/*! 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 3 A pencil that is \(9.0 \mathrm{~... [FREE SOLUTION] | 91Ó°ÊÓ

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A pencil that is \(9.0 \mathrm{~cm}\) long is held perpendicular to the surface of a plane mirror with the tip of the pencil lead \(12.0 \mathrm{~cm}\) from the mirror surface and the end of the eraser \(21.0 \mathrm{~cm}\) from the mirror surface. What is the length of the image of the pencil that is formed by the mirror? Which end of the image is closer to the mirror surface: the tip of the lead or the end of the eraser?

Short Answer

Expert verified
The length of the image of the pencil in the mirror is \(9.0 \mathrm{~cm}\). The end of the pencil that is closer to the mirror surface in the image is the tip of the lead.

Step by step solution

01

Determine Object Size

Given the length of the pencil, which is the object, as \(9.0 \mathrm{~cm}\). This will be used to find the size of the image.
02

Find Image Size

In a plane mirror, the image size is equal to the object size. As per this principle, the image size or the length of the image of the pencil in the mirror will be equal to the length of the pencil, which is \(9.0 \mathrm{~cm}\).
03

Determine Image Orientation

The image formed by a plane mirror is a virtual, upright image, which means the orientation of objects is preserved as it is. This means if the pencil lead was closer to the mirror, then in the image also, the pencil lead will be closer to the mirror. Comparing the given distance of pencil lead (\(12.0 \mathrm{~cm}\)) and eraser end (\(21.0 \mathrm{~cm}\)) from the mirror, it's evident that pencil lead is closer to the mirror. Thus, in the image, the pencil lead will also be closer to the mirror.

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Key Concepts

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

Optics
Optics is a branch of physics that involves the study of light and its interactions with matter. It includes understanding various phenomena such as reflection, refraction, dispersion, and diffraction. The behavior of light when it encounters different media is fundamental to many aspects of science and technology, including vision, optical devices, and various forms of imaging. In the context of plane mirrors, optics explores how light reflects off the surface to create images.

When light rays strike a smooth surface like that of a plane mirror, they reflect in such a way that the angle of incidence is equal to the angle of reflection. This predictable behavior is key to determining how images are formed by mirrors and is a core principle in the study of optics.
Reflection
Reflection is the change in direction of a wavefront, such as light, when it bounces off a surface. According to the law of reflection, the angle at which the light rays hit a reflective surface (the angle of incidence) is the same as the angle at which they reflect away (the angle of reflection). This principle applies perfectly to plane mirrors, where the surface is flat and smooth, causing light rays to spread out uniformly after reflecting.

In terms of forming images, when the rays from an object, such as a pencil, are reflected off a plane mirror, the brain interprets the rays as though they are emerging from behind the mirror. This process leads to the creation of an image that appears to be the same distance behind the mirror as the object is in front of it.
Virtual Image
A virtual image is a type of image that is formed by diverging light rays. In contrast to a real image, which can be projected onto a screen, a virtual image cannot be captured because the light rays do not actually converge at the image's location. Instead, virtual images seem to be located 'behind' the mirror surface from which the light rays are reflecting. These images are called virtual because they are formed by extensions of the reflected light rays backward until they appear to intersect.

Virtual images are upright and have the same size as the object they represent. This is demonstrated by the pencil exercise, where the image of the pencil appears upright and is the same length as the actual pencil. The virtual image is a fundamental concept in optics and is essential to understanding the nature of images produced by plane mirrors.
Mirror Image Characteristics
When a mirror creates an image, there are several characteristics that define its appearance. Firstly, the image formed by a plane mirror is always virtual, meaning it cannot be projected onto a screen. Secondly, it is upright relative to the object. This makes plane mirrors unique in that they do not invert images as some other reflective surfaces might. Thirdly, the size of the image is equal to the size of the object, which is evident in the pencil example provided where the image is the same length as the pencil itself.

Another characteristic is the laterally inverted nature of the image. Text and other asymmetric objects appear reversed from left to right in a mirror image. Finally, the distance of the image from the mirror is equal to the distance of the object from the mirror. This equidistance is key to solving problems related to mirror image formation, such as determining which end of an object, like the pencil lead or eraser, is closer to the mirror surface.

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Most popular questions from this chapter

Figure P34.99 shows a simple version of a zoom lens. The converging lens has focal length \(f_{1}\) and the diverging lens has focal length \(f_{2}=-\left|f_{2}\right| .\) The two lenses are separated by a variable distance \(d\) that is always less than \(f_{1}\). Also, the magnitude of the focal length of the diverging lens satisfies the inequality \(\left|f_{2}\right|>\left(f_{1}-d\right) .\) To determine the effective focal length of the combination lens, consider a bundle of parallel rays of radius \(r_{0}\) entering the converging lens. (a) Show that the radius of the ray bundle decreases to \(r_{0}^{\prime}=r_{0}\left(f_{1}-d\right) / f_{1}\) at the point that it enters the diverging lens. (b) Show that the final image \(I^{\prime}\) is formed a distance \(s_{2}^{\prime}=\left|f_{2}\right|\left(f_{1}-d\right) /\left(\left|f_{2}\right|-f_{1}+d\right)\) to the right of the diverging lens. (c) If the rays that emerge from the diverging lens and reach the final image point are extended backward to the left of the diverging lens, they will eventually expand to the original radius \(r_{0}\) at some point \(Q .\) The distance from the final image \(I^{\prime}\) to the point \(Q\) is the effective focal length \(f\) of the lens combination; if the combination were replaced by a single lens of focal length \(f\) placed at \(Q\), parallel rays would still be brought to a focus at \(I^{\prime} .\) Show that the effective focal length is given by \(f=f_{1}\left|f_{2}\right| /\left(\left|f_{2}\right|-f_{1}+d\right) .\) (d) If \(f_{1}=12.0 \mathrm{~cm}, f_{2}=-18.0 \mathrm{~cm},\) and the separation \(d\) is adjustable between 0 and \(4.0 \mathrm{~cm}\), find the maximum and minimum focal lengths of the combination. What value of \(d\) gives \(f=30.0 \mathrm{~cm} ?\)

An insect \(3.75 \mathrm{~mm}\) tall is placed \(22.5 \mathrm{~cm}\) to the left of a thin planoconvex lens. The left surface of this lens is flat, the right surface has a radius of curvature of magnitude \(13.0 \mathrm{~cm},\) and the index of refraction of the lens material is \(1.70 .\) (a) Calculate the location and size of the image this lens forms of the insect. Is it real or virtual? Erect or inverted? (b) Repeat part (a) if the lens is reversed.

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?

=A converging lens forms an image of an \(8.00-\mathrm{mm}\) -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?

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.

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