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Chapter 34: Problem 26

A converging lens with a focal length of 90.0 \(\mathrm{cm}\) forms an image of a 3.20 -cm-tall real object that is to the left of the lens. The image is 4.50 \(\mathrm{cm}\) tall and inverted. Where are the object and image located in relation to the lens? Is the image real or virtual?

Short Answer

Expert verified
Object is at 219.38 cm, image at 308.54 cm opposite the object; image is real.

Step by step solution

01

Understand the lens formula

The lens formula is given as \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length, \( d_o \) is the object distance, and \( d_i \) is the image distance. We know \( f = 90.0 \, \text{cm} \), we just need to find \( d_o \) and \( d_i \).
02

Use the magnification formula

The magnification \( m \) is given by \( m = \frac{h_i}{h_o} = -\frac{d_i}{d_o} \). We know the height of the object \( h_o = 3.20 \, \text{cm} \) and the height of the image \( h_i = -4.50 \, \text{cm} \) (since it is inverted), hence we can calculate \( m = \frac{-4.50}{3.20} \).
03

Calculate magnification

Calculate the magnification using the formula: \( m = \frac{-4.50}{3.20} = -1.40625 \). This indicates the image is inverted and larger than the object.
04

Relate magnification to distances

Using the magnification relation \( m = -\frac{d_i}{d_o} \), plug in the value of \( m \) from Step 3: \( -1.40625 = -\frac{d_i}{d_o} \), which simplifies to \( d_i = 1.40625 \times d_o \).
05

Solve the system of equations

Substitute \( d_i = 1.40625 \times d_o \) into the lens formula: \( \frac{1}{90} = \frac{1}{d_o} + \frac{1}{1.40625 \times d_o} \). This simplifies to a single equation with \( d_o \) as the unknown.
06

Solve for object distance \(d_o\)

Combine and solve the equation: \( \frac{1}{d_o} + \frac{1}{1.40625 d_o} = \frac{1}{90} \), which leads to \( d_o = 219.38 \, \text{cm} \) after simplification.
07

Calculate image distance \(d_i\)

Plug the value of \( d_o \) into \( d_i = 1.40625 \times d_o \): \( d_i = 1.40625 \times 219.38 \approx 308.54 \, \text{cm} \).
08

Determine the nature of the image

Since \( d_i \) is positive, the image is real and located on the opposite side of the light source.

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Key Concepts

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

Converging Lens
A converging lens is a type of lens that brings parallel rays of light to a single point of focus, known as the focal point.
This is why it's also referred to as a convex lens.
Converging lenses are widely used in all sorts of optical devices, such as cameras, magnifying glasses, and glasses for correcting farsightedness. When we place an object in front of a converging lens, it can form either a real or virtual image.
  • Real images are formed when light actually converges at a point. These images are inverted.
  • Virtual images, on the other hand, are formed when light rays only appear to converge. These images are upright.
In our original exercise, since the image formed is inverted and the image distance turned out to be positive, it confirms that we have a real image.
Lens Formula
The lens formula is a crucial equation in geometrical optics that relates the object distance (\( d_o \)), image distance (\( d_i \)), and the focal length (\( f \)) of a lens.
It is given by: \[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]This formula helps us understand how a lens will behave with given objects and conditions. To solve the original problem, we needed to apply this formula to find both the object and image distances.
Initially, we had the focal length of 90.0 cm.
After calculating other pieces of the equation, we found out that the object distance (\( d_o \)) is 219.38 cm, and the image distance (\( d_i \)) is 308.54 cm.Understanding the lens formula allows us not only to find out where images will form but also to determine what type of image will form, such as real or virtual.
Magnification
Magnification in optics helps us understand how much larger or smaller the image is compared to the actual object.
It is a dimensionless number given by the formula:\[m = \frac{h_i}{h_o} = -\frac{d_i}{d_o}\]where \( h_i \) is the image height and \( h_o \) is the object height.
In the original problem, the image height was negative because the image was inverted, thus, \( m \) was negative.The magnitude of magnification tells us the size comparison:
  • If \( |m| > 1 \), the image is larger than the object.
  • If \( |m| < 1 \), the image is smaller.
  • If \( m > 0 \), the image is upright, hence virtual.
  • If \( m < 0 \), the image is inverted, hence real.
Calculating it, we found that the magnification was approximately -1.40625, meaning the image was inverted and larger than the object.

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Most popular questions from this chapter

Three-Dimensional Image. The longitudinal magnification is defined as \(m^{\prime}=d s^{\prime} / d s .\) It relates the longitudinal dimension of a small object to the longitudinal dimension of its image. (a) Show that for a spherical mirror, \(m^{\prime}=-m^{2}\) . What is the significance of the fact that \(m^{\prime}\) is always negative? (b) A wire frame in the form of a small cube 1.00 \(\mathrm{mm}\) on a side is placed with its center on the axis of a concave mirror with radius of curvature 150.0 \(\mathrm{cm}\) . The sides of the cube are all either parallel or perpendicular to the axis. The cube face toward the mirror is 200.0 \(\mathrm{cm}\) to the left of the mirror vertex. Find (i) the location of the image of this face and of the opposite face of the cube; (ii) the lateral and longitudinal magnifications; (iii) the shape and dimensions of each of the six faces of the image.

A concave mirror is to form an image of the filament of a headlight lamp on a screen 8.00 \(\mathrm{m}\) from the mirror. The filament is 6.00 \(\mathrm{mm}\) tall, and the image is to be 36.0 \(\mathrm{cm}\) tall. (a) How far in front of the vertex of the mirror should the filament be placed? (b) What should be the radius of curvature of the mirror?

A solid glass hemisphere of radius 12.0 \(\mathrm{cm}\) and index of refraction \(n=1.50\) is placed with its flat face downward on a table. A parallel beam of light with a circular cross section 3.80 \(\mathrm{mm}\) in diameter travels straight down and enters the hemisphere at the center of its curved surface. (a) What is the diameter of the circle of light formed on the table? (b) How does your result depend on the radius of the hemisphere?

You want to view an insect 2.00 \(\mathrm{mm}\) in length through a magnifier. If the insect is to be at the focal point of the magnifier. what focal length will give the image of the insect an angular size of 0.025 radian?

Choosing a Camera Lens. The picture size on ordinary \(35-\mathrm{mm}\) camera film is \(24 \mathrm{mm} \times 36 \mathrm{mm}\) . Focal lengths of lenses available for \(35-\mathrm{mm}\) cameras typically include \(28,35,50\) (the "normal" lens), \(85,100,135,200,\) and \(300 \mathrm{mm},\) among others. Which of these lenses should be used to photograph the following objects, assuming that the object is to fill most of the picture area? (a) a building 240 \(\mathrm{m}\) tall and 160 \(\mathrm{m}\) wide at a distance of 600 \(\mathrm{m}\) , and (b) a mobile home 9.6 \(\mathrm{m}\) in length at a distance of 40.0 \(\mathrm{m}\) .
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