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 90a

Chapter 31, Problem 90a

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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Verified step by step guidance

Step 1: Identify the key variables in the problem. The diameter of the Sun is \(1.4 \times 10^6\, \text{km}\), the distance to the Sun is \(1.5 \times 10^8\, \text{km}\), and the image diameter on the film is \(15\, \text{mm}\). We are tasked with estimating the focal length of the camera lens (or telescope).

Step 2: Use the thin lens formula for magnification. The magnification \(M\) is given by \(M = \frac{h_i}{h_o}\), where \(h_i\) is the height of the image (15 mm or \(0.015\, \text{m}\)) and \(h_o\) is the height of the object (the Sun's diameter, \(1.4 \times 10^6\, \text{km}\)).

Step 3: Relate the magnification to the focal length. For distant objects, the magnification can also be expressed as \(M = \frac{f}{d_o}\), where \(f\) is the focal length of the lens and \(d_o\) is the distance to the object (\(1.5 \times 10^8\, \text{km}\)).

Step 4: Combine the two expressions for magnification. Set \(\frac{h_i}{h_o} = \frac{f}{d_o}\). Rearrange this equation to solve for the focal length \(f\): \(f = \frac{h_i \cdot d_o}{h_o}\).

Step 5: Substitute the known values into the equation. Use \(h_i = 0.015\, \text{m}\), \(d_o = 1.5 \times 10^8\, \text{km}\), and \(h_o = 1.4 \times 10^6\, \text{km}\). Ensure all units are consistent (e.g., convert kilometers to meters if necessary) before performing the calculation.

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Key Concepts

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

Similar Triangles

The concept of similar triangles is fundamental in optics and geometry, where two triangles are similar if their corresponding angles are equal and their sides are in proportion. In this scenario, the relationship between the size of the Sun's image on the film and its actual size can be analyzed using similar triangles, allowing us to set up a proportion to find the focal length of the lens.

Lens Formula

The lens formula relates the object distance (distance from the lens to the object), image distance (distance from the lens to the image), and the focal length of the lens. It is expressed as 1/f = 1/d_o + 1/d_i, where f is the focal length, d_o is the object distance, and d_i is the image distance. This formula is crucial for calculating the focal length based on the distances involved in the given problem.

Magnification

Magnification in optics refers to the ratio of the height of the image to the height of the object, and it can also be expressed in terms of distances as the ratio of the image distance to the object distance. In this context, understanding magnification helps relate the size of the Sun's image on the film to its actual size, which is essential for determining the focal length of the camera lens.

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Related Practice

Textbook Question

Light from a laser (in air) strikes the exact center of one face of a solid glass cube (n = 1.40) at an angle θ relative to the normal. The refracted beam travels inside the glass until it strikes an adjacent face of the cube. The original angle of incidence θ is such that no light exits the cube where the beam strikes the second face. What is the maximum value θ can have?

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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

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.]

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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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