Problem 45 - College Physics

Question

An object \(4.0 \mathrm{~cm}\) tall is in front of a converging lens of focal length \(22 \mathrm{~cm}\). The object is \(15 \mathrm{~cm}\) away from the lens. (a) Use a ray diagram to determine whether the (2) upright or inverted, and image is (1) real or virtual, (3) magnified or reduced. (b) Calculate the image distance and lateral magnification.

Step-by-Step Solution

Answer

The image is virtual, upright, and magnified; the image distance is -47.14 cm and magnification is 3.14.

Step 1: Understand the Given Problem

We are given a converging lens with focal length \( f = 22 \; \mathrm{cm} \). The object's height is \( 4.0 \; \mathrm{cm} \) and placed at \( 15 \; \mathrm{cm} \) from the lens. We need to find the image characteristics and perform calculations for image distance and magnification using these values.

Step 2: Ray Diagram Analysis

A ray diagram for a converging lens can be drawn to determine the image characteristics. First, draw the principal axis, lens, and focal points on both sides of the lens. Place the object at \( 15 \; \mathrm{cm} \) from the lens. Draw rays from the top of the object: one parallel to the principal axis, bending through the focal point on the opposite side; one through the lens center, continuing straight. The rays diverge on the opposite side of the lens, confirming a virtual image formed there. This indicates the image is virtual, upright, and magnified because it appears larger.

Step 3: Use the Lens Formula

The lens formula is \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( d_o \) is object distance (\( 15 \; \mathrm{cm} \)) and \( d_i \) is image distance. Substitute the known values:\[ \frac{1}{22} = \frac{1}{15} + \frac{1}{d_i} \] Solve for \( \frac{1}{d_i} \): \[ \frac{1}{d_i} = \frac{1}{22} - \frac{1}{15} \]\[ = \frac{15 - 22}{330} \]\[ = \frac{-7}{330} \]Thus, \( d_i = -47.14 \; \mathrm{cm} \). The negative sign indicates that the image is virtual.

Step 4: Calculate Lateral Magnification

The magnification formula is \( M = -\frac{d_i}{d_o} \).Substitute \( d_i = -47.14 \; \mathrm{cm} \) and \( d_o = 15 \; \mathrm{cm} \):\[ M = -\frac{-47.14}{15} \]\[ M = 3.14 \]. Therefore, the image is magnified by a factor of 3.14.

Key Concepts

Ray Diagram

A ray diagram is a visual tool used to determine the image characteristics formed by a lens. To construct one for a converging lens, begin by drawing the lens with its principal axis and mark the lens's focal points on both sides.

Place the object on the axis, 15 cm from the lens, corresponding to the given object distance. Draw a ray from the top of the object parallel to the principal axis. This ray will refract and pass through the focal point on the other side of the lens. Another ray goes straight through the center of the lens without deviating.

• Where these rays appear to diverge on the same side as the object post-refraction, a virtual image is indicated.
• Such an image is upright, appearing right-side-up compared to the object.
• Additionally, if the rays appear to converge on the same side, it means the image is magnified.

Constructing a proper ray diagram helps us understand the nature of the image formed without detailed calculations.

Lens Formula

The lens formula is crucial in determining the image distance for a lens. The formula is given by:

\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]

where \( f \) is the focal length of the lens, \( d_o \) is the object distance, and \( d_i \) is the image distance.

By rearranging the formula, you'll calculate the image distance \( d_i \) when the object distance and focal length are known. Using the provided values, substitute \( f = 22 \; \mathrm{cm} \) and \( d_o = 15 \; \mathrm{cm} \) into the formula to find \( d_i \).
The process results in:
\[ \frac{1}{d_i} = \frac{1}{22} - \frac{1}{15} \]
\[ d_i = -47.14 \; \mathrm{cm} \]
A negative \( d_i \) signifies that the image is virtual, as it forms on the same side as the object. Understanding and using the lens formula is fundamental in optics to predict image locations accurately.

Image Characteristics

Image characteristics describe the properties of the image formed by a lens

• Real or Virtual: A real image occurs on the opposite side to the object, while a virtual image forms on the same side.
• Upright or Inverted: Images that appear right-side-up are upright, whereas those upside down are inverted.
• Magnified or Reduced: This depends on the size of the image relative to the object size.

In our exercise, the virtual and upright nature of the image denotes it is on the same side as the object and appears right-side-up.
The magnification, discussed below, determines how enlarged or diminished the image is compared to its object.

Lateral Magnification

Lateral magnification describes how much larger or smaller the image is compared to the object. It's represented with the formula:

\[ M = -\frac{d_i}{d_o} \]

where \( d_i \) is the image distance and \( d_o \) is the object distance.

By plugging in the values for this problem:
\[ M = -\frac{-47.14}{15} = 3.14 \]
This positive magnification implies the image is upright and greater than \(1\) indicating magnification. Here, the image is larger by a factor of 3.14 compared to the object.

• Positive magnification values explain the upright nature of the image.
• Values greater than 1 suggest the image is larger or magnified.

Understanding lateral magnification lets us measure how much an image enlarges or shrinks in size relative to the actual object.