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Two converging lenses are separated by \(24.00 \mathrm{cm} .\) The focal length of each lens is \(12.00 \mathrm{cm} .\) An object is placed \(36.00 \mathrm{cm}\) to the left of the lens that is on the left. Determine the final image distance relative to the lens on the right.

Short Answer

Expert verified
The final image distance is 4 cm to the right of the second lens.

Step by step solution

01

Understand Lens Setup

We have two converging lenses and need to analyze the movement of light through these lenses. The object is first placed before the first lens. We will measure all distances from the lenses to determine the location of images formed by each lens.
02

Calculate the Image Distance for the First Lens

Using the lens formula \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\), where \(f = 12\, \text{cm}\) is the focal length of the lens, \(d_o = 36\, \text{cm}\) is the object distance, solve for \(d_i\), the image distance for the first lens. Plug in the values: \[\frac{1}{12\, \text{cm}} = \frac{1}{36\, \text{cm}} + \frac{1}{d_i}\]Solving this equation gives \(d_i = 18\, \text{cm}\). This means the image formed by the first lens is 18 cm to the right of the first lens.
03

Determine the Object Distance for the Second Lens

The image from the first lens becomes the object for the second lens. Since the lenses are separated by 24 cm, and the image is 18 cm to the right of the first lens, the object distance for the second lens is \(24 \text{ cm} - 18 \text{ cm} = 6 \text{ cm}\) to the left of the second lens.
04

Calculate the Final Image Distance from the Second Lens

Again use the lens formula for the second lens: \[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]where the focal length \(f = 12\, \text{cm}\) and the object distance \(d_o = -6\, \text{cm}\) (negative because the object is on the same side of the lens).Plug in the values: \[\frac{1}{12} = \frac{1}{-6} + \frac{1}{d_i}\]Solve for \(d_i\) to get \(d_i = 4\, \text{cm}\). This means the final image is 4 cm to the right of the second lens.

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Key Concepts

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

Converging Lenses
Converging lenses, also known as convex lenses, are designed to focus light rays that pass through them. They are thicker in the center and thinner at the edges. A key characteristic of these lenses is their ability to take parallel incoming light rays and bend them so that they meet at a specific point. This point is called the focal point.

These lenses are commonly used in various optical devices like cameras, eyeglasses, and telescopes. The focal point's distance from the lens is called the focal length, and it's a crucial factor in determining how a lens refracts light. With converging lenses, objects placed beyond the focal length will have their light rays focused to form a real image on the opposite side of the lens. This property is essential in applications involving image formation.
Image Distance
The image distance is an important concept in understanding how lenses work. It refers to the distance from the lens to the point where the image is formed. For converging lenses, the image distance (\(d_i\)) can be calculated using the lens formula:

\[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]

Here, \(f\) is the focal length of the lens, and \(d_o\) is the object distance. By rearranging the formula, you can solve for \(d_i\) when you know the other values.

In practical terms, a positive image distance indicates that the image is formed on the opposite side of the lens from the object. A negative image distance would mean the image is on the same side as the object, often resulting in a virtual image. Understanding and calculating image distance helps us predict where an image will appear after passing through a lens.
Focal Length
The focal length of a lens is one of its defining characteristics. It is the distance from the center of the lens to its focal point, where converging light rays meet. A lens with a shorter focal length bends light rays more sharply than one with a longer focal length.

The focal length is crucial in determining the power of a lens. A lens with a short focal length has a higher optical power and can focus light rays to a point more quickly, which is why shorter focal lengths are used for applications requiring strong magnification, such as in microscopes.

When working with multiple lenses, such as in the original problem, each lens retains its focal length characteristic, and you must consider each lens's focal length to accurately determine image position and magnification.
Object Distance
Object distance refers to how far an object is placed from the lens. It's measured from the object's position to the center of the lens. In the context of lenses, the object distance (\(d_o\)) is a critical value used in the lens formula to calculate either the image distance or confirm the type of image formed.

The sign and magnitude of the object distance provide insight into the nature of the image formed. A positive object distance means the object is positioned on the side from which light enters the lens, leading to a real image on the other side.

In our example exercise, calculating the correct object distance for each lens was key in determining where the image formed. Understanding this concept can help in configuring optical devices to focus and display images accurately.

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Most popular questions from this chapter

A woman can read the large print in a newspaper only when it is at a distance of \(65 \mathrm{cm}\) or more from her eyes. (a) Is she nearsighted (myopic) or farsighted (hyperopic), and what kind of lens is used in her glasses to correct her eyesight? (b) What should be the refractive power (in diopters) of her glasses (worn \(2.0 \mathrm{cm}\) from the eyes), so she can read the newspaper at a distance of \(25 \mathrm{cm}\) from her eyes?

Visitors at a science museum are invited to sit in a chair to the right of a full-length diverging lens \(\left(f_{1}=-3.00 \mathrm{m}\right)\) and observe a friend sitting in a second chair, \(2.00 \mathrm{m}\) to the left of the lens. The visitor then presses a button and a converging lens \(\left(f_{2}=+4.00 \mathrm{m}\right)\) rises from the floor to a position \(1.60 \mathrm{m}\) to the right of the diverging lens, allowing the visitor to view the friend through both lenses at once. Find (a) the magnification of the friend when viewed through the diverging lens only and (b) the overall magnification of the friend when viewed through both lenses. Be sure to include the algebraic signs \((+\) or \(-)\) with your answers.

An object is in front of a converging lens \((f=0.30 \mathrm{m})\). The magnification of the lens is \(m=4.0 .\) (a) Relative to the lens, in what direction should the object be moved so that the magnification changes to \(m=-4.0 ?\) (b) Through what distance should the object be moved?

The equation $$\frac{1}{d_{0}}+\frac{1}{d_{i}}=\frac{1}{f}$$ is called the Gaussian form of the thin-lens equation. The drawing shows the variables \(d_{0}, d_{i},\) and \(f .\) The drawing also shows the distances \(x\) and \(x^{\prime},\) which are, respectively, the distance from the object to the focal point on the left of the lens and the distance from the focal point on the right of the lens to the image. An equivalent form of the thin-lens equation, involving \(x, x^{\prime},\) and \(f,\) is called the Newtonian form. Show that the Newtonian form of the thin-lens equation can be written as \(x x^{\prime}=f^{2}\).

An object is placed to the left of a lens, and a real image is formed to the right of the lens. The image is inverted relative to the object and is one- half the size of the object. The distance between the object and the image is \(90.0 \mathrm{cm} .\) (a) How far from the lens is the object? (b) What is the focal length of the lens?

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