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Chapter 26: Problem 105

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).

Short Answer

Expert verified
The image distance is -75.0 cm, and the magnification is 2.5.

Step by step solution

01

Understand the Problem

We have an object situated 30.0 cm to the left of a converging lens with a focal length of 50.0 cm. We need to find the image distance and magnification using a ray diagram and verify with equations.
02

Sketch the Ray Diagram

Draw a horizontal line to represent the lens's principal axis. Mark the lens's position and locate the object on the axis at 30.0 cm to the left of the lens. Use the focal length (50.0 cm) to mark the focal points on both sides of the lens. Draw the following rays: a parallel ray from the top of the object that refracts and passes through the focal point on the opposite side, a ray through the center of the lens that continues straight, and a third ray passing through the lens's focal point towards the lens that exits parallel to the principal axis. The intersection of these rays on the other side of the lens indicates the image position.
03

Analyze the Ray Diagram

Observe where the rays converge to determine the image's position and height. This convergence point gives us the image's location relative to the lens. Since the focal length is longer than the object distance, the rays should appear to diverge on the same side as the object, indicating a virtual image.
04

Apply the Thin-Lens Equation

Use the thin-lens equation \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f = 50.0 \) cm, \( d_o = 30.0 \) cm, and \( d_i \) is the image distance we need to find. Rearrange to solve for \( d_i \): \( \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} = \frac{1}{50} - \frac{1}{30} \). Calculate \( d_i \) to find the image distance.
05

Calculate the Image Distance

Substitute the focal length and object distance into the equation: \( \frac{1}{d_i} = \frac{1}{50.0} - \frac{1}{30.0} \). This evaluates to \( \frac{1}{d_i} = \frac{30 - 50}{1500} = -\frac{20}{1500} = -\frac{1}{75} \), giving \( d_i = -75.0 \) cm, indicating a virtual image 75 cm to the left of the lens.
06

Calculate the Magnification

The magnification equation is \( m = -\frac{d_i}{d_o} \). Substitute \( d_i = -75.0 \) cm and \( d_o = 30.0 \) cm into the formula to get \( m = -\frac{-75.0}{30.0} = 2.5 \). This means the image is magnified by a factor of 2.5 and is upright.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ray Diagram
A ray diagram is a visual tool used to locate an image formed by a lens. It relies on drawing specific rays from the top of the object to determine where they meet (or appear to meet) after refracting through a lens. Here are the key steps to draw a ray diagram:
  • Identify the principal axis, which is a horizontal line passing through the center of the lens.
  • Mark the object position to the left of the lens and the focal points on each side of the lens (50 cm from the lens in this case).
  • Draw a parallel ray from the top of the object towards the lens. Upon reaching the lens, this ray refracts and passes through the focal point on the opposite side.
  • Draw a ray through the object and the center of the lens, continuing in a straight line as this ray is undeviated.
  • Finally, draw a ray passing through the focal point on the object's side towards the lens; after refraction, this ray travels parallel to the principal axis.
The point where these refracted rays intersect gives the location and size of the image. If the rays diverge, extend them backward to find the point where they appear to originate, indicating a virtual image.
Image Distance
The image distance is the distance from the lens to the image formed. It's determined through both the ray diagram and the thin-lens equation. Using the thin-lens equation:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]where:
  • \( f \) is the focal length of the lens (50 cm).
  • \( d_o \) is the object distance from the lens (30 cm).
  • \( d_i \) is the image distance. Solving it gives:
\[ \frac{1}{d_i} = \frac{1}{50} - \frac{1}{30}\]\[ \frac{1}{d_i} = \frac{30 - 50}{1500} = -\frac{1}{75} \]So, the image distance \( d_i \) is -75 cm, indicating a virtual image on the same side as the object, 75 cm from the lens. The negative sign denotes that the image is virtual.
Magnification
Magnification tells us how much larger or smaller the image is compared to the object. It also indicates the image orientation: whether it is upright or inverted. The magnification \( m \) is calculated using:\[ m = -\frac{d_i}{d_o} \]In our scenario:
  • The image distance \( d_i \) is -75 cm (from previous calculation).
  • The object distance \( d_o \) is 30 cm.
Substituting these values gives:\[ m = -\frac{-75}{30} = 2.5\]This means the image is magnified by a factor of 2.5, indicating the image is 2.5 times the size of the object. Since the magnification is positive, the image is upright and virtual.

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Most popular questions from this chapter

In a compound microscope, the focal length of the objective is 3.50 \(\mathrm{cm}\) and that of the eyepiece is \(6.50 \mathrm{cm} .\) The distance between the lenses is 26.0 cm. (a) What is the angular magnification of the microscope if the person using it has a near point of \(35.0 \mathrm{cm} ?\) (b) If, as usual, the first image lies just inside the focal point of the eyepiece (see Figure 26.32 ), how far is the object from the objective? (c) What is the magnification (not the angular magnification) of the objective?

Horizontal rays of red light \((\lambda=660 \mathrm{nm},\) in vacuum) and violet light \((\lambda=410 \mathrm{nm},\) in vacuum) are incident on the flint-glass prism shown in the drawing. The indices of refraction for the red and violet light are \(n_{\text {red }}=1.662\) and \(n_{\text {violet}}=1.698 .\) The prism is surrounded by air. What is the angle of refraction for each ray as it emerges from the prism?

A layer of liquid \(B\) floats on liquid \(A\). A ray of light begins in liquid \(A\) and undergoes total internal reflection at the interface between the liquids when the angle of incidence exceeds \(36.5^{\circ} .\) When liquid \(B\) is replaced with liquid \(C,\) total internal reflection occurs for angles of incidence greater than \(47.0^{\circ} .\) Find the ratio \(n_{B} / n_{C}\) of the refractive indices of liquids \(B\) and \(C\).

A ray of sunlight is passing from diamond into crown glass; the angle of incidence is \(35.00^{\circ} .\) The indices of refraction for the blue and red components of the ray are: blue \(\left(n_{\text {diamond }}=2.444, n_{\text {crown glass }}=1.531\right),\) and red \(\left(n_{\text {diamond }}=2.410, n_{\text {crown glass }}=1.520\right) .\) Determine the angle between the refracted blue and red rays in the crown glass.

Two converging lenses \(\left(f_{1}=9.00 \mathrm{cm}\right.\) and \(\left.f_{2}=6.00 \mathrm{cm}\right)\) are separated by \(18.0 \mathrm{cm} .\) The lens on the left has the longer focal length. An object stands \(12.0 \mathrm{cm}\) to the left of the left-hand lens in the combination. (a) Locate the final image relative to the lens on the right. (b) Obtain the overall magnification. (c) Is the final image real or virtual? With respect to the original object, (d) is the final image upright or inverted and (e) is it larger or smaller?
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