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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 87
Chapter 31, Problem 87

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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Step 1: Identify the type of mirror and the nature of the image. Since the image is real, inverted, and smaller than the object, the mirror must be a concave mirror. The magnification (M) is given as -1/2 (negative because the image is inverted).
Step 2: Use the magnification formula to find the image distance (v). The formula is: M = -v/u, where u is the object distance (21 cm). Substitute M = -1/2 and u = 21 cm to solve for v.
Step 3: Use the mirror equation to relate the object distance, image distance, and focal length. The mirror equation is: 1f = 1u + 1v, where f is the focal length. Substitute the values of u and v to solve for f.
Step 4: Relate the focal length to the radius of curvature of the mirror. The radius of curvature (R) is given by the formula: R = 2f. Use the value of f obtained in the previous step to calculate R.
Step 5: Summarize the results. The image distance (v) and the radius of curvature (R) are now determined based on the calculations from the previous steps.

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Key Concepts

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

Mirror Formula

The mirror formula relates the object distance (u), image distance (v), and focal length (f) of a mirror. It is expressed as 1/f = 1/v + 1/u. This formula is essential for determining the position of the image formed by a mirror based on the object's position.
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Magnification

Magnification (m) is the ratio of the height of the image (h') to the height of the object (h), given by m = h'/h = -v/u. In this case, the image is half the height of the object and inverted, indicating a magnification of -0.5, which helps in calculating the image distance.
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Radius of Curvature

The radius of curvature (R) of a mirror is related to its focal length (f) by the equation R = 2f. This concept is crucial for understanding the geometry of the mirror and how it affects the formation of images, particularly in determining the curvature based on the focal length derived from the image and object distances.
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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

A triangular prism made of crown glass (n = 1.52) with base angles of 26.0° is surrounded by air. If parallel rays are incident normally on its base as shown in Fig. 32–66, what is the angle Φ between the two emerging rays?

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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

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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Textbook Question

When light passes through a prism, the angle that the refracted ray makes relative to the incident ray is called the deviation angle δ, Fig. 32–64. Show that this angle is a minimum when the ray passes through the prism symmetrically, perpendicular to the bisector of the apex angle Φ, and show that the minimum deviation angle, δm, is related to the prism’s index of refraction n by


n=sin12(ϕ+δm)sinϕ/2.n = \(\frac{\sin \frac{1}{2}\)(\(\phi\) + \(\delta\)_m)}{\(\sin\) \(\phi\)/2}.


[Hint: For θ in radians, (d/dθ)(sin−1θ)=1/1−θ2(d/d\(\theta\)) (\(\sin\)^{-1}\(\theta\)) = 1/\(\sqrt{1 - \theta^2}\).]

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