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Draw diagrams to indicate qualitatively the position, nature, and size of the image formed by a converging lens of focal length \(f\) for the following object distances: ( \(a\) ) infinity, (b) greater than \(2 f\), \((c)\) equal to \(2 f,(d)\) between \(2 f\) and \(f\), (e) equal to \(f\), \((f)\) less than \(f\).

Short Answer

Expert verified
For a converging lens, image is real and inverted unless object is within focal length, then it's virtual and upright.

Step by step solution

01

Understanding image formation in converging lens

In a converging lens, the nature of the image formed depends on the object distance compared to the focal length \( f \). The image can be real or virtual and can vary in size.
02

Object at infinity (Case a)

When an object is at infinity, rays of light entering the lens are parallel. These rays converge at the focal point on the opposite side of the lens. Hence, the image forms at the focal point \( f \), being a real point-image, very small, and inverted.
03

Object distance greater than \(2f\) (Case b)

When the object is placed beyond \(2f\), the image is formed between \(f\) and \(2f\) on the opposite side. It is real, inverted, and smaller than the object.
04

Object distance equal to \(2f\) (Case c)

When the object is placed at \(2f\), the image is formed at the same distance \(2f\) on the opposite side. The image is real, inverted, and has the same size as the object.
05

Object distance between \(f\) and \(2f\) (Case d)

For an object located between \(f\) and \(2f\), the image is formed beyond \(2f\). The image is real, inverted, and larger than the object.
06

Object distance equal to \(f\) (Case e)

When the object is placed at \(f\), the rays emerging from the lens are parallel on the other side indicating that the image forms at infinity. Thus, there is no real image formed on the screen (considered undefined for practical purposes).
07

Object distance less than \(f\) (Case f)

If the object is within the focal length \(f\), the image is formed on the same side as the object. The image is virtual, upright, and larger than the object.

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Key Concepts

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

Focal Length
The focal length of a lens, often represented as \(f\), is a measure of how strongly a lens converges or diverges light. For a converging lens, it is the distance between the lens's center and the focal point on the opposite side where parallel rays of light converge after passing through the lens.
A shorter focal length indicates a lens with a stronger converging power, meaning it bends light rays more sharply towards the focal point. On the contrary, a longer focal length suggests a gentler bending of the rays, resulting in them converging further away from the lens.
Knowing the focal length is essential for understanding how a lens will form images and is a key factor in calculating the image's position and size.
Image Formation
Image formation is the process by which lenses create visual representations of objects. In a converging lens, the image's characteristics—such as its location, size, and nature—are determined by the object's distance from the lens relative to the focal length \(f\).
When an object is at infinity, the image forms precisely at the focal point. For objects positioned beyond \(2f\), images form between \(f\) and \(2f\), while objects at \(2f\) result in an image of equal size on the opposite side.
As the object approaches the focal length, the image becomes larger and moves further away, becoming undefined at \(f\). Inside the focal length, the lens can no longer project a real image; instead, a virtual image appears on the same side as the object, providing a magnified view.
Real and Virtual Images
In optics, the terms 'real' and 'virtual' describe the nature of the images formed by lenses. A real image occurs when light rays actually converge at a point, meaning it can be projected onto a screen. This is the case with objects beyond the focal length \(f\). Such images are typically inverted.
Conversely, a virtual image forms when light rays diverge, creating the illusion of an image by tracing rays backward to a point behind the object. Virtual images cannot be directly projected because they exist only as perceptions. They are found when objects are placed within the focal length \(f\) of a converging lens, resulting in upright appearances that seem larger.
Inverted and Upright Images
The orientation of an image, whether inverted or upright, is an important aspect of image formation by converging lenses. Inverted images are those that appear upside down compared to the object, resulting from the light rays crossing over as they pass through the lens.
These inverted images occur when the object is placed beyond the focal length \(f\), indicating that they are real and can be captured on a screen. As the object is moved closer to the lens, eventually within the focal length, the image becomes upright.
Upright images are virtual, meaning they can't be projected but can be seen directly through the lens. These images appear right side up and are often magnified, offering a larger view of the object seen through magnifying glasses or similar lenses.

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Most popular questions from this chapter

A double convex glass lens \((n=1.50)\) has faces of radius \(8 \mathrm{~cm}\) each. Compute its focal length in air and when immersed in water \((n=1.33)\).

In what two positions will a converging thin lens of focal length \(+9.00 \mathrm{~cm}\) form images of a luminous object on a screen located \(40.0 \mathrm{~cm}\) from the object? Given \(s_{0}+s_{i}=40.0 \mathrm{~cm}\) and \(f=+9.00 \mathrm{~cm}\), we have $$\frac{1}{s_{o}}+\frac{1}{40.0 \mathrm{~cm}-s_{o}}=\frac{1}{9.0 \mathrm{~cm}} \quad \text { or } \quad s_{o}^{2}-40.0 s_{o}+360=0$$ The use of the quadratic formula gives $$s_{o}=\frac{40.0 \pm \sqrt{1600-1440}}{2}$$ from which \(s_{0}=13.7 \mathrm{~cm}\) and \(s_{0}=26.3 \mathrm{~cm}\). The two lens positions are \(13.7 \mathrm{~cm}\) and \(26.3 \mathrm{~cm}\) from the object.

An achromatic lens is formed from two thin lenses in contact, having powers of \(+10.0\) diopters and \(-6.0\) diopters. Determine the power and focal length of the combination. Since reciprocal focal lengths add, $$\text { Power }=+10.0-6.0=+4.0 \text { diopters and } \quad \text { Focal length }=\frac{1}{\text { Power }}=\frac{1}{+4.0 \mathrm{~m}^{-1}}=+25 \mathrm{~cm}$$

A thin glass lens \((n=1.50)\) has a focal length of \(+10 \mathrm{~cm}\) in air. Compute its focal length in water \((n=1.33)\). Using $$\frac{1}{f}=\left(\frac{n_{1}}{n_{2}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$$ we get $$\begin{array}{ll} \text { For air: } & \frac{1}{10}=(1.50-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \\ \text { For water: } & \frac{1}{f}=\left(\frac{1.50}{1.33}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \end{array}$$ Divide one equation by the other to obtain \(f=5.0 / 0.128=39 \mathrm{~cm}\).

We have a thin negative lens with a focal length of \(-1.40 \mathrm{~m}\). An object is placed on the central axis \(200 \mathrm{~cm}\) from the lens. If the object is \(2.00 \mathrm{~cm}\) tall, how tall is the image? Locate and describe the image. [Hint: Find \(s_{i}\); then find the magnification. Check with Table 38-1.]

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