/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 A candle \(4.85 \mathrm{~cm}\) t... [FREE SOLUTION] | 91Ó°ÊÓ

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A candle \(4.85 \mathrm{~cm}\) tall is \(39.2 \mathrm{~cm}\) to the left of a plane mirror. Where is the image formed by the mirror, and what is the height of this image?

Short Answer

Expert verified
The image of the candle formed by the mirror is \(39.2 \mathrm{ ~cm}\) behind the mirror and has a height of \(4.85 \mathrm{ ~cm}\). It's a virtual and upright image.

Step by step solution

01

Identifying the location of the image

By the properties of the plane mirror, the image of the object will be located at the same distance behind the mirror as the object is in front of it. Therefore, the image of the candle stands \(39.2 \mathrm{ ~cm}\) behind the mirror.
02

Identifying the height of the image

The image formed by a plane mirror is always of the same size as the object. Hence, the height of the image is the same as the original candle, that is, \(4.85 \mathrm{ ~cm}\).
03

Displaying the Image

The image is virtual (since it appears to be located at a position where light does not reach), upright (the same orientation as the object), and located \(39.2 \mathrm{ ~cm}\) behind the mirror with a height of \(4.85 \mathrm{ ~cm}\).

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Key Concepts

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

Image Formation
A key concept in physics, particularly in the study of optics, is how an image is formed in a plane mirror. When an object is placed in front of a plane mirror, light rays from the object reflect off the surface of the mirror. This reflection follows the law of reflection, where the angle of incidence equals the angle of reflection. The human brain naturally extends these reflected rays behind the mirror, leading to the perception of an image located there.

This process results in an image that seems to "exist" behind the mirror at the same distance as the object from the mirror. Hence, for a candle placed 39.2 cm in front of the mirror, its image will appear 39.2 cm behind the mirror.

Understanding image formation involves grasping that these images are not physically present; they are visual projections generated by the way light interacts with the mirror and interpreted by the observer's eyes.
Optics
In optics, the study of light and its interactions with different surfaces is fundamental. Plane mirrors are flat, smooth surfaces that reflect light in a consistent direction, directly related to the law of reflection. Optics explores various phenomena such as reflection, refraction, and diffraction.

When dealing with plane mirrors, remember:
  • The angle of incidence is equal to the angle of reflection.
  • Mirrors do not alter the speed of light; thus, distances in front and behind the mirror are perceived equally by an observer.
These principles allow us to predict the behavior of light and the appearance of images accurately. By understanding basic optics, such as those illustrated with the behavior of a plane mirror, we come closer to comprehending more complex optical systems.
Virtual Image
In the context of plane mirrors, the image formed is known as a "virtual image." Unlike a real image, which can be projected onto a screen, a virtual image cannot be caught on a surface because it does not actually exist in space.

Characteristics of virtual images include:
  • They appear in locations where light does not physically reach.
  • These images maintain the same orientation as the object and are upright.
  • Both the size and shape of the image are identical to those of the object.
This type of image helps in visual interpretation and is crucial in applications like dressing mirrors, where you need an accurate, undistorted reflection.
Mirror Properties
Plane mirrors possess distinctive properties that differentiate them from other types of mirrors. These properties directly affect how images are perceived.

Noteworthy mirror properties include:
  • Image distance: In plane mirrors, the distance of the image behind the mirror is the same as the object's distance in front of it.
  • Size consistency: The size of the image is equal to the size of the object.
  • Orientation: The image remains upright.
  • Reversibility: Images are laterally inverted, meaning left and right are swapped.
These properties make plane mirrors unique tools for creating clear and straightforward reflections. They help individuals understand basic reflective behaviors before tackling more complex surfaces such as curved mirrors.

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Most popular questions from this chapter

A spherical, concave shaving mirror has a radius of curvature of \(32.0 \mathrm{~cm}\). (a) What is the magnification of a person's face when it is \(12.0 \mathrm{~cm}\) to the left of the vertex of the mirror? (b) Where is the image? Is the image real or virtual? (c) Draw a principal-ray diagram showing the formation of the image.

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.

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 \(1.75 \mathrm{~m}\).

Given that frogs are nearsighted in air, which statement is most likely to be true about their vision in water? (a) They are even more nearsighted; because water has a higher index of refraction than air, a frog's ability to focus light increases in water. (b) They are less nearsighted, because the cornea is less effective at refracting light in water than in air. (c) Their vision is no different, because only structures that are internal to the eye can affect the eye's ability to focus. (d) The images projected on the retina are no longer inverted, because the eye in water functions as a diverging lens rather than a converging lens.

It is your first day at work as a summer intern at an optics company. Your supervisor hands you a diverging lens and asks you to measure its focal length. You know that with a converging lens, you can measure the focal length by placing an object a distance \(s\) to the left of the lens, far enough from the lens for the image to be real, and viewing the image on a screen that is to the right of the lens. By adjusting the position of the screen until the image is in sharp focus, you can determine the image distance \(s^{\prime}\) and then use Eq. (34.16) to calculate the focal length \(f\) of the lens. But this procedure won't work with a diverging lens - by itself, a diverging lens produces only virtual images, which can't be projected onto a screen. Therefore, to determine the focal length of a diverging lens, you do the following: First you take a converging lens and measure that, for an object \(20.0 \mathrm{~cm}\) to the left of the lens, the image is \(29.7 \mathrm{~cm}\) to the right of the lens. You then place a diverging lens \(20.0 \mathrm{~cm}\) to the right of the converging lens and measure the final image to be \(42.8 \mathrm{~cm}\) to the right of the converging lens. Suspecting some inaccuracy in measurement, you repeat the lenscombination measurement with the same object distance for the converging lens but with the diverging lens \(25.0 \mathrm{~cm}\) to the right of the converging lens. You measure the final image to be \(31.6 \mathrm{~cm}\) to the right of the converging lens. (a) Use both lens-combination measurements to calculate the focal length of the diverging lens. Take as your best experimental value for the focal length the average of the two values. (b) Which position of the diverging lens, \(20.0 \mathrm{~cm}\) to the right or \(25.0 \mathrm{~cm}\) to the right of the converging lens, gives the tallest image?

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