/*! 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 12 For a concave spherical mirror t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

For a concave spherical mirror that has focal length \(f\) = \(+\)18.0 cm, what is the distance of an object from the mirror's vertex if the image is real and has the same height as the object?

Short Answer

Expert verified
The object is 36.0 cm from the mirror's vertex.

Step by step solution

01

Understand the Mirror Equation

The mirror equation relates the object distance \(d_o\), the image distance \(d_i\), and the focal length \(f\) of the mirror. It is given by: \[\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}.\] In this problem, the focal length \(f\) is given as \(+18.0\) cm.
02

Analyze the Given Conditions

The problem states that the image is real and has the same height as the object. For a real image with the same height, the magnification \(m\), which is \(\frac{h_i}{h_o} = \frac{-d_i}{d_o}\), must be \(-1\). Hence, \(d_i = -d_o\).
03

Substitute and Solve

Substitute \(d_i = -d_o\) into the mirror equation: \[\frac{1}{d_o} + \frac{1}{-d_o} = \frac{1}{18.0}.\] Simplify and solve: \[\frac{1}{d_o} - \frac{1}{d_o} = \frac{1}{18.0}\rightarrow 0 = \frac{1}{18.0}.\] However, this approach indicates a mistake in substitution. Instead, solve keeping \(d_i = d_o\): \[2\left(\frac{1}{d_o}\right) = \frac{1}{18.0}.\] Thus, \[\frac{2}{d_o} = \frac{1}{18.0}.\] Solving for \(d_o\), we find \[d_o = 36.0 \text{ cm}.\]
04

Confirm the Solution

Verify the solution by checking if the conditions are satisfied. Substituting \(d_o = 36.0\) cm back gives \(d_i = 36.0\) cm, and using the magnification \(m = \frac{h_i}{h_o} = \frac{-d_i}{d_o}\), confirming \(-1\). Therefore, the solution is consistent.

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.

Mirror Equation
The mirror equation is a fundamental formula used to relate the key elements of a concave mirror's setup. This equation links the object distance ( d_o ), the image distance ( d_i ), and the focal length ( f ) of the mirror. It is expressed as:
  • \( \frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f} \)
This equation is crucial because it helps determine how an image will form based on where an object is placed relative to the mirror. In the context of concave mirrors, a positive focal length value indicates that the mirror converges light rays, allowing for real image formation. By substituting the known values into the mirror equation, we can solve for unknown distances, providing a mathematical basis to predict image placement.
Focal Length
The focal length ( f ) is a critical characteristic of any spherical mirror, particularly concave ones. It is the distance from the mirror's surface to its focal point, where parallel rays of light converge. The focal length's sign is vital:
  • A positive focal length, as in this exercise with f = +18.0 cm, indicates a concave mirror that focuses light to a point.
  • A negative focal length would indicate a convex mirror, which diverges light rays.
Understanding the focal length is essential for predicting how mirrors affect light pathways and image projection. In this problem, the positive focal length establishes that the mirror can produce a real image, emphasizing its role in shaping light behavior.
Real Image
A real image is one that can be projected onto a surface, existing where light rays physically converge. This occurs when the image distance ( d_i ) is positive, marking the image as on the opposite side of the object relative to the mirror. In this exercise, the image height equaling the object height implies a magnification ( m ) of -1 , satisfying the condition for a real image of the same size but inverted:
  • \( m = \frac{-d_i}{d_o} = -1 \)
This equation shows that the distances are equal in magnitude but opposite in sign, giving physical significance to the image's realism and orientation. Understanding real images aids in applications ranging from optical instruments to everyday reflections.
Magnification
Magnification describes how much larger or smaller an image appears compared to the object itself. It is calculated with the formula:
  • \( m = \frac{h_i}{h_o} = \frac{-d_i}{d_o} \)
Where \( h_i \) is the image height and \( h_o \) is the object height. In this scenario, the magnification is -1, indicating that the image is the same size but inverted relative to the object. This negative sign is characteristic of real images formed by concave mirrors. Magnification is essential in determining not only the size but also the nature of the image (real or virtual). By applying these principles, the relationships between object placement, image characteristics, and mirror properties become clearer, enhancing our understanding of optical systems.

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

An object 0.600 cm tall is placed 16.5 cm to the left of the vertex of a concave spherical mirror having a radius of curvature of 22.0 cm. (a) Draw a principal-ray diagram showing the formation of the image. (b) Determine the position, size, orientation, and nature (real or virtual) of the image.

The image formed by a microscope objective with a focal length of 5.00 mm is 160 mm from its second focal point. The eyepiece has a focal length of 26.0 mm. (a) What is the angular magnification of the microscope? (b) The unaided eye can distinguish two points at its near point as separate if they are about 0.10 mm apart. What is the minimum separation between two points that can be observed (or resolved) through this microscope?

A transparent rod 30.0 cm long is cut flat at one end and rounded to a hemispherical surface of radius 10.0 cm at the other end. A small object is embedded within the rod along its axis and halfway between its ends, 15.0 cm from the flat end and 15.0 cm from the vertex of the curved end. When the rod is viewed from its flat end, the apparent depth of the object is 8.20 cm from the flat end. What is its apparent depth when the rod is viewed from its curved end?

When an object is placed at the proper distance to the left of a converging lens, the image is focused on a screen 30.0 cm to the right of the lens. A diverging lens is now placed 15.0 cm to the right of the converging lens, and it is found that the screen must be moved 19.2 cm farther to the right to obtain a sharp image. What is the focal length of the diverging lens?

A small tropical fish is at the center of a water-filled, spherical fish bowl 28.0 cm in diameter. (a) Find the apparent position and magnification of the fish to an observer outside the bowl. The effect of the thin walls of the bowl may be ignored. (b) A friend advised the owner of the bowl to keep it out of direct sunlight to avoid blinding the fish, which might swim into the focal point of the parallel rays from the sun. Is the focal point actually within the bowl?

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.