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Chapter 34: Problem 68

Where must you place an object in front of a concave mirror with radius \(R\) so that the image is erect and 2\(\frac{1}{2}\) times the size of the object? Where is the image?

Short Answer

Expert verified
Place the object at \(\frac{7R}{10}\) from the mirror. The image is at \(\frac{7R}{4}\).

Step by step solution

01

Understanding Mirror Formula and Magnification

For a concave mirror, the mirror formula is \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \), where \(f\) is the focal length, \(v\) is the image distance, and \(u\) is the object distance. The magnification \(m\) is given by \( m = -\frac{v}{u} \). An erect image implies that \(m\) is positive, so \(m = \frac{v}{u} \). We are given that \(m = 2.5\).
02

Determine the Focal Length

The radius of curvature \(R\) of a mirror is twice the focal length \(f\), so \( f = \frac{R}{2} \).
03

Set Up the Given Magnification Condition

We know that \(m = 2.5\), hence \( \frac{v}{u} = 2.5 \). This means that \( v = 2.5u \).
04

Substitute and Solve the Mirror Equation

Substitute \(v = 2.5u\) into the mirror formula: \( \frac{1}{f} = \frac{1}{2.5u} + \frac{1}{u} \). Simplify to find \(u\): \( \frac{1}{f} = \frac{3.5}{2.5u} \), leading to \( u = \frac{3.5f}{2.5} \).
05

Express Object Distance in Terms of Radius of Curvature

Replace \(f\) with \(\frac{R}{2}\) in the found expression for \(u\): \( u = \frac{3.5}{2.5} \cdot \frac{R}{2} = \frac{7R}{10} \). Thus, the object is placed at a distance of \(\frac{7R}{10}\) from the mirror.
06

Calculate Image Distance

Substitute \( u = \frac{7R}{10} \) into \( v = 2.5u \) to find the image distance: \( v = 2.5 \times \frac{7R}{10} = \frac{7R}{4} \). Thus, the image is located at a distance of \(\frac{7R}{4}\) from the mirror.

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Key Concepts

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

Mirror Formula
The mirror formula is essential for understanding how images are formed by mirrors, particularly for concave mirrors. This formula is expressed as: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] where:
  • \( f \) is the focal length,
  • \( v \) is the image distance,
  • \( u \) is the object distance.
A concave mirror allows parallel rays of light to converge, forming images. By applying the mirror formula, one can precisely calculate where an image will appear in relation to the object and the mirror. Understanding this relationship helps predict the nature and position of the image formed.
Magnification
Magnification describes how much larger or smaller an image appears compared to the actual object. It is given by: \[ m = \frac{v}{u} \] For lenses and mirrors:
  • If \( m > 1 \), the image is magnified larger than the object.
  • If \( m < 1 \), the image is smaller than the object.
  • If \( m = 1 \), the image size is equal to the object size.
For erect images formed by concave mirrors, magnification is positive (\( m = \frac{v}{u} \)). This translates to an upright image. In our exercise, a magnification of 2.5 indicates the image is not only upright but also 2.5 times larger than the object.
Image Distance
Image distance refers to the distance between the image and the mirror along the principal axis. This is represented by \( v \) in the mirror formula. Knowing the image distance is crucial for determining where the final image will form relative to the mirror and the object. In the given exercise, using the relationship \( v = 2.5u \), it was determined that the image distance \( v \) is \( \frac{7R}{4} \), which tells us precisely where to look for the image.
Focal Length
The focal length is a vital characteristic of mirrors that determines how they converge or diverge light rays. It is half the radius of curvature, so for a mirror with radius \( R \), the focal length \( f \) is: \[ f = \frac{R}{2} \] A concave mirror focuses light to a point called the focal point, which is located at this distance from the mirror itself. Understanding the focal length is essential when calculating how objects and images relate spatially. In the exercise, this value helped find the correct placements of both object and image along the principal axis.

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Most popular questions from this chapter

Curvature of the Cornea. In a simplified model of the human eye, the aqueous and vitreous humors and the lens all have a refractive index of 1.40 , and all the refraction occurs at the cornea, whose vertex is 2.60 \(\mathrm{cm}\) from the retina. What should be the radius of curvature of the cornea such that the image of an object 40.0 \(\mathrm{cm}\) from the cornea's vertex is focused on the retina?

A transparent rod 30.0 \(\mathrm{cm}\) long is cut flat at one end and rounded to a hemispherical surface of radius 10.0 \(\mathrm{cm}\) at the other end. A small object is embedded within the rod along its axis and halfway between its ends, 15.0 \(\mathrm{cm}\) from the flat end and 15.0 \(\mathrm{cm}\) from the vertex of the curved end. When viewed from the flat end of the rod, the apparent depth of the object is 9.50 \(\mathrm{cm}\) from the flat end. What is its apparent depth when viewed from the curved end?

A thin lens with a focal length of 6.00 \(\mathrm{cm}\) is used as a simple magnifier. (a) What angular magnification is obtainable with the lens if the object is at the focal point? (b) When an object is examined through the lens, how close can it be brought to the lens? Assume that the image viewed by the eye is at the near point, 25.0 \(\mathrm{cm}\) from the eye, and that the lens is very close to the eye.

Three thin lenses, each with a focal length of \(40.0 \mathrm{cm},\) are aligned on a common axis; adjacent lenses are separated by 52.0 \(\mathrm{cm} .\) Find the position of the image of a small object on the axis, 80.0 \(\mathrm{cm}\) to the left of the first lens.

A pencil that is 9.0 \(\mathrm{cm}\) long is held perpendicular to the surface of a plane mirror with the tip of the pencil lead 12.0 \(\mathrm{cm}\) from the mirror surface and the end of the eraser 21.0 \(\mathrm{cm}\) from the mirror surface. What is the length of the image of the pencil that is formed by the mirror? Which end of the image is closer to the mirror surface: the tip of the lead or the end of the eraser?
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