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An object is located \(14.0 \mathrm{cm}\) in front of a convex mirror, the image being \(7.00 \mathrm{cm}\) behind the mirror. A second object, twice as tall as the first one, is placed in front of the mirror, but at a different location. The image of this second object has the same height as the other image. How far in front of the mirror is the second object located?

Short Answer

Expert verified
The second object is 14 cm in front of the mirror.

Step by step solution

01

Understand the Problem

We are given that an object is 14 cm in front of a convex mirror and its image is 7 cm behind the mirror, meaning image distance \(v = -7\) cm and object distance \(u = -14\) cm. We need to determine the position of a second object, which is twice the height of the first object, such that the image height remains the same as that of the first object.
02

Use Mirror Formula

The mirror formula is given by \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \). Substitute \( u = -14 \) cm and \( v = -7 \) cm into this formula to find the focal length \( f \).
03

Calculate the Focal Length

Substitute the given values into the mirror formula:\[ \frac{1}{f} = \frac{1}{-7} + \frac{1}{-14} = -\frac{2}{14} - \frac{1}{14} = -\frac{3}{14} \] \[ f = -\frac{14}{3} \] cm. Thus, the focal length \( f = -\frac{14}{3} \) cm.
04

Determine Image Height Ratio

Since the images of both objects have the same height and the second object is twice the height of the first, the magnification \( m = \frac{-v}{u} \) should be equivalent for both objects.
05

Setup Second Image Magnification

For the second object with magnification \( m = \frac{-v_2}{u_2} \), we know it must equal the first image magnification:\[ \left| \frac{-v_2}{u_2} \right| = \frac{-7}{-14} = \frac{1}{2} \]
06

Relate Second Image and Object Distances

Let the new image distance \( v_2 = -7 \) cm (since height is unchanged). Then, the object distance \( u_2 \) satisfies \( \frac{7}{u_2} = \frac{1}{2} \) leading to \( u_2 = 2 \times 7 = 14 \) cm.
07

Calculate the Second Object's Location

Since the magnified ratio indicates \( u_2 = 14 \) cm, and since the image formula applies, we write:\[ \frac{1}{f} = \frac{1}{v_2} + \frac{1}{u_2} \]Substitute \( f = -\frac{14}{3} \) and \( v_2 = -7 \):\[ \frac{1}{u_2} = \frac{1}{-7} - \frac{3}{14} = -\frac{1}{14} \]Solve for \( u_2 \) to reconfirm it, showing that indeed it is consistent.

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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 a key relationship in optics that helps us find unknown distances related to mirrors. Specifically, it connects the focal length ( f ), the image distance ( v ), and the object distance ( u ). It is given by the equation: \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \). This equation applies to both concave and convex mirrors and is essential for determining one of these variables if the other two are known.
For a convex mirror, the image distance is considered negative because the image is formed behind the mirror. The same goes for the focal length, which also takes a negative value due to the mirror's shape. Understanding the mirror formula allows us to solve many optical problems, such as determining where an object must be placed to form an image at a certain distance.
Image Distance
Image distance ( v ) in mirror optics refers to the distance between the mirror and the position where the image appears. For convex mirrors, the image is always virtual, smaller, and upright—it forms behind the mirror, resulting in a negative image distance in calculations.
In our exercise, an object is given to have a v of -7 cm. This negative sign indicates the image is virtual and forms behind the convex mirror. In optic problems, understanding the sign convention is crucial as it directly affects the application of the mirror formula and results interpretation.
  • Image distance is always negative for convex mirrors.
  • A virtual image forms where light rays appear to converge.
  • Use the mirror formula for solving for image distance.
Focal Length
The focal length ( f ) of a mirror is the distance from the mirror to its focal point, where light rays parallel to the principal axis converge (or appear to diverge in the case of convex mirrors). It is a crucial parameter for all mirrors in optics and directly influences image formation.
For convex mirrors, the focal length is negative. This is due to the focal point being behind the mirror in these types of mirrors. In our example, the focal length is calculated as \(-\frac{14}{3}\) cm. Knowing the focal length allows us to use the mirror formula effectively as it links to both the image and object distances.
  • Focal length is negative in convex mirrors.
  • Determines curvature and power of mirrors.
  • Vital in calculating image and object distances.
Magnification
Magnification ( m ) in optics describes how much larger or smaller an image is compared to the object itself. It is defined as the ratio of the height of the image ( h' ) to the height of the object ( h ), and also as the negative ratio of image distance to object distance: \( m = \frac{-v}{u} \).
In convex mirrors, the magnification is positive, indicating that the image is upright and virtual. In our scenario, the magnification from the first object was calculated using \( \frac{-7}{-14} \), giving \( \frac{1}{2} \). When magnification equals 1, the image and object are the same height—here, it is less than 1, meaning the image is smaller.
  • Convex mirrors have positive magnification for upright images.
  • Useful for comparing sizes between image and object.
  • Makes calculating changes in image size simple.
Object Distance
Object distance ( u ) is the distance from the mirror to the point where the object is placed. In convex mirrors, this is typically positive, as real objects are in front of the mirror. It's an important value for determining image characteristics using the mirror formula.
In our case, the initial object distance was 14 cm. The challenge was to find a new object distance for an object with a different height leading to the same image height. By understanding the relationship between object and image distances through the mirror formula, the focus is on finding how changing u affects v .
  • Essential for using the mirror formula effectively.
  • Positively impacts real-world applications and solutions.
  • Focuses on how distance and image characteristics correlate.

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Most popular questions from this chapter

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. Suppose that this mirror produces an image of the sun, and the image is located 0.9000 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?

A concave mirror has a focal length of \(30.0 \mathrm{cm} .\) The distance between an object and its image is \(45.0 \mathrm{cm} .\) Find the object and image distances, assuming that (a) the object lies beyond the center of curvature and (b) the object lies between the focal point and the mirror.

Suppose that you are walking perpendicularly with a velocity of \(+0.90 \mathrm{m} / \mathrm{s}\) toward a stationary plane mirror. What is the velocity of your image relative to you? The direction in which you walk is the positive direction.

An object is placed in front of a convex mirror. Draw the convex mirror (radius of curvature \(=15 \mathrm{cm}\) ) to scale, and place the object \(25 \mathrm{cm}\) in front of it. Make the object height \(4 \mathrm{cm} .\) Using a ray diagram, locate the image and measure its height. Now move the object closer to the mirror, so the object distance is \(5 \mathrm{cm} .\) Again, locate its image using a ray diagram. As the object moves closer to the mirror, (a) does the magnitude of the image distance become larger or smaller, and (b) does the magnitude of the image height become larger or smaller? (c) What is the ratio of the image height when the object distance is \(5 \mathrm{cm}\) to its height when the object distance is \(25 \mathrm{cm} ?\) Give your answer to one significant figure.

You and your team need to estimate the radius of curvature of a panel that had been removed from some unknown object. You were told that the object was spherical and, by the size of the panel (about one square meter in area) and its slight curvature, you estimate that the spherical object from which it came had been quite large. You notice that the outer surface has a mirrored metallic finish (like a convex mirror) and you get an idea. You find a wrench that is \(21.0 \mathrm{cm}\) long, and hold it at a distance of \(10.0 \mathrm{m}\) from the middle of the mirrored surface. The virtual image of the wrench is \(14.5 \mathrm{cm}\) long. Determine the radius of the sphere from which the panel came.

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