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Chapter 26: Problem 56

An object is placed in front of a converging lens in such a position that the lens \((f=12.0\) \(\mathrm{cm}\) ) creates a real image located \(21.0 \mathrm{~cm}\) from the lens. Then, with the object remaining in place, the lens is replaced with another converging lens \((f=16.0 \mathrm{~cm})\). A new, real image is formed. What is the image distance of this new image?

Short Answer

Expert verified
The new image distance is approximately 37.3 cm.

Step by step solution

01

Understand the Lens Formula

The lens formula relates the object distance \(d_o\), the image distance \(d_i\), and the focal length \(f\) of a lens. It is given by the equation \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\). We will use this formula to find the initial object distance with the first lens.
02

Apply the Lens Formula for the First Lens

Given the focal length \(f_1 = 12.0\) cm and the image distance \(d_{i1} = 21.0\) cm, we can find the object distance \(d_o\) using the lens formula: \[\frac{1}{12.0} = \frac{1}{d_o} + \frac{1}{21.0}\]. Rearranging gives \[\frac{1}{d_o} = \frac{1}{12.0} - \frac{1}{21.0}\]. Calculate \(d_o\).
03

Calculate Object Distance

Computing \(\frac{1}{d_o} = \frac{1}{12.0} - \frac{1}{21.0} = \frac{7-4}{84} = \frac{3}{84}\). Thus, \(d_o = \frac{84}{3} = 28.0\) cm. This is the object distance for both lenses since the object remains in the same place.
04

Apply the Lens Formula for the Second Lens

Now using the second lens with focal length \(f_2 = 16.0\) cm and the same object distance \(d_o = 28.0\) cm: \[\frac{1}{16.0} = \frac{1}{28.0} + \frac{1}{d_{i2}}\]. Rearrange and solve for \(d_{i2}\).
05

Solve for the New Image Distance

Calculating \(\frac{1}{d_{i2}} = \frac{1}{16.0} - \frac{1}{28.0} = \frac{7 - 4}{112} = \frac{3}{112}\). Hence, \(d_{i2} = \frac{112}{3} \approx 37.3\) cm. The image distance for the new lens is \(37.3\) 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, also known as a convex lens, has the unique ability to focus parallel beams of light to a point known as the focal point. It is thicker in the middle and thinner at the edges, shaping the light rays to converge, or come together. This feature makes it highly useful for forming images in devices like cameras, microscopes, and even the human eye.
For educational problems like the one we are solving, understanding the converging lens is critical as it directly affects how we calculate image formation using the lens formula. When an object is placed in front of a converging lens, depending on the distance, it can form a real and inverted image on the other side of the lens.
Key characteristics of a converging lens include:
  • Ability to form real images.
  • Convergence of parallel light rays to a focus point.
  • Variable effects on images depending on object distance and focal length.
Focal Length
The focal length of a lens is the distance between the center of the lens and its focal point. It indicates how strongly the lens converges or diverges light. A shorter focal length denotes a stronger lens, bending rays more sharply to focus them at a closer point.
In our exercise, we work with two different focal lengths, 12.0 cm and 16.0 cm, to understand how changing lenses affects image formation. Knowing the focal length allows us to use the lens formula to determine the image and object distances.
Important points about focal length:
  • A smaller focal length means a stronger converging effect.
  • It affects the size and position of the resulting image.
  • Ideal for calculating the behavior of lenses in problems.
Real Image
A real image is formed when light rays converge at a point and can be projected on a screen. This contrasts with a virtual image, which cannot be projected as it forms on the same side as the object. In the context of a converging lens, as used in our problem, the real image is inverted and located on the opposite side of the object.
In practice, a real image is essential in applications like cameras and projectors, where capturing and displaying an image is necessary.
Characteristics of a real image formed by a converging lens include:
  • Formed when light rays converge.
  • Can be captured on a screen.
  • Inverted in contrast to the object's orientation.
Object Distance
Object distance refers to the distance between the object and the lens. It is a critical variable in the lens formula, directly influencing the formation and properties of the image.
In the given problem, we calculate object distance for both lenses, which remains unchanged at 28.0 cm. This distance is used to find the details of the real image, such as its position.
Important considerations about object distance:
  • Varies according to the position of the object.
  • Impacts the size and position of the image formed.
  • Integral to solving lens-related problems using the lens formula.

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Most popular questions from this chapter

Light is incident from air onto the surface of a liquid. The angle of incidence is \(53.0^{\circ}\), and the angle of refraction is \(34.0^{\circ} .\) At what angle of incidence would the reflected light be \(100 \%\) polarized?

A spectator, seated in the left field stands, is watching a baseball player \(1.9 \mathrm{~m}\) tall who is \(75 \mathrm{~m}\) away. On a TV screen, located \(3.0\mathrm{~m}\) from a person watching the game at home, the same player has a 0.12 -m image. Find the angular size of the player as seen by (a) the spectator watching the game live and (b) the TV viewer. (c) To whom does the player appear to be larger?

A tourist takes a picture of a mountain \(14 \mathrm{~km}\) away using a camera that has a lens with a focal length of \(50 \mathrm{~mm}\). She then takes a second picture when she is only \(5.0 \mathrm{~km}\) away. What is the ratio of the height of the mountain's image on the film for the second picture to its height on the film for the first picture?

The back wall of a home aquarium is a mirror that is a distance \(L\) away from the front wall. The walls of the tank are negligibly thin. A fish, swimming midway between the front and back walls, is being viewed by a person looking through the front wall. (a) Does the fish appear to be at a distance greater than, less than, or equal to \(\frac{1}{2} L\) from the front wall? (b) The mirror forms an image of the fish. How far from the front wall is this image located? Express your answer in terms of \(L\). (c) Assume that your answer to Question (b) is a distance \(D\). Does the image of the fish appear to be at a distance greater than, less than, or equal to \(D\) from the front wall? (d) Could the image of the fish appear to be in front of the mirror if the index of refraction of water were different than it actually is, and, if so, would the index of refraction have to be greater than or less than its actual value? Explain each of your answers. The distance between the back and front walls of the aquarium is \(40.0 \mathrm{~cm} .\) (a) Calculate the apparent distance between the fish and the front wall. (b) Calculate the apparent distance between the image of the fish and the front wall. Verify that your answers are consistent with your answers to the Concept Questions.

The angular magnification of a telescope is 32800 times as large when you look through the correct end of the telescope as when you look through the wrong end. What is the angular magnification of the telescope?
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