Problem 31 - College Physics

Question

A 15-cm-long pencil is placed with its eraser on the optic axis of a concave mirror and its point directed upward at a distance of \(20 \mathrm{~cm}\) in front of the mirror. The radius of curvature of the mirror is \(30 \mathrm{~cm}\). Use (a) a ray diagram and (b) the mirror equation to locate the image and determine the image characteristics.

Step-by-Step Solution

Answer

The image is located 60 cm in front of the mirror, real, inverted, and magnified by 3 times.

Step 1: Understanding the Problem

The problem involves a concave mirror with a pencil placed in front of it. We need to find the image position, height, and characteristics using a ray diagram and mirror equation. Given values are: object distance \(u = -20\mathrm{~cm}\), object height \(h_o = 15\mathrm{~cm}\), and mirror radius of curvature \(R = 30\mathrm{~cm}\). Compute the focal length using \(f = \frac{R}{2}\).

Step 2: Calculate the Focal Length

The focal length \(f\) of the mirror is given by \(f = \frac{R}{2}\). Thus, \(f = \frac{30\mathrm{~cm}}{2} = 15\mathrm{~cm}\).

Step 3: Draw the Ray Diagram

1. Draw the principal axis and locate the mirror's center of curvature and principal focus. 2. Render a horizontal line from the pencil's tip running parallel to the principal axis. 3. Reflect it through the focal point. 4. Sketch a line straightly from the pencil's tip through the focal point, reflecting parallel to the principal axis. 5. The image is formed where these reflected rays meet. Reflective analysis shows the image is 15 cm in front of the mirror, inverted, and reduced.

Step 4: Use the Mirror Equation

The mirror equation is \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\). Substitute \(f = 15\mathrm{~cm}\) and \(u = -20\mathrm{~cm}\):\[ \frac{1}{15} = \frac{1}{v} - \frac{1}{20} \]Clear fractions and solve for \(v\):\(v = 60\mathrm{~cm}\).

Step 5: Determine Image Characteristics

The magnification \(m\) is given by \(m = -\frac{v}{u} = -\frac{60}{-20} = 3\). The image height \(h_i = m \times h_o = 3 \times 15\mathrm{~cm} = 45\mathrm{~cm}\). The image is larger, inverted, and real.

Key Concepts

Focal Length

The focal length is an essential parameter when working with mirrors, especially concave ones. It describes the distance from the mirror to the focal point, where parallel rays of light either converge or appear to diverge. To find the focal length, we often rely on the radius of curvature, the distance from the mirror's surface to its center of curvature. For a concave mirror, the focal length (\(f\)) is half of the radius of curvature (\(R\)). Hence, the formula for focal length is \(f = \frac{R}{2}\).
This means if a mirror has a radius of curvature of \(30\text{ cm}\), the focal length will be \(15\text{ cm}\).
Understanding focal length is critical, as it helps predict where light rays will focus and the nature of the images formed.

Ray Diagram

Drawing a ray diagram is a visual tool to understand how a mirror forms an image. It involves tracing light rays from the object to see where they converge after reflecting off the mirror. Here’s a simplified process:

• Begin by drawing a horizontal principal axis from the mirror through the center of curvature and focal point.
• Place the object (like a pencil) above the principal axis, with its base touching the axis.
• Trace a ray from the top of the object parallel to the principal axis. After hitting the mirror, this ray reflects and passes through the focal point.
• Draw a second ray from the top through the focal point; this ray will reflect back parallel to the principal axis.
• The intersection of these reflected rays shows where the image forms.

The image in our exercise forms in front of the mirror, indicating it is real, inverted, and reduced. This visualization aids in the physical understanding of object-image relationships.

Mirror Equation

The mirror equation is a mathematical formula that links the focal length, object distance, and image distance for spherical mirrors. It is given by:\[\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\]where \(f\) is the focal length, \(v\) is the image distance, and \(u\) is the object distance.
To find the image distance, you rearrange the equation to solve for \(v\), considering the object distance and focal length.
For example, with \(f = 15 \text{ cm}\) and \(u = -20 \text{ cm}\), substituting these values gives:\[\frac{1}{15} = \frac{1}{v} - \frac{1}{20}\]Solution to these equations helps find the image distance as \(v = 60 \text{ cm}\).
This calculation not only determines where the image forms but also aids in analyzing the image's nature, whether real or virtual, inverted or upright.