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A coin is placed next to the convex side of a thin spherical glass shell having a radius of curvature of 18.0 cm. Reflection from the surface of the shell forms an image of the 1.5-cm-tall coin that is 6.00 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 3.6 cm from the mirror; the image is 2.5 cm tall, upright, and virtual.

Step by step solution

01

Understand the given values and formula needed

We are given the radius of curvature \( R = 18.0 \) cm and the image distance \( i = -6.0 \) cm (since the image is located behind the glass shell, the distance is negative). We need to find the object distance \( o \) and use the mirror equation: \[ \frac{1}{f} = \frac{1}{o} + \frac{1}{i} \] where \( f \) is the focal length, \( f = \frac{R}{2} \) for a spherical mirror, which gives \( f = 9.0 \) cm.
02

Substitute known values to find object distance

Substitute \( f = 9.0 \) cm and \( i = -6.0 \) cm into the mirror equation: \[ \frac{1}{9} = \frac{1}{o} + \frac{1}{-6} \]. Solve for \( \frac{1}{o} \): \[ \frac{1}{o} = \frac{1}{9} + \frac{1}{6} \]. Convert to a common denominator to get \( \frac{1}{o} = \frac{2}{18} + \frac{3}{18} = \frac{5}{18} \). Therefore, \( o = \frac{18}{5} = 3.6 \) cm.
03

Calculate the magnification and image size

The magnification \( m \) is given by the ratio \( m = -\frac{i}{o} \). Substitute \( i = -6.0 \) cm and \( o = 3.6 \) cm: \[ m = -\frac{-6}{3.6} = \frac{6}{3.6} = 1.67 \]. The image height is the object height multiplied by the magnification: \( 1.5 \times 1.67 = 2.5 \) cm.
04

Determine the orientation and nature of the image

Since the magnification is positive \( (1.67) \), the image is upright. The image distance is negative, indicating the image is virtual.

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Key Concepts

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

Convex Surfaces
Convex surfaces are outward-curving surfaces found on objects like spherical mirrors or lenses. In the context of spherical mirrors, a convex mirror reflects light outward. Convex mirrors are unique because they always form virtual images, meaning you cannot project these images onto a screen simply because the light rays spread out and appear to originate from a single point behind the mirror.
These types of mirrors give a wide field of view, which is why they are commonly used as rearview mirrors in vehicles. Key characteristics of images formed by convex mirrors include:
  • Images are always smaller than the actual object.
  • Images are virtual, meaning they cannot be projected.
  • Images are upright, appearing the same way up as the object.
All these definitions and properties apply due to the mirror's shape and the way it interacts with light.
Mirror Equation
The mirror equation is a fundamental formula used to determine the relationship between the object distance, image distance, and the focal length of a spherical mirror. The formula is:\[\frac{1}{f} = \frac{1}{o} + \frac{1}{i}\]where:
  • \( f \) is the focal length,
  • \( o \) is the object distance (distance from the object to the mirror),
  • \( i \) is the image distance (distance from the image to the mirror).
For a convex mirror, the focal length \( f \) is half of the radius of curvature, and it's always positive, but the image distance \( i \) is negative because the image forms on the opposite side of the mirror. This is crucial in defining where images form relative to the object and the mirror itself.
Virtual Images
In optics, a virtual image is one that cannot be captured directly on a screen because the light rays do not actually meet but appear to diverge from a common point. Convex mirrors are perfect for creating virtual images due to their shape.

These images are always formed on the side opposite to where the object exists. They appear to originate from a point behind the mirror where the reflected rays seem to diverge. For our specific problem with the coin and the spherical glass shell, the image is virtual because:
  • The image distance is negative.
  • The light rays are diverging.
  • It appears upright and smaller.
This concept is critical when considering mirrors in practical applications like vehicle mirrors, providing a direct visualization that helps in safe driving.
Magnification
Magnification is a measure of how much larger or smaller the image is compared to the object itself. It is determined by the formula:\[m = -\frac{i}{o}\]Where:
  • \( m \) is the magnification,
  • \( i \) is the image distance, and
  • \( o \) is the object distance.
In our convex mirror example, the magnification is calculated to be positive, which indicates that the image is upright relative to the object. If the magnification is greater than 1, as in this exercise where it's 1.67, the image appears larger compared to the object, though it remains virtual and upright. Understanding magnification is essential in applications where image size needs to be more accurately perceived, such as in telescopes and cameras.
Optics
Optics is the branch of physics that studies the behavior and properties of light, including its interactions with matter. It encompasses everything from the design of lenses and mirrors to understanding complex light phenomena. Within the context of convex mirrors and our exercise example, optics helps explain why images form the way they do. It involves:
  • Understanding light reflection and refraction processes.
  • Applying formulas like the mirror equation to practical tasks.
  • Exploring how lenses and mirrors alter visual perception.
Studying optics allows us to create technologies that enhance human vision, develop precise scientific instruments, and design the cameras and microscopes used in various scientific and everyday applications. It's a foundational aspect of understanding how we perceive the world visually.

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Most popular questions from this chapter

(a) For a lens with focal length f, find the smallest distance possible between the object and its real image. (b) Graph the distance between the object and the real image as a function of the distance of the object from the lens. Does your graph agree with the result you found in part (a)

The radii of curvature of the surfaces of a thin converging meniscus lens are \(R_1\) = +12.0 cm and \(R_2\) = +28.0 cm. The index of refraction is 1.60. (a) Compute the position and size of the image of an object in the form of an arrow 5.00 mm tall, perpendicular to the lens axis, 45.0 cm to the left of the lens. (b) A second converging lens with the same focal length is placed 3.15 m to the right of the first. Find the position and size of the final image. Is the final image erect or inverted with respect to the original object? (c) Repeat part (b) except with the second lens 45.0 cm to the right of the first.

You wish to project the image of a slide on a screen 9.00 m from the lens of a slide projector. (a) If the slide is placed 15.0 cm from the lens, what focal length lens is required? (b) If the dimensions of the picture on a 35-mm color slide are 24 mm \(\times\) 36 mm, what is the minimum size of the projector screen required to accommodate the image?

The left end of a long glass rod 8.00 cm in diameter, with an index of refraction of 1.60, is ground and polished to a convex hemispherical surface with a radius of 4.00 cm. An object in the form of an arrow 1.50 mm tall, at right angles to the axis of the rod, is located on the axis 24.0 cm to the left of the vertex of the convex surface. Find the position and height of the image of the arrow formed by paraxial rays incident on the convex surface. Is the image erect or inverted?

A layer of benzene (\(n\) = 1.502) that is 4.20 cm deep floats on water (\(n\) = 1.332) that is 5.70 cm deep. What is the apparent distance from the upper benzene surface to the bottom of the water when you view these layers at normal incidence?

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