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Ch 34: Ray Optics
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 34: Ray OpticsProblem 76
Chapter 34, Problem 76

A 2.0-cm-tall object is placed in front of a mirror. A 1.0-cm-tall upright image is formed behind the mirror, 150 cm from the object. What is the focal length of the mirror?

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Identify the type of mirror: Since the image is upright and formed behind the mirror, it is a virtual image, indicating that the mirror is a convex mirror.
Use the magnification formula to find the magnification (M): \( M = \frac{h_i}{h_o} \), where \( h_i \) is the height of the image (1.0 cm) and \( h_o \) is the height of the object (2.0 cm). Substitute the values to calculate \( M \).
Relate magnification to the object distance (\( d_o \)) and image distance (\( d_i \)) using the formula \( M = -\frac{d_i}{d_o} \). Substitute \( M \) and \( d_i = -150 \; \text{cm} \) (negative because the image is virtual) to solve for \( d_o \).
Apply the mirror equation: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length. Substitute the values of \( d_o \) and \( d_i \) to solve for \( f \).
Simplify the equation to isolate \( f \) and calculate its value. Remember that for a convex mirror, the focal length \( f \) will be positive.

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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 solving problems involving mirrors, as it allows us to find the focal length when the object and image distances are known.
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Sign Convention in Optics

In optics, a sign convention is used to determine the signs of distances in mirror equations. For mirrors, the object distance is considered negative if the object is in front of the mirror, while the image distance is positive if the image is formed behind the mirror. Understanding this convention is crucial for correctly applying the mirror formula.
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Upright Image Formation

An upright image is one that maintains the same orientation as the object. In the context of mirrors, an upright image can be formed by a convex mirror or by a concave mirror when the object is placed within the focal length. Recognizing the conditions under which upright images are formed helps in analyzing the characteristics of the mirror involved.
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Related Practice
Textbook Question

The mirror in FIGURE CP34.79 is covered with a piece of glass whose thickness at the center equals the mirror's radius of curvature. A point source of light is outside the glass. How far from the mirror is the image of this source?

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Textbook Question

CALC FIGURE CP34.81 shows a light ray that travels from point A to point B. The ray crosses the boundary at position x, making angles θ1 and θ2 in the two media. Suppose that you did not know Snell's law. You've proven that Snell's law is equivalent to the statement that 'light traveling between two points follows the path that requires the shortest time.' This interesting way of thinking about refraction is called Fermat's principle. Write an expression for the time t it takes the light ray to travel from A to B. Your expression should be in terms of the distances a, b, and w; the variable x; and the indices of refraction n1 and n2.

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Textbook Question

An object is 60 cm from a screen. What are the radii of a symmetric converging plastic lens (i.e., two equally curved surfaces) that will form an image on the screen twice the height of the object?

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Textbook Question

Some electro-optic materials can change their index of refraction in response to an applied voltage. Suppose a plano-convex lens (flat on one side, a 15.0 cm radius of curvature on the other), made from a material whose normal index of refraction is 1.500, is creating an image of an object that is 50.0 cm from the lens. By how much would the index of refraction need to be increased to move the image 5.0 cm closer to the lens?

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Textbook Question

CALC A wildlife photographer with a 200-mm-focal-length telephoto lens on his camera is taking a picture of a rhinoceros that is 100 m away. Suddenly, the rhino starts charging straight toward the photographer at a speed of 5.0 m/s. What is the speed, in μm/s, image of the of the rhinoceros? Is the image moving toward or away from the lens?

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Textbook Question

Consider a lens having index of refraction n₂ and surfaces with radii R₁ and R₂. The lens is immersed in a fluid that has index of refraction n₁. A symmetric converging glass lens (i.e., two equally curved surfaces) has two surfaces with radii of 40 cm. Find the focal length of this lens in air and the focal length of this lens in water.

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