Problem 103 - Physics

Question

An object is located \(30.0 \mathrm{~cm}\) to the left of a converging lens whose focal length is \(50.0 \mathrm{~cm} .\) (a) Draw a ray diagram to scale and from it determine the image distance and the magnification, (b) Use the thin-lens and magnification equations to verify your answers to part (a).

Step-by-Step Solution

Answer

Image distance is \(18.75\,\mathrm{cm}\) to the left; magnification is 0.625.

Step 1: Understanding the Problem

We have a converging lens with a focal length of \(f = 50.0\,\mathrm{cm}\) and an object placed \(30.0\,\mathrm{cm}\) to the left of the lens. Our task is to determine the image distance and magnification using a ray diagram and verify these with the thin-lens and magnification equations.

Step 2: Ray Diagram Construction

First, sketch a principal axis, mark the optical center of the lens, and place the focal points on either side at \(50.0 \mathrm{~cm}\). Position the object \(30.0 \mathrm{~cm}\) to the left of the lens. Draw principal rays: (1) a ray parallel to the axis that refracts through the focal point on the opposite side, (2) a ray through the lens center that continues straight, and (3) a ray through the focal point on the object's side becoming parallel post lens. These rays should converge at a point on the object's right that indicates the image location.

Step 3: Analyzing the Ray Diagram

From the ray diagram, estimate the distance from the lens to the image formed where the rays meet. You may find the image is formed on the same side as the object, suggesting a virtual image. Measure the distance from the lens to this intersection. Also, measure the image height to estimate magnification.

Step 4: Using the Thin-Lens Equation

Apply the thin-lens equation: \[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]where \(f = 50.0 \mathrm{~cm}\) and \(d_o = -30.0 \mathrm{~cm}\) (negative as it's to the left). Solve for \(d_i\) to find the image distance.

Step 5: Solving for Image Distance

Rearrange the thin-lens equation to:\[\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} = \frac{1}{50} - \frac{1}{-30}\]Calculate \(\frac{1}{d_i} = \frac{1}{50} + \frac{1}{30}\), which gives \(\frac{1}{d_i} = \frac{3+5}{150} = \frac{8}{150}\). Thus, \(d_i = \frac{150}{8} = 18.75\,\mathrm{cm}\), indicating a virtual image \(18.75\,\mathrm{cm}\) to the left.

Step 6: Calculating Magnification

Use the magnification equation:\[m = -\frac{d_i}{d_o}\]Substituting \(d_i = -18.75 \mathrm{~cm}\) and \(d_o = -30.0 \mathrm{~cm}\), find\[m = -\frac{-18.75}{-30.0} = 0.625\].This suggests the image is smaller and upright, with a magnification of 0.625.

Key Concepts

Ray Diagram

A ray diagram is a graphical representation used to determine the size and position of an image formed by a lens. In the context of a converging lens, this diagram involves a few key steps:
First, draw a horizontal line to represent the principal axis, and mark the optical center of the lens. On this line, position the focal points on both sides of the lens at a distance equal to the focal length, which is 50 cm in our case.
Next, place the object 30 cm to the left of the lens. You'll then draw three critical rays from the top of the object:

• A ray parallel to the principal axis, which refracts through the focal point on the opposite side.
• A ray through the center of the lens, which passes straight without deviation.
• A ray towards the focal point on the object's side, that becomes parallel to the principal axis after passing through the lens.

Where these rays converge gives the location of the image. If they extend back to intersect, this indicates a virtual image, often on the same side as the object.

Thin-Lens Equation

The thin-lens equation is a pivotal tool for analyzing lenses' performance. This equation states:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]Where:

• \( f \) is the focal length of the lens.
• \( d_o \) is the object distance from the lens (negative if the object is to the left of the lens).
• \( d_i \) is the image distance from the lens.

By rearranging for \( d_i \), the image distance, we can determine where the image will be located. For example, substituting the exercise values:\[ \frac{1}{d_i} = \frac{1}{50} - \frac{1}{-30} = \frac{8}{150} \]Solving this gives:\[ d_i = \frac{150}{8} = 18.75 \, \text{cm} \]Thus, indicating the position of the image relative to the lens.

Image Distance

The image distance \( (d_i) \) is the measured length from the lens to the image position. It plays a crucial role in image characteristics.In our converging lens problem, this distance was calculated using the thin-lens equation. It resulted in an image distance of 18.75 cm. Since the calculated value is positive, it suggests the image is virtual and appears on the same side of the lens as the object.
Understanding the sign convention is essential:

• A positive \( d_i \) usually implies a real image on the opposite side of the object.
• A negative \( d_i \) suggests a virtual image on the object's side, as seen in this scenario.

This knowledge allows one to predict not just where an image forms but also its nature—real or virtual.

Magnification

Magnification is a measure of how much larger or smaller the image is compared to the object. To calculate magnification \( (m) \), use the formula:\[ m = -\frac{d_i}{d_o} \]Where:

• \( d_i \) is the image distance.
• \( d_o \) is the object distance.

This formula incorporates the negative sign to indicate the image's orientation relative to the object. For our problem:\[ m = -\frac{-18.75}{-30.0} = 0.625 \]The magnification value of 0.625 indicates the image is smaller than the object, specifically 62.5% of its size, and upright since the magnification is positive.By understanding magnification:

• A positive value implies an upright image.
• A negative value indicates an inverted image.
• A value larger than 1 suggests an enlarged image.
• A value between 0 and 1 denotes a reduced image.

This calculation is key to understanding the image's size and orientation.