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Chapter 34: Problem 86

An object is placed \(22.0 \mathrm{~cm}\) from a screen. (a) At what two points between object and screen may a converging lens with a \(3.00 \mathrm{~cm}\) focal length be placed to obtain an image on the screen? (b) What is the magnification of the image for each position of the lens?

Short Answer

Expert verified
The two possible positions for the converging lens to create an image on the screen are calculated by solving the lens formula and the magnifications for each position are found by applying the magnification formula for object and image distance.

Step by step solution

01

Use the lens formula to find the possible lens positions

Start with the lens formula \(1/v - 1/u = 1/f\). We are given \(u = 22 \mathrm{~cm}\) (object distance), \(f = 3 \mathrm{~cm}\) (focal length), and we need to find \(v\) (image distance). The negative root corresponds to the converging lens positioned closer to the object, whereas the positive root refers to the lens positioned closer to the screen.
02

Find the lens positions

Solve the lens formula equation for \(v\) which will give two solutions: one positive and one negative. These will indicate the two possible positions that the converging lens can be placed to focus the image on the screen.
03

Calculate magnification for each lens position

The magnification for each lens position can be calculated using the formula \(m = -v/u\). Substitute the respective \(v\) values obtained from Step 2 to find the magnification for each lens position.

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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, is a lens that causes parallel rays of light to come together at a point, termed the focus. This lens is thicker in the middle than at the edges, giving it its characteristic shape. The primary purpose of a converging lens is to focus light to form a clear image. This type of lens can produce both real and virtual images, depending on the location of the object.
  • Real Images: These are formed when the object is placed beyond the focal point. Real images can be projected onto a screen.
  • Virtual Images: These appear when the object's location is within the focal point. Such images cannot be projected but are seen when looking through the lens.
Choosing the right position for a converging lens can alter the image's clarity and size, which plays a pivotal role in optics and applications like eyeglasses, cameras, and projectors.
Lens Formula
The lens formula is crucial to determining how an image forms through a lens. It is mathematically given by:\[\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\]where:
  • \(v\): Image distance from the lens (measured in the same unit as object distance and focal length).
  • \(u\): Object distance from the lens.
  • \(f\): Focal length of the lens.
To effectively use this formula, it's essential to correctly identify the given values in a problem and substitute them into the equation. Solving the equation provides the image distance, \(v\), which indicates where the image will form in relation to the lens. Understanding and applying the lens formula is a fundamental skill in optics, as it holds the key to achieving desired image qualities.
Magnification
Magnification in optics refers to how much larger or smaller the image is compared to the actual object. It is determined by the magnification formula:\[ m = -\frac{v}{u}\]Here:
  • \(m\): Magnification factor.
  • \(v\): Image distance.
  • \(u\): Object distance.
The negative sign indicates that the image formed by a converging lens is usually inverted relative to the object. Consequently, if \(m > 1\), the image appears larger than the object, and if \(m < 1\), the image appears smaller. Calculating magnification helps in applications like telescopes and microscopes where clarity and size of the image are pivotal.
Image Distance
In optics, image distance is a crucial parameter that indicates how far the image forms from a lens. This distance is symbolized by \(v\). Finding the correct image distance helps in determining how an image is perceived and focused. In practical applications:
  • Projection: Image distance is critical when using lenses to project images, like in projectors and film production.
  • Corrective Lenses: Proper calculation can improve clarity for vision aids such as glasses.
The image distance varies based on object distance and focal length, as shown by the lens formula. It helps evaluate whether images are real (positive \(v\)) or virtual (negative \(v\)), which dictates how they appear to the observer.

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Most popular questions from this chapter

A coin is placed next to the convex side of a thin spherical glass shell having a radius of curvature of \(18.0 \mathrm{~cm} .\) Reflection from the surface of the shell forms an image of the \(1.5-\mathrm{cm}\) -tall coin that is \(6.00 \mathrm{~cm}\) behind the glass shell. Where is the coin located? Determine the size, orientation, and nature (real or virtual) of the image.

Where must you place an object in front of a concave mirror with radius \(R\) so that the image is erect and \(2 \frac{1}{2}\) times the size of the object? Where is the image?

The focal points of a thin diverging lens are \(25.0 \mathrm{~cm}\) from the center of the lens. An object is placed to the left of the lens, and the lens forms an image of the object that is \(18.0 \mathrm{~cm}\) from the lens. (a) Is the image to the left or right of the lens? (b) How far is the object from the center of the lens? (c) Is the height of the image less than, greater than, or the same as the height of the object?

Given that frogs are nearsighted in air, which statement is most likely to be true about their vision in water? (a) They are even more nearsighted; because water has a higher index of refraction than air, a frog's ability to focus light increases in water. (b) They are less nearsighted, because the cornea is less effective at refracting light in water than in air. (c) Their vision is no different, because only structures that are internal to the eye can affect the eye's ability to focus. (d) The images projected on the retina are no longer inverted, because the eye in water functions as a diverging lens rather than a converging lens.

(a) For a lens with focal length \(f,\) find the smallest distance possible between the object and its real image. (b) Graph the distance between the object and the real image as a function of the distance of the object from the lens. Does your graph agree with the result you found in part (a)?
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