/*! 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 11 A \(\mathrm{A}\) coin is placed ... [FREE SOLUTION] | 91Ó°ÊÓ

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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.

Short Answer

Expert verified
The coin is located at 3.6 cm from the glass. The image is 2.5 cm tall, erect, and virtual.

Step by step solution

01

Understand the Problem

We need to find where the coin is located and determine the properties of the image formed (size, orientation, and type). This involves using concepts and formulas related to spherical mirrors or lenses.
02

Identify the Given Data

We are given:- The radius of curvature of the glass shell: \( R = 18.0 \text{ cm} \).- The image distance behind the glass shell: \( q = 6.0 \text{ cm} \).- The height of the coin: \( h_o = 1.5 \text{ cm} \).
03

Use the Mirror Equation

The mirror equation is given by: \[ \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \]First, calculate the focal length \( f \) using \( f = \frac{R}{2} = \frac{18.0}{2} = 9.0 \text{ cm} \).
04

Calculate the Object Distance

Rearrange the mirror equation to solve for the object distance \( p \):\[ \frac{1}{p} = \frac{1}{f} - \frac{1}{q} \]Substitute \( f = 9.0 \text{ cm} \) and \( q = -6.0 \text{ cm} \) (since the image is behind the mirror, this is negative):\[ \frac{1}{p} = \frac{1}{9.0} + \frac{1}{6.0} \]\[ \Rightarrow p = 3.6 \text{ cm} \]
05

Determine the Size of the Image

Use the magnification formula: \[ m = -\frac{q}{p} = \frac{h_i}{h_o} \]Calculate:\[ m = -\frac{-6.0}{3.6} = 1.67 \]So, the image height \( h_i \) is:\[ h_i = m \times h_o = 1.67 \times 1.5 = 2.5 \text{ cm} \]
06

Determine the Orientation of the Image

Since the magnification \( m = 1.67 \) is positive, the image is erect compared to the object.
07

Determine the Nature of the Image

Since the image distance \( q = -6.0 \text{ cm} \) is negative, the image is virtual as it is formed behind the spherical glass shell.

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Key Concepts

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

Understanding the Radius of Curvature
The radius of curvature is a fundamental property of spherical mirrors that determines their focusing power. It refers to the radius of the sphere from which the mirror segment is taken. For a spherical mirror, this radius of curvature is denoted by \( R \). In the context of our exercise, the radius of curvature is given as 18.0 cm. This means that if you imagine the mirror as part of a complete sphere, the sphere’s radius would be 18.0 cm. This property is crucial because it helps in calculating the focal length of the mirror, which is essential for image formation calculations. The focal length \( f \), a key parameter, is directly related to the radius of curvature by the equation \( f = \frac{R}{2} \).
  • For a convex mirror like the one in the exercise, \( f \) is positive when considering traditional sign conventions, meaning it converges light.
  • The radius of curvature helps in determining the precise point where parallel light rays reflect off and converge or appear to diverge from behind the mirror.
Understanding this concept helps in effectively applying the mirror equation to find out where images will form.
Image Formation from Spherical Mirrors
Image formation in spherical mirrors, such as the one in this exercise, follows basic principles of geometrical optics and involves using key parameters like object distance, image distance, and focal length. Convex mirrors, like the one in our problem, diverge incident light rays making them appear to originate from a common point behind the mirror, forming a virtual image.
  • The image is formed by extrapolating the reflected rays backward until they meet. In real scenarios, the image is upright and smaller than the object.
  • In this situation, a 1.5 cm tall coin placed at a certain distance from the mirror has its image formed at 6.0 cm behind the mirror. This tells us we are dealing with a virtual image as this distance bears a negative sign according to the mirror convention.
Understanding these basic concepts better prepares us to explore how the mirror equation connects the distances involved.
Exploring the Mirror Equation
The mirror equation is a foundational formula in optics that connects the focal length \( f \), object distance \( p \), and image distance \( q \). The equation is given by:\[ \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \]For the problem at hand, a convex mirror requires adjustments in sign conventions. Here, since the image forms behind the mirror, \( q \) takes a negative value. By substituting the values: \( f = 9.0 \, \text{cm} \) and \( q = -6.0 \, \text{cm} \) into the equation, you can solve for \( p \) (the object distance).
  • This yields \( p = 3.6 \, \text{cm} \), pinpointing where the coin should be relative to the mirror’s reflective surface.
  • The equation helps measure how image characteristics, such as distance and size, change when the object position changes.
Hence, becoming comfortable with the mirror equation and its variable role is essential for analyzing similar optics problems.
Understanding Virtual Images
A virtual image, in optics, is one that cannot be projected onto a screen as it forms from divergent light rays. These rays, when extended backward, appear to meet behind the mirror. In the context of this exercise:
  • The image is virtual because it forms at a location where light doesn't actually converge. It just appears like this to an observer.
  • For the convex mirror used here, the virtual image is upright and appears smaller than the actual object, such as a coin.
  • Virtual images occur with mirrors that have light rays diverging, making them distinct from real images which occur at the convergence of light rays and can be captured on a screen.
The key takeaway is that virtual images occur when an image distance is negative based on the mirror convention, signaling that the image forms behind the mirror, reinforcing their hallmark differences from real images.

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Most popular questions from this chapter

If you place a concave glass lens into a tank of a liquid that has an index of refraction that is greater than that of the lens, what will happen? A. The lens will no longer be able to create any images. B. The focal length of the lens will become longer. C. The focal length of the lens will become shorter. D. The lens will become a converoing lens

\(\cdot\) 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?

\(\cdot\) A concave mirror has a radius of curvature of 34.0 \(\mathrm{cm}\) . (a) What is its focal length? (b) A ladybug 7.50 \(\mathrm{mm}\) tall is located 22.0 \(\mathrm{cm}\) from this mirror along the principal axis. Find the location and height of the image of the insect. (c) If the mirror is immersed in water (of refractive index \(1.33 ),\) what is its focal length?

\(\cdot\) The lens of the eye. The crystalline lens of the human eye is a double-convex lens made of material having an index of refraction of 1.44 (although this varies). Its focal length in air is about 8.0 \(\mathrm{mm}\) , which also varies. We shall assume that the radii of curvature of its two surfaces have the same magnitude. (a) Find the radii of curvature of this lens. (b) If an object 16 \(\mathrm{cm}\) tall is placed 30.0 \(\mathrm{cm}\) from the eye lens, where would the lens focus it and how tall would the image be? Is thisimage real or virtual? Is it erect or inverted? (Note: The results.obtained here are not strictly accurate, because the lens is embedded in fluids having refractive indexes different from that of air.)

\(\cdot\) 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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