/*! 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 71 An object is placed \(20.0 \math... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An object is placed \(20.0 \mathrm{cm}\) to the left of a diverging lens \((f=-8.00 \mathrm{cm}) .\) A concave mirror \((f=12.0 \mathrm{cm})\) is placed \(30.0 \mathrm{cm}\) to the right of the lens. (a) Find the final image distance, measured relative to the mirror. (b) Is the final image real or virtual? (c) Is the final image upright or inverted with respect to the original object?

Short Answer

Expert verified
(a) 48.6 cm to the right of the mirror. (b) The final image is real. (c) The image is inverted.

Step by step solution

01

Find Image Distance from the Lens

For a diverging lens, the lens formula is given by \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \),where \( f = -8.00 \, \text{cm} \) and \( d_o = 20.0 \, \text{cm} \). Substituting values:\[ \frac{1}{-8} = \frac{1}{20} + \frac{1}{d_i} \]Solving for \( d_i \):\[ \frac{1}{d_i} = \frac{1}{-8} - \frac{1}{20} \]\[ \frac{1}{d_i} = -\frac{5}{40} - \frac{2}{40} = -\frac{7}{40} \]\[ d_i = -\frac{40}{7} \approx -5.71 \, \text{cm} \]The negative sign indicates the image is on the same side as the object, which is characteristic for virtual images formed by diverging lenses.
02

Determine the Object Distance for the Mirror

The image formed by the lens acts as the virtual object for the mirror. Calculating the position: Image distance from the lens is \(-5.71 \, \text{cm}\) to the left of the lens. Since the lens is 30.0 cm from the mirror, the virtual object's distance from the mirror is:\[ d_{o, \, ext{mirror}} = 30.0 \, \text{cm} - 5.71 \, \text{cm} = 24.29 \, \text{cm} \]This value will be used as the object distance for the mirror in the next step.
03

Find Image Distance from the Mirror

For a concave mirror, the mirror formula is:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]Given \( f = 12.0 \, \text{cm} \) and \( d_o = 24.29 \, \text{cm} \):\[ \frac{1}{12} = \frac{1}{24.29} + \frac{1}{d_i} \]Solving for \( d_i \):\[ \frac{1}{d_i} = \frac{1}{12} - \frac{1}{24.29} \]\[ \frac{1}{d_i} = \frac{2}{24} - \frac{1}{24.29} \]\[ \frac{1}{d_i} = \frac{2 \times 24.29 - 12}{12 \times 24.29} \]\[ d_i \approx 48.58 \text{ cm} \] So, the final image distance from the mirror is approximately 48.6 cm to the right of the mirror.
04

Determine Nature of the Final Image

The image formed is on the same side as the actual object for the mirror. Since the image distance \( d_i \) is positive, this indicates that the image is real.Using the magnification equations, we can check the orientation to see if it's inverted or upright. A concave mirror inverts real images formed.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Diverging Lens
A diverging lens, also known as a concave lens, has a unique shape that is thinner in the center and thicker at the edges. This design causes parallel rays of light that pass through it to spread out or diverge. A crucial feature of diverging lenses is their negative focal length, which determines how they refract light rays. In the case of a diverging lens, the focal point where the refracted rays appear to diverge from is on the same side as the incoming light.

For example, if an object is placed 20 cm from a diverging lens with a focal length of -8 cm, the lens will cause the light rays coming from the object to diverge. By using the lens formula \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), we can find out more about where the image will form. This nature often results in the formation of a virtual image, which appears on the same side of the lens as the object.
Concave Mirror
A concave mirror is a type of mirror with a reflecting surface that curves inward, much like the inside of a sphere. This curvature allows the mirror to converge light rays that hit its surface. The focal length of a concave mirror is positive, and it indicates the distance from the mirror to the focal point where all parallel rays meet.

When an object is placed in front of a concave mirror, the positioning relative to the focal point dictates the type of image that forms. In the scenario at hand, the concave mirror has a focal length of 12 cm. If the virtual image formed by the lens is 24.29 cm away from the mirror, it acts as a new object. The mirror formula \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \) helps predict the position and nature of the image.
Image Formation
Image formation involves the process by which lenses or mirrors create images, based on the path lights rays traverse. For diverging lenses, the image is usually formed by tracing the apparent origin of the diverged rays backward. In concave mirrors, images can be formed in different positions depending upon the object's location relative to the focal point.

The combination of a diverging lens and a concave mirror in this problem means light rays are refracted and reflected, respectively, creating complex paths. By calculating where these paths intersect or appear to originate, we can figure out where the image forms. Image formation involves understanding both the character of the lens or mirror and their optical equations.
Virtual Image
A virtual image is formed when diverging light rays appear to meet when projected backward. This means, unlike real images, which can be projected onto a screen, virtual images cannot be physically captured as they do not result from actual light convergence. In the case of the diverging lens, the virtual image is positioned on the same side as the object.

This image appears to be formed by the extension of the diverging rays coming out of a lens, and cannot be traced directly onto a surface. When dealing with virtual images, they are often seen as right-side-up, with some magnification or size distortion based on the exact placement of both object and optical device.
Real Image
Real images are a contrast to virtual ones because they are formed where light rays actually converge. Concave mirrors are excellent at forming real images due to their ability to focus light. These images can be cast onto surfaces because they correspond with actual light paths.

In this exercise, the real image is formed by the concave mirror. As calculated, this real image is approximately 48.6 cm from the mirror, positioning it on the opposite side from the object, typically inverted compared to the original orientation of the object. Real images can be affected by factors such as object distance and focal length, which determine their size and position.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The near point of a naked eye is \(32 \mathrm{cm} .\) When an object is placed at the near point and viewed by the naked eye, it has an angular size of 0.060 rad. A magnifying glass has a focal length of \(16 \mathrm{cm},\) and is held next to the eye. The enlarged image that is seen is located \(64 \mathrm{cm}\) from the magnifying glass. Determine the angular size of the image.

An object has an angular size of 0.0150 rad when placed at the near point \((21.0 \mathrm{cm})\) of an eye. When the eye views this object using a magnifying glass, the largest possible angular size of the image is 0.0380 rad. What is the focal length of the magnifying glass?

In a compound microscope, the focal length of the objective is 3.50 \(\mathrm{cm}\) and that of the eyepiece is \(6.50 \mathrm{cm} .\) The distance between the lenses is 26.0 cm. (a) What is the angular magnification of the microscope if the person using it has a near point of \(35.0 \mathrm{cm} ?\) (b) If, as usual, the first image lies just inside the focal point of the eyepiece (see Figure 26.32 ), how far is the object from the objective? (c) What is the magnification (not the angular magnification) of the objective?

The angular magnification of a telescope is 32800 times as large when you look through the correct end of the telescope as when you look through the wrong end. What is the angular magnification of the telescope?

Two converging lenses \(\left(f_{1}=9.00 \mathrm{cm}\right.\) and \(\left.f_{2}=6.00 \mathrm{cm}\right)\) are separated by \(18.0 \mathrm{cm} .\) The lens on the left has the longer focal length. An object stands \(12.0 \mathrm{cm}\) to the left of the left-hand lens in the combination. (a) Locate the final image relative to the lens on the right. (b) Obtain the overall magnification. (c) Is the final image real or virtual? With respect to the original object, (d) is the final image upright or inverted and (e) is it larger or smaller?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.