 = 1.00 \) (refractive index of air), and \( R = 32.0 \ \text{cm} \) into the formula. Solve for \( v \), the image distance.
For part (b): The fish is 20.0 cm from the side of the bowl. This means the object distance \( u \) is 20.0 cm. Use the same formula \( \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \), substituting \( u = 20.0 \ \text{cm} \), \( n_1 = 1.33 \), \( n_2 = 1.00 \), and \( R = 32.0 \ \text{cm} \). Solve for \( v \), the image distance.
Interpret the results: For both parts (a) and (b), the sign of \( v \) will indicate whether the image is real or virtual. A positive \( v \) means the image is on the same side as the observer (real image), while a negative \( v \) means the image is on the opposite side (virtual image).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Refraction of Light

Refraction is the bending of light as it passes from one medium to another, such as from water to glass. This bending occurs due to a change in the speed of light in different materials. Understanding refraction is crucial for determining how the image of the fish will appear to an observer, as it affects the path of light rays emanating from the fish.
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Index of Refraction

Spherical Mirrors and Lenses

Spherical mirrors and lenses are curved surfaces that can converge or diverge light rays. In this scenario, the glass bowl acts like a lens, and its curvature influences the formation of the image. The radius of curvature is essential for calculating the focal length, which helps in locating the image of the fish based on its position relative to the bowl.
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Image Formation

Image formation involves the process by which light rays converge to create a visual representation of an object. In optics, the position, size, and nature (real or virtual) of the image depend on the object's distance from the lens or mirror and its curvature. For the fish in the bowl, understanding how to apply the lens formula will allow us to find the image location for different positions of the fish.
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Related Practice
Textbook Question

(II) A ray of light, after entering a light fiber, reflects at an angle of 14.5° with the long axis of the fiber, as in Fig. 32–57. Calculate the distance along the axis of the fiber that the light ray travels between successive reflections off the sides of the fiber. Assume that the fiber has an index of refraction of 1.55 and is 1.60 x 10-4 m in diameter.

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Textbook Question

(III) A beam of light enters the end of an optic fiber as shown in Fig. 32–59. (a) Show that we can guarantee total internal reflection at the side surface of the material (at point A), if the index of refraction is greater than about 1.42. In other words, regardless of the angle α , the light beam reflects back into the material at point A, assuming air outside. (b) What if the fiber is immersed in water?

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Textbook Question

Two identical concave mirrors are set facing each other 1.0 m apart. A small lightbulb is placed halfway between the mirrors. A small piece of paper placed just to the left of the bulb prevents light from the bulb from directly shining on the left mirror, but light reflected from the right mirror still reaches the left mirror. A good image of the bulb appears on the left side of the piece of paper. What is the focal length of the mirrors?

Textbook Question

(c) Determine the magnification of a plane mirror in this same limit.

(d) Are your results in parts (b) and (c) consistent with the discussion of Section 32–2 on plane mirrors?

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Textbook Question

A beam of light is emitted 7.7 cm beneath the surface of a liquid and strikes the surface 7.2 cm from the point directly above the source. If total internal reflection occurs, what can you say about the index of refraction of the liquid?

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Textbook Question

The critical angle of a certain piece of plastic in air is θC = 35.8°. What is the critical angle of the same plastic if it is immersed in water?

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