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Chapter 23: Problem 48

A real object's distance from a converging lens is five times the focal length. (a) Determine the location of the image \(q\) in terms of the focal length \(f .\) (b) Find the magnification of the image. (c) Is the image real or virtual? Is it upright or inverted? Is the image on the same side of the lens as the object or on the opposite side?

Short Answer

Expert verified
The location of the image is \(\frac{5}{4}f\). The magnification of the image is \(-\frac{1}{4}\). Thus, the image is real and inverted, and located on the opposite side of the lens relative to the object.

Step by step solution

01

Determine the location of the image

First, we plug in the given values into the lens equation. The object's distance from the lens \(p\) is five times the focal length, or \(p = 5f\). So, we get: \\[\frac{1}{f} = \frac{1}{5f} + \frac{1}{q}\\] Solving for \(q\), we find that \(q = \frac{5}{4}f\). Hence, the image's location is \(\frac{5}{4}\) times the focal length.
02

Find the magnification

Using the formula for magnification \(m = -\frac{q}{p}\), substituting the values we calculated above, we get: \\[m = -\frac{ \frac{5}{4}f }{5f} = -\frac{1}{4}\\] This implies, the image is reduced and one-fourth the size of the object.
03

Determine the characteristics of the image

Given \(m = -\frac{1}{4}\), which is less than 0, it indicates that the image is real and inverted. Since \(q > 0\), the image is on the opposite side of the lens relative to the object.

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Key Concepts

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

Lens Equation
The lens equation is crucial in geometrical optics to find relationships between an object's distance, the image's distance, and the lens's focal length. It is expressed as:
  • \( \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \)
Here, \( f \) is the focal length, \( p \) is the object distance, and \( q \) is the image distance.
For a converging lens, all distances are positive. Using this formula helps us determine where the image of an object placed in front of a lens will focus.
  • By rearranging the equation, we can solve for \( q \) (image distance) when \( p \) (object distance) and \( f \) are known.
  • The lens equation is the foundation to finding image properties such as its size and orientation.
Image Formation
Image formation through a lens depends significantly on the object's position relative to the lens. A converging lens focuses light rays to form an image.
When the object is outside the focal length, as in this exercise, the lens forms a real image.
  • This real image is located on the opposite side of the incoming light.
  • It can be projected onto a screen, as opposed to a virtual image which cannot.
The image's position can be calculated using the lens equation, which gives us insight into whether the image is real, its size, and its orientation.
Knowing these details allows us to predict how lenses in items like glasses and cameras behave.
Magnification
Magnification measures how much larger or smaller the image is compared to the object. For lenses, it is calculated using the formula:
  • \( m = -\frac{q}{p} \)
In this context, \( m \) represents magnification, \( q \) is the image distance, and \( p \) is the object distance.
The negative sign indicates inversion.
  • If \( |m| > 1 \), the image is larger than the object.
  • If \( |m| < 1 \), the image is smaller.
In the exercise, the magnification was -1/4, meaning the image is inverted and one-fourth the size of the object.
Realization that an image is reduced in size is essential in optical system design.
Converging Lens
A converging lens, often referred to as a convex lens, bends incoming parallel light rays to converge at a focal point. This characteristic makes it highly useful for a variety of applications like in magnifying glasses and cameras.
The shape and refractive index of the lens determine the focal point location.
  • Converging lenses cause light to focus, forming an image.
  • They are essential in vision correction and imaging technologies.
Understanding how converging lenses function helps in making precise predictions about image formation. They are central in devices such as telescopes by focusing distant object's images.
Focal Length
The focal length of a lens is the distance from the lens to the focal point, where light rays converge.
  • It's a crucial parameter determining how strongly a lens converges or diverges light.
  • A smaller focal length implies a stronger lens with more pronounced curvature, suitable for focusing on closer objects.
In experiments, once the focal length is known, calculations for image distance and magnification become possible using the lens equation.
This concept appears everywhere, from simple lens exams to complex optical systems engineering. With focal length, lenses reveal their capability to magnify or reduce image size.

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Most popular questions from this chapter

A dentist uses a mirror to examine a tooth that is \(1.00 \mathrm{~cm}\) in front of the mirror. The image of the tooth is formed \(10.0 \mathrm{~cm}\) behind the mirror. Determine (a) the mirror's radius of curvature and (b) the magnification of the image.

A \(2.00-\mathrm{cm}\) -high object is placed \(3.00 \mathrm{~cm}\) in front of a concave mirror. If the image is \(5.00 \mathrm{~cm}\) high and virtual, what is the focal length of the mirror?

Two plane mirrors stand facing each other, \(3.0 \mathrm{~m}\) apart, and a woman stands between them. The woman faces one of the mirrors from a distance of \(1.0 \mathrm{~m}\), with the palm of her left hand facing the closer mirror. (a) What is the apparent position of the closest image of her left hand, measured from the surface of the mirror in front of her? Does it show the palm of her hand or the back of her hand? (b) What is the position of the next image? Does it show the palm of her hand or the back of her hand? (c) Repeat for the third image. (d) Which of the images are real and which are virtual?

The top of a swimming pool is at ground level. If the pool is \(2 \mathrm{~m}\) deep, how far below ground level does the bottom of the pool appear to be located when (a) the pool is completely filled with water and (b) the pool is filled halfway with water?

A person walks into a room that has, on opposite walls, two plane mirrors producing multiple images. Find the distances from the person to the first three images seen in the left-hand mirror when the person is \(5.00 \mathrm{ft}\) from the mirror on the left wall and \(10.0 \mathrm{ft}\) from the mirror on the right wall.
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