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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 \(1.5-\) 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 \(18.0\, \mathrm{cm}\) from the mirror, the image is \(0.5\, \mathrm{cm}\) tall, upright, and virtual.

Step by step solution

01

Analyze the Given Information

We are provided the following details: the radius of curvature \(R = 18.0\, \mathrm{cm}\) and the image distance \(v = -6.0\, \mathrm{cm}\). The negative sign indicates that the image is formed behind the mirror. The height of the object (coin) is \(1.5\, \mathrm{cm}\). We need to find the object distance \(u\) and analyze the image.
02

Use Mirror Formula

The mirror formula is given by \(\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\). Using the mirror formula for spherical mirrors, we find the focal length \(f\) as \(f = \frac{R}{2} = \frac{18.0}{2} = 9.0\, \mathrm{cm}\).
03

Solve for Object Distance

Rearrange the mirror formula to solve for \(u\): \[ \frac{1}{u} = \frac{1}{f} - \frac{1}{v} \]Substitute the known values: \[ \frac{1}{u} = \frac{1}{9.0} - \frac{1}{-6.0} \]Calculate:\[ \frac{1}{u} = \frac{2 - 3}{18} = \frac{-1}{18} \]Thus, \(u = -18.0\, \mathrm{cm}\).
04

Determine Image Size and Orientation

Use the magnification formula, \(m = \frac{v}{u} = \frac{\text{image height}}{\text{object height}}\).Calculate the magnification:\\[ m = \frac{-6.0}{-18.0} = \frac{1}{3} \]Thus, the image height is:\[ \text{Image height} = m \times \text{Object height} = \frac{1}{3} \times 1.5 = 0.5\, \mathrm{cm} \].The positive sign indicates the image is upright.
05

Determine the Nature of the Image

The image is virtual because it forms behind the mirror, as indicated by the negative image distance \(v = -6.0\, \mathrm{cm}\). Virtual images cannot be projected onto a screen and are always upright when formed by a convex surface.

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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 an essential tool in understanding how spherical mirrors create images. It relates the focal length (\( f \)), object distance (\( u \)), and image distance (\( v \)) through the equation: \[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]. This formula allows us to calculate unknown distances by rearranging and substituting known values.
  • The focal length (\( f \)) is half the radius of curvature (\( R \)) for spherical mirrors: \( f = \frac{R}{2} \).
  • In convex mirrors, the focal length and image distance are considered negative due to their geometric nature. Images form behind the mirror, indicating a virtual image location.
By rearranging the formula to solve for \( u \), we can determine the object distance when focal length and image distance are known. This helps in locating where an object must be placed relative to a spherical mirror to achieve a specific image formation.
Virtual Image
A virtual image in optics is what you see in a mirror鈥攊t appears to stand behind the mirror surface. Unlike real images, virtual images cannot be captured on a screen because they don't actually converge in real space.
  • In convex mirrors, images are always virtual because they form where light rays appear to diverge from.
  • These images retain the same orientation as the object, hence they are upright.
  • Convex mirrors, with a computed negative image distance, always give rise to virtual images鈥攕uggesting the image is on the same side of the mirror as the object.
Virtual images provide important benefits in practical applications too. For example, in car side mirrors, virtual images allow drivers to view a wider field of vision, enhancing safety.
Magnification
Magnification describes the degree to which an image is enlarged or reduced compared to the actual object. The magnification (\( m \)) formula is given by: \[ m = \frac{v}{u} = \frac{\text{image height}}{\text{object height}} \].
  • A magnification value (\( m \)) greater than 1 suggests that the image is larger than the object.
  • A value less than 1 indicates reduction; in our case, \( m = \frac{1}{3} \), showing the image is one-third the object's height.
  • The sign of magnification indicates image orientation鈥攑ositive for upright, negative for inverted.
In convex mirrors, the magnification is often less than 1, denoting a minimized view. This property makes convex mirrors invaluable for situations needing wide perspective views, such as in security or rear-view mirrors.

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Most popular questions from this chapter

cp calc You are in your car driving on a highway at 25 \(\mathrm{m} / \mathrm{s}\) when you glance in the passenger-side mirror (a convex mirror with radius of curvature 150 \(\mathrm{cm}\) ) and notice a truck approaching. If the image of the truck is approaching the vertex of the mirror at a speed of 1.9 \(\mathrm{m} / \mathrm{s}\) when the truck is 2.0 \(\mathrm{m}\) from the mirror, what is the speed of the truck relative to the highway?

A converging lens forms an image of an 8.00 -mm-tall real object. The image is 12.0 \(\mathrm{cm}\) to the left of the lens, 3.40 \(\mathrm{cm}\) tall, and erect. What is the focal length of the lens? Where is the object located?

What Is the Smallest Thing We Can See? The smallest object we can resolve with our eye is limited by the size of the light receptor cells in the retina. In order for us to distinguish any detail in an object, its image cannot be any smaller than a single retinal cell. Although the size depends on the type of cell (rod or cone), a diameter of a few microns \((\mu \mathrm{m})\) is typical near the center of the eye. We shall model the eye as a sphere 2.50 \(\mathrm{cm}\) in diameter with a single thin lens at the front and the retina at the rear, with light receptor cells 5.0\(\mu \mathrm{m}\) in diameter. (a) What is the smallest object you can resolve at a near point of 25 \(\mathrm{cm}\) ? (b) What angle is subtended by this object at the eye? Express your answer in units of minutes \(\left(1^{\circ}=60 \mathrm{min}\right),\) and compare it with the typical experimental value of about 1.0 min. (Note: There are other limitations, such as the bending of light as it passes through the pupil, but we shall ignore them here.)

A person with a near point of \(85 \mathrm{cm},\) but excellent distant vision, normally wears corrective glasses. But he loses them while traveling. Fortunately, he has his old pair as a spare. (a) If the lenses of the old pair have a power of \(+2.25\) diopters, what is his near point (measured from his eye) when he is wearing the old glasses if they rest 2.0 \(\mathrm{cm}\) in front of his eye? (b) What would his near point be if his old glasses were contact lenses instead?

Recall that the intensity of light reaching film in a camera is proportional to the effective area of the lens. Camera A has a lens with an aperture diameter of 8.00 \(\mathrm{mm} .\) It photographs an object using the correct exposure time of \(\frac{1}{30} \mathrm{s}\) . What exposure time should be used with camera \(\mathrm{B}\) in photographing the same object with the same film if this camera has a lens with an aperture diameter of 23.1 \(\mathrm{mm}\) ?
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