Problem 11 - College Physics

Question

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.

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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: \( h_o = 1.5 \text{ cm} \).

Step 3: 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 \( f = \frac{R}{2} = \frac{18.0}{2} = 9.0 \text{ cm} \).

Step 4: 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):\[ \frac{1}{p} = \frac{1}{9.0} + \frac{1}{6.0} \]\[ \Rightarrow p = 3.6 \text{ cm} \]

Step 5: 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:\[ h_i = m \times h_o = 1.67 \times 1.5 = 2.5 \text{ cm} \]

Step 6: Determine the Orientation of the Image

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

Step 7: 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.

Key Concepts

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.