/*! 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 73 A concave mirror is to form an i... [FREE SOLUTION] | 91Ó°ÊÓ

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A concave mirror is to form an image of the filament of a headlight lamp on a screen 8.00 \(\mathrm{m}\) from the mirror. The filament is 6.00 \(\mathrm{mm}\) tall, and the image is to be 24.0 \(\mathrm{cm}\) tall. (a) How far in front of the vertex of the mirror should the filament be placed? (b) What should be the radius of curvature of the mirror?

Short Answer

Expert verified
(a) Place the filament 0.20 m from the mirror. (b) The radius of curvature is about 0.410 m.

Step by step solution

01

Understand the Given Information

We are given a concave mirror and need to find two things: (a) the distance of the filament from the mirror, and (b) the radius of curvature of the mirror. The screen is 8.00 m away from the mirror, the image is 24.0 cm tall, and the filament is 6.00 mm tall.
02

Relate Image and Object Distances

Use the magnification formula for a mirror: \( m = \frac{h_i}{h_o} = \frac{-d_i}{d_o} \), where \( h_i = 24.0 \, \mathrm{cm} \) is the image height, \( h_o = 6.00 \, \mathrm{mm} = 0.60 \, \mathrm{cm} \) is the object height, \( d_i = 8.00 \, \mathrm{m} = 800 \, \mathrm{cm} \) is the image distance, and \( d_o \) is the object distance. Set up the equation \( \frac{24.0}{0.60} = \frac{-800}{d_o} \).
03

Solve for Object Distance

Solve the equation from Step 2: \( \frac{24.0}{0.60} = 40 \), thus \( -\frac{800}{d_o} = 40 \). Solving this gives \( d_o = \frac{-800}{40} = -20 \, \mathrm{cm} \). Convert to meters: \( d_o = -0.20 \, \mathrm{m} \).
04

Use Mirror Equation to Find Focal Length

Apply the mirror equation: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). Substituting \( d_o = -0.20 \, \mathrm{m} \) and \( d_i = 8.00 \, \mathrm{m} \) gives \( \frac{1}{f} = \frac{1}{-0.20} + \frac{1}{8} \). Solve for \( f \).
05

Solve for Focal Length

Calculate \( \frac{1}{f} = \frac{1}{-0.20} + \frac{1}{8} = -5 + 0.125 = -4.875 \). Thus, \( f = -\frac{1}{4.875} \approx -0.205 \) meters.
06

Find Radius of Curvature

Use the relation between the focal length and the radius of curvature: \( R = 2f \). Substitute \( f \approx -0.205 \) meters into the equation to find \( R = 2 \times -0.205 \approx -0.410 \) meters.

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Key Concepts

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

Mirror Equation
The mirror equation is fundamental when dealing with concave mirrors. It allows us to link the focal length, image distance, and object distance all in one neat formula, making it easy to solve problems involving mirror reflections. The mirror equation is expressed as:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]Where:- \( f \) is the focal length of the mirror.- \( d_o \) is the object distance.- \( d_i \) is the image distance.In our exercise, you start by substituting known values into the equation. This connection helps us find the missing variable, in this case, the focal length \( f \). You carefully work through each component to understand how the concave mirror bends light to form images at different distances.
Magnification Formula
The magnification formula is a tool to describe how much larger or smaller the image is compared to the actual object. For mirrors, it's denoted by:\[ m = \frac{h_i}{h_o} = \frac{-d_i}{d_o} \]Where:- \( m \) is the magnification of the image.- \( h_i \) is the height of the image.- \( h_o \) is the height of the object.- \( d_i \) is the image distance (negative for concave mirrors if the image is real).- \( d_o \) is the object distance.In the exercise, the problem tells us how tall the image and object are, which allows us to directly use this equation to find the object distance. By rearranging the formula, you solve for \( d_o \), revealing how the mirror's curvature affects the image's size compared to the object's actual dimensions.
Radius of Curvature
Understanding the radius of curvature is key in the design and application of concave mirrors. This radius is the distance from the mirror's surface to its center of curvature. In simpler terms, it indicates how "curved" the mirror is. The relationship between focal length \( f \) and the radius of curvature \( R \) of a spherical mirror is:\[ R = 2f \]This equation shows that the radius of curvature is twice the focal length. By determining \( f \), you can quickly find \( R \), which helps in designing mirrors that need precise focal lengths. In solving the exercise, once the focal length was calculated, calculating the radius of curvature was straightforward using this direct relationship.
Image and Object Distance
Image and object distances are crucial for locating where the image forms in relation to where the object is placed before the mirror. For concave mirrors:- The object distance \( (d_o) \) is usually measured from the object to the mirror.- The image distance \( (d_i) \) indicates where the image appears, and is considered positive if the image is formed on the same side as the object (real image) and negative otherwise.In solving for \( d_o \), you use the mirror equation in relation with the known \( d_i \) — in this case, given in the problem as the distance from the mirror to the screen. Combining this knowledge, you derive where the object (the filament) must be placed to achieve the given image size, illustrating how changing positions affects image formation.

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Most popular questions from this chapter

A camera lens has a focal length of 180.0 \(\mathrm{mm}\) and an aperture diameter of 16.36 \(\mathrm{mm}\) . (a) What is the \(f\) -number of the lens? (b) If the correct exposure of a certain scene is \(\frac{1}{30} \mathrm{s}\) at \(f / 11\) what is the correct exposure at \(f / 2.8 ?\)

An object is placed 18.0 \(\mathrm{cm}\) from a screen. (a) At what two points between object and screen may a converging lens with a 3.00 -cm focal length be placed to obtain an image on the screen? (b) What is the magnification of the image for each position of the lens?

The Lens of the Eye. The crystalline lens of the human eye is a double-convex lens made of material having an index of refraction of 1.44 (although this varies). Its focal length in air is about 8.0 \(\mathrm{mm}\) , which also varies. We shall assume that the radii of curvature of its two surfaces have the same magnitude. (a) Find the radii of curvature of this lens. (b) If an object 16 \(\mathrm{cm}\) tall is placed 30.0 \(\mathrm{cm}\) from the eye lens, where would the lens focus it and how tall would the image be? Is this image real or virtual? Is it erect or inverted? (Note: The results obtained here are not strictly accurate because the lens is embedded in fluids having refractive indexes different from that of air.)

A camera with a 90 -mm-focal-length lens is focused on an object 1.30 \(\mathrm{m}\) from the lens. To refocus on an object 6.50 \(\mathrm{m}\) from the lens, by how much must the distance between the lens and the film be changed? To refocus on the more distant object, is the lens moved toward or away from the film?

Photography. A 35 -mm camera has a standard lens with focal length 50 \(\mathrm{mm}\) and can focus on objects between 45 \(\mathrm{cm}\) and infinity. (a) Is the lens for such a camera a concave or a convex lens? (b) The camera is focused by rotating the lens, which moves it on the camera body and changes its distance from the film. In what range of distances between the lens and the film plane must the lens move to focus properly over the 45 \(\mathrm{cm}\) to infinity range?

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