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Chapter 34: Problem 29

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?

Short Answer

Expert verified
The object's location is 2.82 cm left of the lens; the focal length is -1.69 cm.

Step by step solution

01

Identify the given quantities and symbols

We have a real object with an actual height \( h = 8.00 \text{ mm} = 0.80 \text{ cm} \). The image height is \( h' = 3.40 \text{ cm} \), and it is erect, which indicates a positive value. The image distance \( d_i = -12.0 \text{ cm} \) because it is on the same side as the object for a converging lens.
02

Use the magnification formula

The magnification formula is \( m = \frac{h'}{h} = \frac{d_i}{d_o} \), where \( d_o \) is the object distance we need to find. Substituting the known values gives \( m = \frac{3.40}{0.80} = \frac{-12.0}{d_o} \).
03

Calculate the object distance \( d_o \)

From \( m = \frac{-12.0}{d_o} = \frac{3.40}{0.80} \), solve for \( d_o \): \[ d_o = \frac{-12.0 \times 0.80}{3.40} = -2.82 \text{ cm} \]. This means the object is positioned 2.82 cm to the left of the lens.
04

Use the lens formula to find focal length

The lens formula is \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). Substitute known values: \( \frac{1}{f} = \frac{1}{-2.82} + \frac{1}{-12.0} \). Calculating this gives its reciprocal: \[ \frac{1}{f} \approx -0.509 + (-0.083) = -0.592 \Rightarrow f \approx \frac{1}{-0.592} \approx -1.69 \text{ cm} \].

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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, commonly known as a convex lens, is a lens that focuses incoming parallel light rays to a single point known as the focus. These lenses are thicker in the middle than at the edges and can be made of various materials like glass or plastic. They are used in a variety of applications, including cameras, eyeglasses, and in scientific instruments like microscopes and telescopes.
A key characteristic of the converging lens is its ability to produce real and virtual images. A real image is formed when light rays physically converge, whereas a virtual image appears when light rays seem to diverge from a point behind the lens. The real image produced by a converging lens can be projected onto a screen, and its properties are dictated by the relative distances of the object and lens.
Focal Length Calculation
Determining the focal length of a converging lens is crucial in understanding its optical nature. The focal length (\( f \)) is the distance between the lens and its focus. It reflects the "strength" of the lens in converging or diverging light.
To find the focal length, we use the lens formula: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \);where \( d_o \) is the object distance and \( d_i \) is the image distance.
In the given problem, the image distance is negative because the image is on the same side of the lens as the object, indicating a virtual image. After solving the formula with the given values, we find that the focal length is approximately -1.69 cm. This negative value indicates that the lens forms a virtual image.
Magnification Formula
The magnification of a lens tells us how much larger or smaller the image is compared to the object. The magnification formula is \( m = \frac{h'}{h} = \frac{d_i}{d_o} \),where \( h' \) is the image height, \( h \) is the object height, \( d_i \) is the image distance, and \( d_o \) is the object distance.
This formula helps in understanding how the image is affected by the positioning of the object. If \( m > 1 \), the image is larger than the object; \( m < 1 \) indicates the image is smaller.
In the problem, substituting the known heights and solving for the object distance helps establish the relative positions of the object and lens, showing \( m = 4.25 \), which means the image is 4.25 times taller than the object.
Image Formation
The formation of images by a lens depends on several factors, including the type of lens, object distance, and light path. For a converging lens like the one in the exercise, an image can be real or virtual, inverted or erect, enlarged or reduced.
In this particular example, the image was erect and taller than the object, implying a virtual image, as virtual images formed by converging lenses are usually upright.
  • A virtual image cannot be projected, as it appears on the same side of the lens as the object.
  • The orientation and size provide essential clues about the image type and the relative distances from the lens.
The calculated values in the exercise confirm the virtual nature and the relative positions of the object and image in relation to the converging lens, offering insights into the practical behavior of optical systems.

