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Chapter 34: Problem 9

A coin is placed next to the convex side of a thin spherical glass shell having a radius of curvature of \(18.0 \mathrm{~cm} .\) Reflection from the surface of the shell forms an image of the \(1.5-\mathrm{cm}\) -tall coin that is \(6.00 \mathrm{~cm}\) behind the glass shell. Where is the coin located? Determine the size, orientation, and nature (real or virtual) of the image.

Short Answer

Expert verified
The coin is located 4.5 cm in front of the mirror and the image produced is virtual, upright, and 2.0 cm in size.

Step by step solution

01

Use the given values and mirror equation

The mirror equation is given as \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\), where \(f\) is the focal length, \(d_o\) is the object distance and \(d_i\) is the image distance. We are given that the image is 6.0 cm behind the mirror, which is negative for a spherical mirror; therefore, \(d_i = -6.0 \, cm\). For a spherical mirror, the focal length is half the radius of curvature, so the focal length \(f = \frac{-18 \, cm}{2} = -9 \, cm\). Using this value in the mirror equation, we can calculate the object distance \(d_o\).
02

Calculate object distance

Plugging the values of \(f\) and \(d_i\) into the mirror equation, we get \(\frac{1}{-9} = \frac{1}{d_o} - \frac{1}{6}\). Solving this equation for \(d_o\), we get \(d_o = -4.5 \, cm\). The negative sign indicates that the coin is placed in front of the mirror.
03

Determine the orientation and nature of the image

The magnification equation is given by \(m = \frac{-d_i}{d_o} = \frac{h_i}{h_o}\), where \(h_i\) is the height of the image and \(h_o\) is the height of the object. We are given \(h_o = 1.5 \, cm\) and we calculated that \(d_i = -6.0 \, cm\) and \(d_o = -4.5 \, cm\). Plugging these values into the magnification equation, we get \(m = \frac{-(-6)}{-4.5} = 1.33\). Therefore, the height of the image is given by \(h_i = m \cdot h_o = 1.33 \times 1.5 = 2.0 \,cm\). This means the image is upright and larger than the coin, and because the image is behind the mirror the image is virtual.

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Key Concepts

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

Understanding Spherical Mirrors
Spherical mirrors are mirrors with surfaces that are part of a sphere. These can be either concave (curving inward) or convex (curving outward).
These mirrors have unique properties:
  • Concave Mirror: Curves inward, like the inside of a bowl. These are often used to focus light and can form real or virtual images depending on the object's position.
  • Convex Mirror: Curves outward, like the back of a spoon. Convex mirrors always form virtual images that are smaller than the object.
The mirror's shape and the distance between the object and the mirror determine the characteristics of the image formed. Key to understanding these mirrors is the radius of curvature, which is the radius of the sphere from which the mirror is a part. The radius determines the focal length, calculated as half of the radius of curvature. This focal length is essential in predicting where an image will be formed and what it will look like.
Image Formation with Mirrors
Image formation in mirrors involves determining the size, orientation, and type of image that results when an object is placed in front of a mirror.
  • Real Image: Real images are formed when light actually converges at a point. These images can be projected onto a screen. They are inverted relative to the object.
  • Virtual Image: Virtual images occur when the light diverges, only appearing to converge. These cannot be projected onto a screen and are upright relative to the object.
For spherical mirrors:
  • Concave mirrors can produce both real and virtual images, depending on the object distance.
  • Convex mirrors always produce virtual images since they diverge light.
The nature of the image also includes its orientation (upright or inverted) and its size. The size is determined by magnification, defined by the ratio of the image's height to the object's height.
Explaining the Mirror Equation
The mirror equation helps us find the relationship between object distance, image distance, and focal length. The equation is:\[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]Here, \(f\) is the focal length, \(d_o\) is the object distance from the mirror, and \(d_i\) is the image distance from the mirror.
The equation helps us in several ways:
  • By rearranging, we can find any of the three variables if the other two are known.
  • The sign convention is vital:
    • Focal length \(f\) is positive for concave mirrors and negative for convex mirrors.
    • Image distance \(d_i\) is positive for real images and negative for virtual images.
    • Object distance \(d_o\) is always positive, indicating a real object in front of the mirror.
This equation is a powerful tool to predict how and where images will form with spherical mirrors. Understanding and applying this equation allows us to solve complex problems involving image formation, like where a coin's reflection appears when placed next to a spherical glass shell.

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Most popular questions from this chapter

(a) You want to use a lens with a focal length of \(35.0 \mathrm{~cm}\) to produce a real image of an object, with the height of the image twice the height of the object. What kind of lens do you need, and where should the object be placed? (b) Suppose you want a virtual image of the same object, with the same magnification-what kind of lens do you need, and where should the object be placed?

A telescope is constructed from two lenses with focal lengths of \(95.0 \mathrm{~cm}\) and \(15.0 \mathrm{~cm},\) the \(95.0 \mathrm{~cm}\) lens being used as the objective. Both the object being viewed and the final image are at infinity. (a) Find the angular magnification for the telescope. (b) Find the height of the image formed by the objective of a building \(60.0 \mathrm{~m}\) tall, \(3.00 \mathrm{~km}\) away. (c) What is the angular size of the final image as viewed by an eye very close to the eyepiece?

An object with height \(4.00 \mathrm{~mm}\) is placed \(28.0 \mathrm{~cm}\) to the left of a converging lens that has focal length \(8.40 \mathrm{~cm} .\) A second lens is placed \(8.00 \mathrm{~cm}\) to the right of the converging lens. (a) What is the focal length of the second lens if the final image is inverted relative to the \(4.00-\mathrm{mm}\) -tall object and has height \(5.60 \mathrm{~mm} ?\) (b) What is the distance between the original object and the final image?

A photographic slide is to the left of a lens. The lens projects an image of the slide onto a wall \(6.00 \mathrm{~m}\) to the right of the slide. The image is 80.0 times the size of the slide. (a) How far is the slide from the lens? (b) Is the image erect or inverted? (c) What is the focal length of the lens? (d) Is the lens converging or diverging?

A converging lens with a focal length of \(12.0 \mathrm{~cm}\) forms a virtual image \(8.00 \mathrm{~mm}\) tall, \(17.0 \mathrm{~cm}\) to the right of the lens. Determine the position and size of the object. Is the image erect or inverted? Are the object and image on the same side or opposite sides of the lens? Draw a principal-ray diagram for this situation.
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