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Chapter 23: Problem 47

An object placed \(10.0 \mathrm{~cm}\) from a concave spherical mirror produces a real image \(8.00 \mathrm{~cm}\) from the mirror. If the object is moved to a new position \(20.0 \mathrm{~cm}\) from the mirror, what is the position of the image? Is the final image real or virtual?

Short Answer

Expert verified
The position of the final image is \(20.0 \mathrm{~cm}\) from the mirror. The final image is real.

Step by step solution

01

Apply the mirror equation to find the focal length

The mirror equation is given by \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \(f\) is the focal length, \(d_o\) is the object distance, and \(d_i\) is the image distance. Given the initial object distance \(d_o = 10.0 \, \mathrm{cm}\) and image distance \(d_i = -8.00 \, \mathrm{cm}\) (we take the image distance as negative as the image is formed on the same side as the light is coming from i.e. it is real), we get \( \frac{1}{f} = \frac{1}{10.0} - \frac{1}{8.00} \). This gives the focal length as \(f = -20.0 \, \mathrm{cm}\).
02

Use the mirror equation again to find the new image distance

Now, we are informed that the object is moved to a new position \(\mathrm{20.0 \, cm}\) from the mirror, or \(d_o = 20.0 \, \mathrm{cm}\). We can apply the mirror equation again to get: \( \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} = \frac{1}{-20.0} - \frac{1}{20.0} \). This gives the image distance as \(d_i = -20.0 \, \mathrm{cm}\).
03

Determine the nature of the image

The negative sign in \(d_i = -20.0 \, \mathrm{cm}\) implies that the image is a real image and is formed on the same side from where the light is coming.

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Key Concepts

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

Concave Mirror
A concave mirror is a type of spherical mirror where the reflective surface is curved inward, resembling the inside of a bowl. Such mirrors are also known as converging mirrors because they can converge light beams to a focal point. This type of mirror is widely used in devices like telescopes, flashlights, and certain types of magnifying tools.

Concave mirrors have unique characteristics:
  • They can focus parallel beams of light to a single point, known as the focal point.
  • The focal length of the mirror (distance from the mirror to the focal point) determines where the image will form.
  • These mirrors can produce both real and virtual images depending on the object's position relative to the mirror's focal point.
Understanding the behavior of concave mirrors helps in predicting how images form and aids in solving physics problems using the mirror equation.
Real Image
In the context of mirrors and lenses, a real image is one where the light rays actually converge and pass through the image point. Real images are typically inverted and can be projected onto a screen, making them practical in applications such as cameras and projectors.

For concave mirrors:
  • If the object is placed outside the focal length, the mirror will form a real image.
  • This image will appear on the same side as the object if the light is directed from that side.
  • In optics equations, real images have a positive focal length, but their image distances are considered negative because they form on the incoming light side.
Recognizing the conditions that lead to a real image helps in understanding their practical uses and addressing problems involving light and mirrors.
Focal Length
The focal length of a mirror or lens is a critical measure in optics, representing the distance from the mirror's surface to its focal point. The focal point is where converging light rays meet after reflecting off a concave mirror.

To find the focal length, one can use the mirror equation:\[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]Where:
  • \(f\) is the focal length,
  • \(d_o\) is the object distance,
  • \(d_i\) is the image distance.
This equation helps in predicting where the image will form when the position of the object changes.

The sign convention in the equation is crucial:
  • For concave mirrors, the focal length is considered negative because they are converging mirrors.
  • A positive focal length usually indicates a convex mirror or a diverging lens.
Understanding focal length and its calculation is essential in solving mirror-related problems accurately.

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Most popular questions from this chapter

A dentist uses a mirror to examine a tooth that is \(1.00 \mathrm{~cm}\) in front of the mirror. The image of the tooth is formed \(10.0 \mathrm{~cm}\) behind the mirror. Determine (a) the mirror's radius of curvature and (b) the magnification of the image.

A convex mirror has a focal length of magnitude \(8.0 \mathrm{~cm}\). (a) If the image is virtual, what is the object location for which the magnitude of the image distance is one third the magnitude of the object distance? (b) Find the magnification of the image and state whether it is upright or inverted.

A "floating strawberry" illusion can be produced by two parabolic mirrors, each with a focal length of \(7.5 \mathrm{~cm}, \mathrm{fac}\) ing each other so that their centers are \(7.5 \mathrm{~cm}\) apart (Fig. P23.58). If a strawberry is placed on the bottom mirror, an image of the strawberry forms at the small opening at the center of the top mirror. Show that the final image forms at that location and describe its characteristics. Note: \(\mathrm{A}\) flashlight beam shone on these images has a very startling effect: Even at a glancing angle, the incoming light beam is seemingly reflected off the images of the strawberry! Do you understand why?

A cubical block of ice \(50.0 \mathrm{~cm}\) on an edge is placed on a level floor over a speck of dust. Locate the image of the speck, when viewed from directly above, if the index of refraction of ice is \(1.309 .\)

The lens-maker's equation for a lens with index \(n_{1}\) immersed in a medium with index \(n_{2}\) takes the form $$ \frac{1}{f}=\left(\frac{n_{1}}{n_{2}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) $$ A thin diverging glass (index \(=1.50\) ) lens with \(R_{1}=\) \(-3.00 \mathrm{~m}\) and \(R_{2}=-6.00 \mathrm{~m}\) is surrounded by air. An arrow is placed \(10.0 \mathrm{~m}\) to the left of the lens. (a) Determine the position of the image. Repeat part (a) with the arrow and lens immersed in (b) water (index \(=1.33\) ) (c) a medium with an index of refraction of \(2.00\). (d) How can a lens that is diverging in air be changed into a converging lens?
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