/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 A concave mirror has a focal len... [FREE SOLUTION] | 91Ó°ÊÓ

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A concave mirror has a focal length of \(40.0 \mathrm{cm} .\) Determine the object position for which the resulting image is upright and four times the size of the object.

Short Answer

Expert verified
The object must be positioned 32.0 cm from the mirror.

Step by step solution

01

Understand the Definitions

Two commonly used equations in mirror optics are the mirror equation (\(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\)), where \(d_o\) is the object distance, \(d_i\) is the image distance, and \(f\) is the focal length, and the magnification equation (\(m = -\frac{d_i}{d_o}\)), where \(m\) is the magnification.
02

Identify Known Information and Needed Variable

In this problem, the given information is that \(f = 40.0 \mathrm{cm}\) and \(m = -4\). The negative sign in \(m\) indicates that the image is upright. The variable to be calculated is \(d_o\), the object position.
03

Solve the Mirror Equation

Insert \(m\) into the mirror equation, \(\frac{1}{f} = \frac{1}{d_o} - \frac{1}{4d_o}\). After combining terms, this becomes \(5d_o=4f\). Solving for \(d_o\), we get \(d_o = \frac{4f}{5}\)
04

Substitute Known Values

Replace \(f\) with its given value of \(40.0 \mathrm{cm}\) to find \(d_o = \frac{4 \cdot 40.0 \mathrm{cm}}{5} = 32.0 \mathrm{cm}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mirror Equation
In optics, the mirror equation provides an essential link between an object's position, the image distance, and the mirror's focal length. Specifically for a concave mirror, the equation is represented as \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \(f\) is the focal length, \(d_o\) is the distance from the mirror to the object, and \(d_i\) is the distance from the mirror to the image.

When using the mirror equation, it's important to understand the sign conventions: positive values of \(d_i\) correspond to real images (formed on the same side as the object), while negative values indicate virtual images (formed on the opposite side). Likewise, the focal length \(f\) is positive for concave mirrors and negative for convex mirrors. In our exercise, with a given focal length of 40 cm, we zero in on the object position to find where an upright and larger image can be formed.
Magnification in Optics
Magnification is a measure of how much larger or smaller an image is compared to the object. In the context of mirrors, it is given by the formula \(m = -\frac{d_i}{d_o}\), where \(m\) represents magnification, \(d_o\) is the object distance, and \(d_i\) is the image distance.

A positive magnification implies that the image is upright relative to the object, while a negative magnification indicates an inverted image. If the magnitude of magnification is greater than one, as is the case in our example with \(m = -4\), the image is larger than the object. The negative sign here tells us that the image is upright. By using this magnification concept and linking it to the mirror equation, one can solve for unknown variables, such as object position or image distance, given certain conditions for the image.
Ray Diagrams for Mirrors
Ray diagrams are a graphical method to determine the properties of an image formed by a mirror. To construct a ray diagram for a concave mirror, follow these steps:
  • Draw a ray parallel to the principal axis, reflecting through the focal point.
  • Draw a ray through the center of curvature which reflects back on itself.
  • Draw a ray through the focal point that reflects parallel to the principal axis.

All reflected rays should meet at a point where the image is formed. By analyzing this point's position relative to the mirror, one can infer whether the image is real or virtual, upright or inverted, and the relative size compared to the object. Through ray diagrams, we visually verify the conclusions drawn from the mathematical solutions provided by the mirror equation and the magnification formula.
Optics in Physics
Optics is a branch of physics that studies the behavior of light, including its interactions with matter and the construction of instruments that use or detect it. Optics is divided into several subfields, one of which is geometric optics - the focus of our concave mirror problem. Geometric optics applies principles like reflection and refraction to solve problems concerning the paths of light rays as they encounter surfaces. Concave mirrors are a classical topic within geometric optics, as they have a mirrored surface that caves inward, like the inner surface of a sphere.

Understanding the principles of optics not only helps in solving textbook problems but is also crucial in real-world applications such as designing telescopes, microscopes, cameras, and even in corrective vision devices like glasses and contact lenses. With a firm grasp of these optics principles, one can unravel the mysteries of light behavior and its versatile applications.

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Most popular questions from this chapter

A converging lens has a focal length of \(20.0 \mathrm{cm} .\) Locate the image for object distances of (a) \(40.0 \mathrm{cm},\) (b) \(20.0 \mathrm{cm},\) and (c) \(10.0 \mathrm{cm} .\) For each case, state whether the image is real or virtual and upright or inverted. Find the magnification in each case.

A camera is being used with a correct exposure at \(f / 4\) and a shutter speed of \((1 / 16)\) s. In order to photograph a rapidly moving subject, the shutter speed is changed to \((1 / 128)\) s. Find the new \(f\) -number setting needed to maintain satisfactory exposure.

The distance between an object and its upright image is \(20.0 \mathrm{cm} .\) If the magnification is \(0.500,\) what is the focal length of the lens that is being used to form the image?

A person sees clearly when he wears eyeglasses that have a power of -4.00 adopters and sit \(2.00 \mathrm{cm}\) in front of his eyes. If the person wants to switch to contact lenses, which are placed directly on the eyes, what lens power should be prescribed?

Assume the intensity of sunlight is \(1.00 \mathrm{kW} / \mathrm{m}^{2}\) at a particular location. A highly reflecting concave mirror is to be pointed toward the Sun to produce a power of at least \(350 \mathrm{W}\) at the image. (a) Find the required radius \(R_{a}\) of the circular face area of the mirror. (b) Now suppose the light intensity is to be at least \(120 \mathrm{kW} / \mathrm{m}^{2}\) at the image. Find the required relationship between \(R_{a}\) and the radius of curvature \(R\) of the mirror. The disk of the Sun subtends an angle of \(0.533^{\circ}\) at the Earth.

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