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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

What should be the index of refraction of a transparent sphere in order for paraxial rays from an infinitely distant object to be brought to a focus at the vertex of the surface opposite the point of incidence?

Dental Mirror. A dentist uses a curved mirror to view teeth on the upper side of the mouth. Suppose she wants an enect image with a magnification of 2.00 when the mirror is 1.25 \(\mathrm{cm}\) from a tooth. (Treat this problem as though the object and image lie along a straight line.) (a) What kind of mirror (concave or convex) is needed? Use a ray diagram to decide, without performing any calculations. (b) What must be the focal length and radius of curvature of this mirror?(c) Draw a principal-ray diagram to check your answer in part (b).

A converging lens forms an image of an 8.00 -mm-tall real object. The image is 12.0 \(\mathrm{cm}\) to the left of the lens, 3.40 \(\mathrm{cm}\) tall, and erect. What is the focal length of the lens? Where is the object located?

A camera with a 90 -mm-focal-length lens is focused on an object 1.30 \(\mathrm{m}\) from the lens. To refocus on an object 6.50 \(\mathrm{m}\) from the lens, by how much must the distance between the lens and the film be changed? To refocus on the more distant object, is the lens moved toward or away from the film?

A photographer takes a photograph of a Boeing 747 airliner (length 70.7 \(\mathrm{m} )\) when it is flying directly overbead an altitude of 9.50 \(\mathrm{km}\) . The lens has a focal length of 5.00 \(\mathrm{m}\) . How long is the image of the airliner on the film?
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