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Chapter 33: Problem 16

(II) A converging lens has focal length \(f\). When an object is placed a distance \(d_{0}>f\) from this lens, a real image with magnification \(m\) is formed. \((a)\) Show that \(m=f /\left(f-d_{0}\right)\). (b) Sketch \(m\) vs, \(d_{\mathrm{o}}\) over the range \(f

Short Answer

Expert verified
(a) Magnification is \( m = \frac{f}{f - d_o} \). (b) As \( d_o \to f \), \( m \to \infty \); as \( d_o \to \infty \), \( m \to 0 \). (c) \( d_o = 2f \) for same size image. (d) Place object at \( d_o > 2f \) for larger image.

Step by step solution

01

Understand the Lens Formula

The lens formula is given by \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \), where \( f \) is the focal length, \( d_o \) is the object distance from the lens, and \( d_i \) is the image distance from the lens.
02

Express Image Magnification

The magnification \( m \) of the lens is given by \( m = \frac{h_i}{h_o} = -\frac{d_i}{d_o} \), where \( h_i \) and \( h_o \) are the heights of the image and object, respectively.
03

Derive the Expression for Magnification

We aim to derive \( m = \frac{f}{f - d_o} \). From the lens formula \( \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} \), rearrange to find \( d_i \):\[ d_i = \frac{d_o f}{d_o - f} \]Substitute this expression for \( d_i \) into the magnification formula:\[ m = -\frac{\frac{d_o f}{d_o - f}}{d_o} = -\frac{f}{d_o - f} \] This simplifies to:\[ m = \frac{f}{f - d_o} \]
04

Analyze Magnification vs. Object Distance

To sketch \( m \) against \( d_o \) with \( f = 0.45 \) cm, analyze the behavior of \( m \) as \( d_o \) approaches \( f \) from above and as \( d_o \) approaches infinity. As \( d_o \to f \), \( m \to \infty \); as \( d_o \to \infty \), \( m \to 0 \).
05

Find Object Distance for Unit Magnification

For the image to have the same size as the object, set \( |m| = 1 \) (unit magnification):\[ 1 = \frac{f}{f - d_o} \]Solving for \( d_o \), we find:\[ f - d_o = f \] Thus, \[ d_o = 2f \]
06

Determine the Object Placement for Large Magnification

To obtain a real image larger than the object, place the object just beyond the focal length, specifically in the region where \( d_o > 2f \). This configuration achieves higher magnification without the image being non-real.

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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 (\( f \)) is a key concept when discussing converging lenses. It is the distance from the lens to the point where parallel rays of light either converge to a single point or appear to diverge from a single point on the axis.
In practical terms, this means:
  • If a lens has a shorter focal length, it means it bends the light more sharply and the rays converge closer to the lens.
  • If the focal length is longer, the convergence point is further from the lens, creating a different form of magnification and image size.
The focal length is crucial when placing an object to form images, as it determines the nature of images鈥攔eal or virtual, magnified or minified.
By understanding how the focal length works in a converging lens, one can foresee whether the image will be erect or inverted, a thumbnail or larger than the object itself.
Lens Formula
The lens formula is fundamental in optics and helps describe the relationship between the object distance (\( d_o \)), the image distance (\( d_i \)), and the focal length (\( f \)). It is expressed as:\[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]This equation is essential for determining where an image will form given a certain object position relative to the lens.
Using the lens formula, you can:
  • Calculate the position of the image formed by the lens based on the object鈥檚 distance from the lens.
  • Predict whether the image is real or virtual, since a real image has positive image distance while a virtual one has negative.
One can deduce that when the object is placed beyond the focal length, the lens formula can be manipulated to show the real image's characteristics, specifically pointing towards magnification factors and other optical behaviors.
Image Magnification
Image magnification deals with the size of an image formed by a lens compared to the actual size of the object. The magnification (\( m \)) is expressed as:\[m = \frac{h_i}{h_o} = -\frac{d_i}{d_o}\]Where \( h_i \) is the image height and \( h_o \) is the object height, the negative sign indicates the image is inverted.
When exploring magnification in the context of a converging lens:
  • A magnification greater than 1 (\( |m| > 1 \)) indicates an enlarged image compared to the object.
  • A magnification less than 1 (\( |m| < 1 \)) signifies the image is smaller than the object.
  • If the magnification equals 1 (\( |m| = 1 \)), the image is of the same size as the object.
Understanding these principles allows one to place objects at specific distances relative to the focal length, to achieve desired image characteristics in size and orientation.

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Most popular questions from this chapter

If a 135-mm telephoto lens is designed to cover object distances from \(1.30 \mathrm{~m}\) to \(\infty\), over what distance must the lens move relative to the plane of the sensor or film?

(III) In the "magnification" method, the focal length \(f\) of a converging lens is found by placing an object of known size at various locations in front of the lens and measuring the resulting real-image distances \(d_{i}\) and their associated magnifications \(m\) (minus sign indicates that image is inverted). The data taken in such an experiment are given here:\(\begin{array}{rrrrrr}{d_{\mathrm{i}}(\mathrm{cm})} & {20} & {25} & {30} & {35} & {40} \\ {m} & {-0.43} & {-0.79} & {-1.14} & {-1.50} & {-1.89}\end{array}\) (a) Show analytically that a graph of \(m\) vs. \(d_{\text { i should }}\) produce a straight line. What are the theoretically expected values for the slope and the \(y\) -intercept of this line? [Hint: \(d_{\mathrm{o}}\) is not constant.] \((b)\) Using the data above, graph \(m\) vs. \(d_{\mathrm{i}}\) and show that a straight line does indeed result. Use the slope of this line to determine the focal length of the lens. Does the \(y\) -intercept of your plot have the expected value? (c) In performing such an experiment, one has the practical problem of locating the exact center of the lens since \(d_{\mathrm{i}}\) must be measured from this point. Imagine, instead, that one measures the image distance \(d\) from the back surface of the lens, which is a distance \(\ell\) from the lens's center. Then, \(d_{i}=d_{1}^{\prime}+\ell .\) Show that, when implementing the magnification method in this fashion, a plot of \(m\) vs.di will still result in a straight line. How can \(f\) be determined from this straight line?

A 50-year-old man uses \(+2.5-\mathrm{D}\) lenses to read a newspaper \(25 \mathrm{~cm}\) away. Ten years later, he must hold the paper \(32 \mathrm{~cm}\) away to see clearly with the same lenses. What power lenses does he need now in order to hold the paper \(25 \mathrm{~cm}\) away? (Distances are measured from the lens.)

(II) A lighted candle is placed 36 \(\mathrm{cm}\) in front of a converging lens of focal length \(f_{1}=13 \mathrm{cm},\) which in turn is 56 \(\mathrm{cm}\) in front of another converging lens of focal length \(f_{2}=16 \mathrm{cm}\) (see Fig. \(47 ) .(a)\) Draw a ray diagram and estimate the location and the relative size of the final image. (b) Calculate the position and relative size of the final image.

A woman can see clearly with her right eye only when objects are between \(45 \mathrm{~cm}\) and \(155 \mathrm{~cm}\) away. Prescription bifocals should have what powers so that she can see distant objects clearly (upper part) and be able to read a book \(25 \mathrm{~cm}\) away (lower part) with her right eye? Assume that the glasses will be \(2.0 \mathrm{~cm}\) from the eye.
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