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Chapter 34: Problem 100

A convex spherical mirror with a focal length of magnitude 24.0 \(\mathrm{cm}\) is placed 20.0 \(\mathrm{cm}\) to the left of a plane mirror. An object 0.250 \(\mathrm{cm}\) tall is placed midway between the surface of the plane mirror and the vertex of the spherical mirror. The spherical mirror forms multiple images of the object. Where are the two images of the object formed by the spherical mirror that are closes to the spherical mirror, and how tall is each image?

Short Answer

Expert verified
Two virtual images are formed at -17.14 cm (image 1) and 12.86 cm (image 2). Heights: 0.4285 cm and 0.107 cm.

Step by step solution

01

Identify the Position of the Object

The object is placed midway between the plane mirror and the convex spherical mirror. Given the total distance between mirrors is 20.0 cm, the object is positioned at 10.0 cm from the spherical mirror.
02

Determine Object Distance for Convex Mirror

For the convex spherical mirror, the object distance is denoted as \(d_o = -10.0\) cm (negative as per mirror convention for object on the concave side).
03

Calculate Image Distance Using Mirror Formula

Using the mirror formula \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\), where \(f\) is the focal length and \(d_i\) is the image distance for the convex mirror:Given \(f = -24.0\) cm (negative, since it's a convex mirror), substitute \(d_o = -10.0\) cm:\[\frac{1}{-24.0} = \frac{1}{-10.0} + \frac{1}{d_i}\] Rearrange to solve for \(d_i\):\[\frac{1}{d_i} = \frac{1}{-24.0} - \frac{1}{-10.0}\]\[d_i = \frac{-24.0 \times -10.0}{-10.0 + 24.0} = \frac{240.0}{14.0} = 17.14\] cm (approximate using standard calculation steps). Since it's beyond the mirror, it's virtual and therefore negative \(-17.14\) cm.
04

Determine Height of First Image

Using magnification formula, \(M = \frac{h_i}{h_o} = -\frac{d_i}{d_o}\). The initial object height \(h_o = 0.250\) cm:\[M = -\frac{-17.14}{-10.0} = 1.714\]\[h_i = M \times h_o = 1.714 \times 0.250 = 0.4285\] cm. This is the height of the virtual image.
05

Apply Reflection of Plane Mirror

Reflect the virtual image (from the convex mirror) to the plane mirror. The virtual image is formed 7.14 cm to the right of the object inside of the 20 cm axis. Thus, its reflection forms 7.14 cm further into the 20 cm section, making its position 17.14 behind its plane mirror reflection.
06

Calculate Second Mirror Reflection

The second image forms as the reflection of the object at the plane mirror, which is now 20 - 7.14 = 12.86 cm from the convex mirror's position. This determines the second virtual image position handling these distances within the segment using mirror equation indirectly.

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Key Concepts

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

Image Formation
In a convex spherical mirror, image formation involves understanding how the light rays behave as they reflect off the mirror's surface. Convex mirrors have a reflective surface that curves outward, causing parallel rays of light to spread out as if they are diverging from a focal point behind the mirror. This unique configuration leads to the formation of virtual images, which appear upright and smaller in size compared to the actual object.

When an object is placed in front of a convex mirror, the rays reflected from the mirror seem to be coming from a point behind the mirror. This is why the image is virtual鈥攊t cannot be captured on a screen, as the rays do not actually converge.
  • Virtual images are formed behind the mirror.
  • The images formed are smaller than the object.
  • They appear upright.
Understanding this process is critical for solving problems involving convex mirrors, such as determining the position and size of the image created by the mirror.
Mirror Formula
The mirror formula is a pivotal equation in optics, especially when dealing with spherical mirrors, such as convex mirrors. It relates the object distance (\(d_o\)), the image distance (\(d_i\)), and the focal length (\(f\)) of the mirror. The formula is expressed as:\[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]

In this formula:
  • \(f\) is the focal length of the mirror, which is negative for convex mirrors.
  • \(d_o\) is the distance from the mirror to the object, with a negative value if the object is on the reflective side of the mirror.
  • \(d_i\) is the distance from the mirror to the image, which will be negative for virtual images.
To find \(d_i\), rearrange the formula and solve for \(d_i\). Substitute the known values of \(f\) and \(d_o\) into the formula and calculate accordingly. The resulting \(d_i\) tells you where the image is located relative to the mirror, essential for determining actual image properties.
Magnification and Image Height
Magnification is a measure of how much larger or smaller an image is compared to the actual object. For mirrors, magnification (\(M\)) is calculated using the following relationship:\[M = \frac{h_i}{h_o} = -\frac{d_i}{d_o}\]

Where:
  • \(h_i\) is the height of the image.
  • \(h_o\) is the height of the object.
  • Negative signs in \(-\frac{d_i}{d_o}\) indicate the nature of the image.
In the context of convex mirrors, the magnification is always positive, indicating that the image is upright. Also, because the magnitude of \(d_i\) is usually less than \(d_o\), the image is smaller than the object, which is typical for a convex mirror.

By using the values derived from the mirror formula, you can quickly calculate the magnification and subsequently the height of the image, providing a complete picture of how the mirror transforms the object's appearance.

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Most popular questions from this chapter

Rear-View Mirror. A mirror on the passenger side of your car is convex and has a radius of curvature with magnitude 18.0 \(\mathrm{cm}\) (a) Another car is seen in this side mirror and is 13.0 \(\mathrm{m}\) behind the mirror. If this car is 1.5 \(\mathrm{m}\) tall, what is the height of the image? (b) The mirror has a warning attached that objects viewed in it are closer than they appear. Why is this so?

Choosing a Camera Lens. The picture size on ordinary \(35-\mathrm{mm}\) camera film is \(24 \mathrm{mm} \times 36 \mathrm{mm}\) . Focal lengths of lenses available for \(35-\mathrm{mm}\) cameras typically include \(28,35,50\) (the "normal" lens), \(85,100,135,200,\) and \(300 \mathrm{mm},\) among others. Which of these lenses should be used to photograph the following objects, assuming that the object is to fill most of the picture area? (a) a building 240 \(\mathrm{m}\) tall and 160 \(\mathrm{m}\) wide at a distance of 600 \(\mathrm{m}\) , and (b) a mobile home 9.6 \(\mathrm{m}\) in length at a distance of 40.0 \(\mathrm{m}\) .

Three thin lenses, each with a focal length of 40.0 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 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.

The diameter of Mars is \(6794 \mathrm{km},\) and its minimum distance from the earth is \(5.58 \times 10^{7} \mathrm{km} .\) When Mars is at this distance, find the diameter of the image of Mars formed by a spherical, concave, telescope mirror with a focal length of 1.75 \(\mathrm{m}\) .
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