Problem 8 - University Physics with Modern Physics

Question

An object is 24.0 \(\mathrm{cm}\) from the center of a silvered spherical glass Christmas tree ornament 6.00 \(\mathrm{cm}\) in diameter. What are the position and magnification of its image?

Step-by-Step Solution

Answer

The image is at 1.60 cm from the mirror and is inverted with a magnification of -0.0667.

Step 1: Understand the Terms

The problem involves finding the position and magnification of an image through a spherical mirror. The diameter of the ornament is given, allowing us to calculate the radius. The object distance is provided.

Step 2: Calculate the Radius of the Ornament

The diameter of the spherical ornament is 6.00 cm. The radius \( R \) is half of the diameter: \( R = \frac{6.00}{2} = 3.00 \) cm.

Step 3: Determine the Focal Length

For a spherical mirror, the focal length \( f \) is half of the radius of curvature, \( f = \frac{R}{2} \). Since this is a silvered ornament, it acts as a concave mirror: \( f = \frac{3.00}{2} = 1.50 \) cm.

Step 4: Use the Mirror Formula

The mirror formula is \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( d_o = 24.0 \) cm (object distance) and \( f = 1.50 \) cm. Thus, \( \frac{1}{1.50} = \frac{1}{24.0} + \frac{1}{d_i} \).

Step 5: Solve for Image Position \( d_i \)

Rearrange the mirror formula to find \( d_i \): \( \frac{1}{d_i} = \frac{1}{1.50} - \frac{1}{24.0} \). Calculate to find \( d_i \).

Step 6: Perform the Calculations for \( d_i \)

\( \frac{1}{1.50} = 0.6667 \) and \( \frac{1}{24.0} = 0.0417 \). Thus, \( \frac{1}{d_i} = 0.6667 - 0.0417 = 0.625 \). Solve for \( d_i \): \( d_i = \frac{1}{0.625} = 1.60 \) cm.

Step 7: Calculate Magnification

Magnification \( m \) is given by \( m = -\frac{d_i}{d_o} \). Substitute \( d_i = 1.60 \) cm and \( d_o = 24.0 \) cm: \( m = -\frac{1.60}{24.0} \).

Step 8: Perform the Calculation for Magnification

Solve the expression for magnification: \( m = -0.0667 \). The negative sign indicates the image is inverted.

Key Concepts

Focal Length

Spherical mirrors come in two types: concave and convex. The focal length is a critical property of these mirrors, which is the distance from the mirror's surface to its focal point.

• The focal point is where light rays parallel to the mirror's principal axis converge (for concave mirrors) or appear to diverge from (for convex mirrors).
• To calculate the focal length (\( f \)), you use the radius of curvature (\( R \)) of the mirror.

For spherical mirrors, the focal length is half of the radius of curvature:\[f = \frac{R}{2}\]In the given problem, the radius (\( R \)) is determined by dividing the diameter of the ornament by 2, which results in 3.00 cm. Thus, the focal length is calculated as 1.50 cm, showing how the focal length links directly to the size of the mirror.

Mirror Formula

The mirror formula is an essential equation that relates an object's distance from the mirror, the image's distance from the mirror, and the mirror's focal length. It is expressed as:\[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]Where:

• \( f \) is the focal length.
• \( d_o \) is the object distance (distance from the object to the mirror).
• \( d_i \) is the image distance (distance from the image to the mirror).

Using the mirror formula in the example, we substitute \( f = 1.50 \) cm and \( d_o = 24.0 \) cm. This setup starts the problem-solving process for finding \( d_i \), giving you insight into the mirror's focal length helping decide where the image will form. This formula is a vital tool for determining both the image position and its nature (real or virtual).

Image Magnification

Magnification in mirrors describes how much larger or smaller the image is compared to the actual object. It also indicates the orientation of the image.

• The magnification (\( m \)) is calculated using the formula:\[m = -\frac{d_i}{d_o}\]
• The negative sign indicates that if the image is inverted, it will have a negative magnification.
• A magnification greater than 1 means the image is larger, while less than 1 suggests it is smaller compared to the object.

In this example, the object is 24.0 cm away, and the image is 1.60 cm on the same side due to the concave nature of the mirror. Thus, the magnification becomes \( m = -0.0667 \), showing that the image is significantly smaller than the object and also inverted. This understanding of magnification provides deeper insight into the nature of images produced by spherical mirrors.