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Consult Interactive Solution \(\underline{25.25}\) at for insight into this problem. An object is placed in front of a convex mirror, and the size of the image is one-fourth that of the object. What is the ratio \(d_{0} / f\) of the object distance to the focal length of the mirror?

Short Answer

Expert verified
The ratio \(d_0 / f = -3\).

Step by step solution

01

Understand the Problem

The problem states that an object is in front of a convex mirror and its image is one-fourth the size of the object. We are asked to find the ratio of the object distance \(d_0\) to the focal length \(f\) of the mirror.
02

Use the Mirror Equation

The mirror equation for mirrors is given by \(\frac{1}{f} = \frac{1}{d_0} + \frac{1}{d_i}\), where \(f\) is the focal length, \(d_0\) is the object distance, and \(d_i\) is the image distance.
03

Define the Magnification

The magnification \(m\) for a mirror is defined as \(m = \frac{h_i}{h_0} = -\frac{d_i}{d_0}\), where \(h_i\) and \(h_0\) are the image and object heights respectively. Given that the image is one-fourth the size of the object, \(m = \frac{1}{4}\).
04

Relate Magnification to Image Distance

Since \(m = -\frac{d_i}{d_0} = \frac{1}{4}\), we have \(-\frac{d_i}{d_0} = \frac{1}{4}\). This implies that \(d_i = -\frac{d_0}{4}\). The negative sign indicates that the image formed by a convex mirror is virtual.
05

Substitute Image Distance into Mirror Equation

Substitute \(d_i = -\frac{d_0}{4}\) into the mirror equation: \[\frac{1}{f} = \frac{1}{d_0} + \frac{1}{-d_0/4} = \frac{1}{d_0} - \frac{4}{d_0} = -\frac{3}{d_0}\].
06

Solve for \(d_0 / f\)

We rearrange the equation \(-\frac{3}{d_0} = \frac{1}{f}\) to find \(d_0 / f\). \[d_0 = -3f\], hence \[\frac{d_0}{f} = -3\].
07

Conclusion

The ratio \(\frac{d_0}{f}\) is \(-3\), indicating that the object is placed three times the focal length from the mirror in the negative direction of the chosen coordinate system due to the use of convex mirrors.

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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 a fundamental concept in optics used to relate the object distance, image distance, and focal length of a mirror. For any mirror, this equation is expressed as:\[\frac{1}{f} = \frac{1}{d_0} + \frac{1}{d_i}\]Where:
  • \(f\) is the focal length of the mirror
  • \(d_0\) is the distance from the object to the mirror
  • \(d_i\) is the distance from the image to the mirror
Convex mirrors create images that differ from those created by concave mirrors due to their divergent reflective surfaces.
This leads to images that are always virtual, upright, and smaller, as suggested by the negative image distance calculated in prior equations. Substituting known values like image distance \(-\frac{d_0}{4}\), provides insight into specific problems featuring convex mirrors.
Magnification
Magnification in optics describes how much larger or smaller an image is compared to the object itself. Its formula for mirrors is:\[ m = \frac{h_i}{h_0} = -\frac{d_i}{d_0} \]Where:
  • \(m\) is the magnification factor
  • \(h_i\) and \(h_0\) are the height of the image and object respectively
  • \(d_i\) and \(d_0\) are the image and object distances from the mirror
In convex mirrors, the resulting images are virtual and upright, producing a negative magnification factor, exemplifying the image's reduced size.
In the given scenario, the image is one-fourth the size of the object, yielding a magnification \(m\) of \(\frac{1}{4}\). This value indicates how considerably smaller and virtual the reflected images become.
Focal Length
The focal length \(f\) of a mirror is a crucial optical property that indicates the distance from the mirror where convergent rays meet, or appear to meet after reflection. In convex mirrors, the focal length is negative, meaning it appears, virtually, behind the mirror.
This intrinsic property of convex mirrors results in rays spreading after reflection, portraying images that are reduced in size.
The understanding of this distance \(f\) becomes even more important when applying the mirror equation, assisting in determining the positioning of objects like in the problem scenario. The negative ratio \(d_0/f\) derived from solving the mirror equation reflects the nature of image formation in convex mirrors.
Virtual Image
A virtual image is one that cannot be projected onto a screen, and is instead located in the perceived or imaginative view where light does not actually converge. Convex mirrors are known for generating such virtual images consistently because of their outward bow.
Virtual images:
  • Are upright and smaller in comparison to the object
  • Appear behind the mirror as the light rays diverge
In mathematics, this is mirrored by a negative image distance which symbolizes its virtual and optical nature.
The task's depiction of a one-fourth image size is an exemplary representation of such virtual imaging, helping consolidate the concept for learners diving into optics fundamentals.

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Most popular questions from this chapter

A spherical mirror is polished on both sides. When the convex side is used as a mirror, the magnification is \(+1 / 4\). What is the magnification when the concave side is used as a mirror, the object remaining the same distance from the mirror?

When viewed in a spherical mirror, the image of a setting sun is a virtual image. The image lies \(12.0 \mathrm{~cm}\) behind the mirror. (a) Is the mirror concave or convex? Why? (b) What is the radius of curvature of the mirror?

A tall tree is growing across a river from you. You would like to know the distance between yourself and the tree, as well as its height, but are unable to make the measurements directly. However, by using a mirror to form an image of the tree, and then measuring the image distance and the image height, you can calculate the distance to the tree, as well as its height. (a) What kind of mirror, concave or convex, must you use? Why? (b) You will need to know the focal length of the mirror. The sun is shining. You aim the mirror at the sun and form an image of it. How is the image distance of the sun related to the focal length of the mirror? (c) Having measured the image distance \(d_{\mathrm{i}}\) and the image height \(h_{\mathrm{i}}\) of the tree, as well as the image distance of the sun, describe how you would use these numbers to determine the distance and height of the tree. Problem A mirror produces an image of the sun, and the image is located \(0.9000 \mathrm{~m}\) from the mirror. The same mirror is then used to produce an image of the tree. The image of the tree is \(0.9100 \mathrm{~m}\) from the mirror. (a) How far away is the tree? (b) The image height of the tree has a magnitude of \(0.12 \mathrm{~m}\). How tall is the tree?

An object that is \(25 \mathrm{~cm}\) in front of a convex mirror has an image located \(17 \mathrm{~cm}\) behind the mirror. How far behind the mirror is the image located when the object is \(19 \mathrm{~cm}\) in front of the mirror?

A small mirror is attached to a vertical wall, and it hangs a distance \(y\) above the floor. A ray of sunlight strikes the mirror, and the reflected ray forms a spot on the floor. (a) From a knowledge of \(y\) and the horizontal distance \(x\) from the base of the wall to the spot, describe how one can determine the angle of incidence of the ray striking the mirror. If it is morning and the mirror is facing due east, would (b) the angle of incidence and (c) the distance \(x\) increase or decrease in time? Why? Problem Suppose the mirror is \(1.80 \mathrm{~m}\) above the floor. The reflected ray of sunlight strikes the floor at a distance of \(3.86 \mathrm{~m}\) from the base of the wall. Later in the morning, the ray is observed to strike the floor at a distance of \(1.26 \mathrm{~m}\) from the wall. The earth rotates at a rate of \(15.0^{\circ}\) per hour. How much time (in hours) has elapsed between the two observations?

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