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Most popular questions from this chapter

An insect 3.75 mm tall is placed 22.5 \(\mathrm{cm}\) to the left of a thin planoconvex lens. The left surface of this lens is flat, the right surface has a radius of curvature of magnitude \(13.0 \mathrm{cm},\) and the index of refraction of the lens material is 1.70. (a) Calculate the location and size of the image this lens forms of the insect. Is it real or virtual? Erect or inverted? (b) Repeat part (a) if the lens is reversed.

A telescope is constructed from two lenses with focal lengths of 95.0 \(\mathrm{cm}\) and \(15.0 \mathrm{cm},\) the 95.0 -cm lens being used as the objective. Both the object being viewed and the final image are at infinity. (a) Find the angular magnification for the telescope. (b) Find the height of the image formed by the objective of a building 60.0 \(\mathrm{m}\) tall, 3.00 \(\mathrm{km}\) away. (c) What is the angular size of the final image as viewed by an eye very close to the eyepiece?

Focal Length of a Zoom Lens. Figure P34.113 shows a simple version of a zoom lens. The converging lens has focal length \(f_{1},\) and the diverging lens has focal length \(f_{2}=-\left|f_{2}\right|\) The two lenses are separated by a variable distance \(d\) that is always less than \(f_{1} .\) Also, the magnitude of the focal length of the diverging lens satisfies the inequality \(\left|f_{2}\right|>\left(f_{1}-d\right) .\) To determine the effective focal length of the combination lens, consider a bundle of parallel rays of radius \(r_{0}\) entering the converging lens. (a) Show that the radius of the ray bundle decreases to \(r_{0}^{\prime}=r_{0}\left(f_{1}-d\right) / f_{1}\) at the point that it enters the diverging lens. (b) Show that the final image \(I^{\prime}\) is formed a distance \(s_{2}^{\prime}=\left|f_{2}\right|\left(f_{1}-d\right) /\left(\left|f_{2}\right|-f_{1}+d\right)\) to the right of the diverging lens. (c) If the rays that emerge from the diverging lens and reach the final image point are extended backward to the left of the diverging lens, they will eventually expand to the original radius \(r_{0}\) at some point \(Q .\) The distance from the final image \(I^{\prime}\) to the point \(Q\) is the effective focal length \(f\) of the lens combination; if the combination were replaced by a single lens of focal length \(f\) placed at \(Q,\) parallel rays would still be brought to a focus at \(I^{\prime}\) . Show that the effective focal length is given by \(f=f_{1}\left|f_{2}\right| /\left(\left|f_{2}\right|-f_{1}+d\right) .\) (d) If \(f_{1}=12.0 \mathrm{cm}\) \(f_{2}=-18.0 \mathrm{cm},\) and the separation \(d\) is adjustable between 0 and \(4.0 \mathrm{cm},\) find the maximum and minimum focal lengths of the combination. What value of \(d\) gives \(f=30.0 \mathrm{cm} ?\)

The Lens of the Eye. The crystalline lens of the human eye is a double-convex lens made of material having an index of refraction of 1.44 (although this varies). Its focal length in air is about 8.0 \(\mathrm{mm}\) , which also varies. We shall assume that the radii of curvature of its two surfaces have the same magnitude. (a) Find the radii of curvature of this lens. (b) If an object 16 \(\mathrm{cm}\) tall is placed 30.0 \(\mathrm{cm}\) from the eye lens, where would the lens focus it and how tall would the image be? Is this image real or virtual? Is it erect or inverted? (Note: The results obtained here are not strictly accurate because the lens is embedded in fluids having refractive indexes different from that of air.)

Three thin lenses, each with a focal length of \(40.0 \mathrm{cm},\) are aligned on a common axis; adjacent lenses are separated by 52.0 \(\mathrm{cm} .\) Find the position of the image of a small object on the axis, 80.0 \(\mathrm{cm}\) to the left of the first lens.
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