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

Short Answer

Expert verified
The image is real, inverted, 1.167 cm tall, and 32.1 cm to the right of the mirror.

Step by step solution

01

Understand the Properties of the Mirror

A concave mirror is a spherical mirror with an inwardly curved surface. It converges light and can form real or virtual images depending on the object's position relative to the focal point.
02

Determine the Focal Length

The radius of curvature (R) is given as 22.0 \mathrm{cm}. The focal length (f) is half of the radius of curvature for a spherical mirror. Thus, \(f = \frac{R}{2} = \frac{22.0\;\mathrm{cm}}{2} = 11.0\;\mathrm{cm}\).
03

Sketch the Principal-Ray Diagram

In a concave mirror, principal rays include: (1) A ray parallel to the axis that reflects through the focus; (2) A ray through the center of curvature that reflects back on the same path; (3) A ray to the vertex that reflects symmetrically. Sketch these rays to locate the image.
04

Use the Mirror Equation

The mirror equation is \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). Here, \(d_o\) is the object distance (16.5 cm), and \(f\) is the focal length (11.0 cm). Substitute to find the image distance \(d_i\):\[\frac{1}{d_i} = \frac{1}{11.0} - \frac{1}{16.5} = \frac{0.0606 - 0.0606}{1} = 0.0212\]. Thus, \(d_i \approx 32.1\;\mathrm{cm}\).
05

Compute the Magnification

The magnification \(m\) is given by \( m = -\frac{d_i}{d_o} \). Substitute \(d_i = 32.1\,\mathrm{cm}\) and \(d_o = 16.5\,\mathrm{cm}\): \(m = -\frac{32.1}{16.5} = -1.945\). The image is inverted since magnification is negative.
06

Determine the Image Size

The image height \(h_i\) is related to the object height \(h_o\) through the magnification: \( h_i = m \times h_o \). Given \(h_o = 0.600\,\mathrm{cm}\), \( h_i = -1.945 \times 0.600 \approx -1.167\,\mathrm{cm} \).
07

Analyze Image Properties

The image is real because it is formed by actual intersection of rays and is inverted due to negative magnification. It is larger than the object since the absolute value of magnification is greater than one.

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Key Concepts

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

Spherical Mirrors
Spherical mirrors are fascinating optical devices that come in two main types: concave and convex. Each serves different purposes based on their unique shapes.
Concave mirrors, like the one in the exercise, feature a curved, inward surface reflecting light so that it converges at a point. This quality is particularly useful in focusing light.
On the other hand, convex mirrors bulge outward and disperse light, making them useful in applications requiring wide-angle views, like security mirrors.
For concave mirrors, key properties include the center of curvature (the center of the sphere of which the mirror forms a part) and the focal point (where parallel rays converge). The distance from the mirror's surface to the focal point is the focal length, crucial in determining how the mirror forms images.
Image Formation
Image formation with concave mirrors depends on the object's position relative to the focal point and center of curvature. An image can be real or virtual.
A real image is formed when reflected rays converge and can be captured on a screen, often inverted and found when the object is beyond the focal point.
Conversely, a virtual image arises from apparently divergent rays that only seem to come from a point behind the mirror. This type of image is upright and can't be projected on a screen.
In our exercise, the image formed by the concave mirror is real and inverted, demonstrating typical properties of a real image from a practical scenario.
Mirror Equation
The mirror equation is an essential tool for calculating image properties in spherical mirrors. Expressed as:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \] it relates the focal length \( f \), object distance \( d_o \), and image distance \( d_i \). Applying this equation helps determine where an image will form and whether it will be real or virtual.
In our problem, by substituting the known values, we determined an image distance of approximately 32.1 cm. This positive \( d_i \) value confirms the real nature of the image, as real images form on the same side as the reflected light.
Ray Diagram
Ray diagrams serve as visual aids in understanding image formation for concave mirrors. They trace key rays that help pinpoint where an image forms.
Three principal rays are used:
  • A ray parallel to the principal axis, reflecting through the focal point.
  • A ray through the center of curvature, reflecting back on the same path.
  • A ray directed at the vertex, reflecting symmetrically through the focal point.
Sketching these rays can visually highlight the image's location, orientation, and size in relation to the object. Such diagrams provide an intuitive grasp of the theoretical calculations derived from the mirror equation.

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Most popular questions from this chapter

What is the size of the smallest vertical plane mirror in which a 10 ft tall giraffe standing erect can see her full-length image? (Hint: Locate the image by drawing a number of rays from the giraffe's body that reflect off the mirror and go to her eye. Then eliminate that part of the mirror for which the reflected rays do not reach her eye.)

\(\cdot\) The left end of a long glass rod 6.00 \(\mathrm{cm}\) in diameter has a convex hemispherical surface 3.00 \(\mathrm{cm}\) in radius. The refractive index of the glass is \(1.60 .\) Determine the position of the image if an object is placed in air on the axis of the rod at the following distances to the left of the vertex of the curved end: (a) infinitely far, (b) \(12.0 \mathrm{cm},\) and \((c) 2.00 \mathrm{cm} .\)

A \(\mathrm{A}\) coin is placed next to the convex side of a thin spherical glass shell having a radius of curvature of 18.0 \(\mathrm{cm} .\) An image of the 1.5 -cm-tall coin is formed 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.

A luminous object is 4.00 \(\mathrm{m}\) from a wall. You are to use a concave mirror to project an image of the object on the wall, with the image 2.25 times the size of the object. How far should the mirror be from the wall? What should its radius of curvature be?

\(\cdot\) The stainless steel rear end of a tanker truck is convex, shiny, and has a radius of curvature of 2.0 \(\mathrm{m} .\) You're tailgating the truck, with the front end of your car only 5.0 \(\mathrm{m}\) behind it. Making the not very realistic assumption that your car is on the axis of the mirror formed by the tank, (a) determine the position, orientation, magnification, and type (real or virtual) of the image of your car's front end that forms in this mirror; (b) draw a principal-ray diagram of the situation to check your answer.

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