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Ch. 33 - Lenses and Optical Instruments
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. 33 - Lenses and Optical InstrumentsProblem 43
Chapter 32, Problem 43

(II) An aquarium filled with water has flat glass sides whose index of refraction is 1.51. A beam of light from outside the aquarium strikes the glass at a 43.5° angle to the perpendicular (Fig. 32–52). What is the angle of this light ray when it enters (a) the glass, and then (b) the water? (c) What would be the refracted angle if the ray entered the water directly?

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Step 1: Identify the given values and the laws of refraction to be used. The index of refraction of the glass is \( n_{glass} = 1.51 \), the angle of incidence in air is \( \theta_{air} = 43.5^{\circ} \), and the index of refraction of air is \( n_{air} = 1.00 \). Use Snell's Law, which states \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \), to calculate the angle of refraction when the light enters the glass.
Step 2: Apply Snell's Law for part (a). Substitute \( n_{air} \), \( \theta_{air} \), and \( n_{glass} \) into the equation \( n_{air} \sin(\theta_{air}) = n_{glass} \sin(\theta_{glass}) \). Solve for \( \sin(\theta_{glass}) \) and then find \( \theta_{glass} \) using the inverse sine function.
Step 3: For part (b), consider the transition of the light ray from the glass into the water. The index of refraction of water is \( n_{water} = 1.33 \). Use Snell's Law again, \( n_{glass} \sin(\theta_{glass}) = n_{water} \sin(\theta_{water}) \), to calculate the angle of refraction \( \theta_{water} \). Substitute the previously calculated \( \theta_{glass} \) and solve for \( \theta_{water} \).
Step 4: For part (c), consider the scenario where the light ray enters the water directly from the air. Use Snell's Law again, \( n_{air} \sin(\theta_{air}) = n_{water} \sin(\theta_{water\_direct}) \). Substitute \( n_{air} \), \( \theta_{air} \), and \( n_{water} \) to solve for \( \sin(\theta_{water\_direct}) \), and then find \( \theta_{water\_direct} \) using the inverse sine function.
Step 5: Summarize the results conceptually. The angles of refraction depend on the indices of refraction of the materials and the angle of incidence. The light bends towards the normal when entering a medium with a higher index of refraction and away from the normal when entering a medium with a lower index of refraction. This behavior is governed by Snell's Law.

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Key Concepts

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

Refraction

Refraction is the bending of light as it passes from one medium to another with a different index of refraction. This phenomenon occurs because light travels at different speeds in different materials. The degree of bending can be described by Snell's Law, which relates the angles of incidence and refraction to the indices of refraction of the two media.
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Snell's Law

Snell's Law provides a mathematical relationship to determine the angle of refraction when light transitions between two media. It is expressed as n1 * sin(θ1) = n2 * sin(θ2), where n1 and n2 are the indices of refraction of the first and second media, respectively, and θ1 and θ2 are the angles of incidence and refraction. This law is essential for calculating how light behaves at the interface of different materials.
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Index of Refraction

The index of refraction is a dimensionless number that describes how much light slows down in a medium compared to its speed in a vacuum. It is defined as n = c/v, where c is the speed of light in a vacuum and v is the speed of light in the medium. A higher index indicates that light travels slower in that medium, which affects the angle at which light is refracted when entering or exiting the medium.
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Related Practice
Textbook Question

(II) A diverging lens is placed next to a converging lens of focal length ƒC , as in Fig. 33–14. If ƒT represents the focal length of the combination, show that the focal length of the diverging lens, ƒD , is given by


1/ƒD = (1/ƒT) - (1/ƒC)

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Textbook Question

"(II) Two plane mirrors meet at a 135° angle, Fig. 32–47. If light rays strike one mirror at 32° as shown, at what angle θ do they leave the second mirror?


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Textbook Question

(II) (a) What is the minimum index of refraction for a glass or plastic prism to be used in binoculars (Fig. 32–34) so that total internal reflection occurs at 45°? (b) Will binoculars work if their prisms (assume n = 1.58) are immersed in water? (c) What minimum n is needed if the prisms are immersed in water?

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Textbook Question

An object is placed 96.0 cm from a glass lens (n = 1.52) with one concave surface of radius 22.0 cm and one convex surface of radius 18.5 cm.

(a) Where is the final image?

(b) What is the magnification?

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Textbook Question

(II) A planoconvex lens (Fig. 33–2a) has one flat surface and the other has R = 15.3 cm. This lens is used to view a red and yellow object which is 62.0 cm away from the lens. The index of refraction of the glass is 1.5106 for red light and 1.5226 for yellow light. What are the locations of the red and yellow images formed by the lens?

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Textbook Question

A series of polarizers are each rotated 10° from the previous polarizer. Unpolarized light is incident on this series of polarizers. How many polarizers does the light have to go through before it is 1/6 of its original intensity?

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