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

Giancoli Douglas 5th edition

Ch. 32 - Light: Reflection and Refraction

Problem 91

Chapter 31, Problem 91

A coin lies at the bottom of a 0.95-m-deep pool. If a viewer sees it at a 45° angle, where is the image of the coin, relative to the coin? [Hint: The image is found by tracing back to the intersection of two rays.]

Verified step by step guidance

Identify the given values: depth of the pool (0.95 m), angle of incidence (45°), and the refractive index of water (approximately 1.33).

Understand that light rays from the coin travel upwards and bend away from the normal as they exit the water due to refraction. This bending makes the coin appear shallower than it actually is.

Use Snell's Law, which is given by \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \), where \( n_1 \) and \( n_2 \) are the refractive indices of water and air respectively, and \( \theta_1 \) and \( \theta_2 \) are the angles of incidence and refraction.

Calculate the angle of refraction using Snell's Law. Since the angle of incidence inside the water is 45°, and knowing the refractive indices, you can find \( \theta_2 \).

Determine the apparent depth using the formula \( d' = d \cdot \frac{\sin(\theta_2)}{\sin(\theta_1)} \), where \( d \) is the actual depth and \( d' \) is the apparent depth. This will give you the position of the image relative to the actual coin.

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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 air. This phenomenon occurs because light travels at different speeds in different materials. The change in speed causes the light rays to change direction, which is crucial for understanding how objects appear when viewed through a medium like water.

Snell's Law

Snell's Law describes the relationship between the angles of incidence and refraction when light passes through different media. It is mathematically expressed as n1 * sin(θ1) = n2 * sin(θ2), where n represents the refractive indices of the two media. This law helps predict how much the light will bend, which is essential for determining the position of the image of the coin.

Ray Diagrams

Ray diagrams are visual representations used to illustrate how light travels and interacts with objects. In this context, tracing back the paths of light rays from the coin to the viewer helps locate the image of the coin. By drawing the incident and refracted rays, one can find the point where they appear to converge, indicating the position of the image relative to the actual object.

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Related Practice

Textbook Question

Suppose Fig. 32–37 shows a cylindrical rod whose end has a radius of curvature R = 2.0 cm, and the rod is immersed in water with index of refraction of 1.33. The rod has index of refraction 1.49. Find the location and height of the image of an object 2.0 mm high located 23 cm away from the rod.

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Textbook Question

The paint used on highway signs often contains small transparent spheres which provide nighttime illumination of the sign’s lettering by retro-reflecting vehicle headlight beams. Consider a light ray from air incident on one such sphere of radius r and index of refraction n. Let θ be its incident angle, and let the ray follow the path shown in Fig. 32–70, so that the ray exits the sphere in the direction exactly antiparallel to its incoming direction. Considering only rays for which sin θ can be approximated as θ, determine the required value for n.

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Textbook Question

You hold a small flat mirror 0.50 m in front of you and can see your reflection twice in that mirror because there is a full-length mirror 1.0 m behind you (Fig. 32–71). Determine the distance of each image from you.

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Textbook Question

An object is placed 21 cm from a certain mirror. The image is half the height of the object, inverted, and real. How far is the image from the mirror, and what is the radius of curvature of the mirror?

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Textbook Question

Figure 33–51 was taken from the NIST Laboratory (National Institute of Standards and Technology) in Boulder, CO, 2.0 km from the hiker in the photo. The Sun’s image was 15 mm across on the film. Estimate the focal length of the camera lens (actually a telescope). The Sun has diameter 1.4 x 10⁶ km, and it is 1.5 x 10⁸ km away.

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