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Chapter 34: Problem 117

A luminous object and a screen are a fixed distance \(D\) apart. (a) Show that a converging lens of focal length \(f\), placed between object and screen, will form a real image on the screen for two lens positions that are separated by a distance \(d=\sqrt{D(D-4 f)}\). (b) Show that $$ \left(\frac{D-d}{D+d}\right)^{2} $$ gives the ratio of the two image sizes for these two positions of the lens.

Short Answer

Expert verified
Position separation: \(d = \sqrt{D(D-4f)}\). Size ratio: \(\left(\frac{D-d}{D+d}\right)^2\).

Step by step solution

01

Understanding the lens equation

The lens equation is given by: \( \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \), where \( f \) is the focal length of the lens, \( u \) is the object distance from the lens, and \( v \) is the image distance from the lens. In this scenario, the total distance between the object and the screen is given as \( u + v = D \).
02

Expressing distances in the lens equation

Using the equation from Step 1 and \( u + v = D \), we can write two equations:1. \( \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \)2. \( v = D - u \)Substituting the second equation into the first one gives: \( \frac{1}{f} = \frac{1}{u} + \frac{1}{D-u} \).
03

Solving for quadratic equation

Converting the expression into a single quadratic equation in terms of \( u \), we get:\( f(D-u) + fu = u(D-u) \)which simplifies to:\( uf^2 = (D-u)u \)Rearrange to form:\( u^2 - Du + Df = 0 \).
04

Finding lens positions using quadratic roots

To find the two possible values of \( u \), solve the quadratic equation:\( u = \frac{D \pm \sqrt{D^2 - 4Df}}{2} \)The separation of these positions is given by the difference \( u_1 - u_2 = d = \sqrt{D^2 - 4Df} \).
05

Verifying expression for position separation

Given: \( d = \sqrt{D(D - 4f)} \). This matches the expression derived from the quadratic formula solution, confirming part (a).
06

Analyzing image size ratio

The magnification (image size) of a lens is given by: \( M = \frac{v}{u} \). For the two positions:- For position 1: \( M_1 = \frac{D-u_1}{u_1} \)- For position 2: \( M_2 = \frac{D-u_2}{u_2} \)The ratio of the two sizes is \( \frac{M_1}{M_2} = \left(\frac{u_2}{u_1}\right)^2 \).
07

Expressing size ratio using given equation

From Step 4, we have:\( u_1 = \frac{D + \sqrt{D(D-4f)}}{2} \) and \( u_2 = \frac{D - \sqrt{D(D-4f)}}{2} \).Thus, \( \frac{u_2}{u_1} = \frac{D-\sqrt{D(D-4f)}}{D+\sqrt{D(D-4f)}} \). This implies the ratio of image sizes is:\( \left(\frac{D-d}{D+d}\right)^2 \), proving part (b).

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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 a fundamental formula used in optics to describe the relationship between the object distance, image distance, and focal length of a lens. This equation is expressed as:\( \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \), where:
  • \( f \) is the focal length of the lens, which is a measure of how strongly the lens converges or diverges light.
  • \( u \) is the distance from the object to the lens, known as the object distance.
  • \( v \) is the distance from the image to the lens, known as the image distance.

In the context of the given exercise, it鈥檚 helpful to acknowledge that the total distance between the object and the screen is \( D \), which equals \( u + v \). Using this information, the lens equation helps us understand how arranging or adjusting the lens between the object and the screen can position the image correctly on the screen. By solving the equation in terms of \( u \), you uncover two potential locations for the lens, illustrated by the roots of the quadratic equation derived from the lens equation.
Image Formation
Image formation through a lens depends on the interaction between light rays traveling from an object and the lens itself. When a lens is positioned correctly, it bends or refracts rays to converge them, resulting in the creation of a real image. This exercise demonstrates how positioning a converging lens precisely between a luminous object and a screen, given the fixed distance \( D \), allows the image to form neatly on the screen.
Bullet points regarding the concepts include:
  • A real image is where light converges to form on the opposite side of the lens from the object, making it visible on the screen.
  • In this exercise, the lens can be positioned in two specific locations due to the focal length \( f \) and the total distance \( D \), which ensures that light converges accurately to project the image.
  • The importance of the equation \( d=\sqrt{D(D-4f)} \), which helps determine the separation between these two potential lens positions, showcases how geometrical optics principles apply.

Solving for the roots of the quadratic equation given by the lens equation reveals these two lens positions, allowing for precise image formation.
Magnification
Magnification in optics refers to how much larger or smaller the image appears compared to the actual object. This is significant for understanding how the lens equation not only helps determine the position of an image but also affects its size. For a lens, the magnification \( M \) can be calculated using the formula:\( M = \frac{v}{u} \)
This means:
  • A magnification greater than 1 implies the image is larger than the object.
  • A magnification less than 1 indicates that the image is smaller than the object.

In this exercise, you calculate the magnification for the two lens positions to derive the ratio of image sizes. The exercise further explains how to express the size ratio using \( \left(\frac{D-d}{D+d}\right)^2 \), which matches the provided mathematical expression. This demonstrates the practical application of the lens equation in verifying the observed image magnification, thereby confirming part (b) of the problem statement. Understanding magnification is crucial as it directly influences how we perceive and utilize imaging in various optical devices.

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Most popular questions from this chapter

stick figure \(O\) (the object) stands on the common central axis of three thin, symmetric lenses, which are mounted in the boxed regions. Lens 1 is mounted within the boxed region closest to \(O\), which is at object distance \(p_{1} .\) Lens 2 is mounted within the middle boxed region, at distance \(d_{12}\) from lens \(1 .\) Lens 3 is mounted in the farthest boxed region, at distance \(d_{23}\) from lens \(2 .\) Each problem in Table \(34-10\) refers to a different combination of lenses and different values for distances, which are given in centimeters. The type of lens is indicated by \(C\) for converging and \(D\) for diverging: the number after \(\mathrm{C}\) or \(\mathrm{D}\) is the distance between a lens and either of the focal points (the proper sign of the focal distance is not indicated). Find (a) the image distance \(i_{2}\) for the (final) image produced by lens 3 (the final image produced by the system) and (b) the overall lateral magnification \(M\) for the system, including signs. Also, determine whether the final image is (c) real (R) or virtual (V), (d) inverted (I) from object \(O\) or noninverted (NI), and (e) on the same side of lens 3 as object \(O\) or on the opposite side. \(\begin{array}{lllllll}95 & +12 & \mathrm{C}, 8.0 & 28 & \mathrm{C}, 6.0 & 8.0 & \mathrm{C}, 6.0\end{array}\)

A moth at about eye level is \(10 \mathrm{~cm}\) in front of a plane mirror; you are behind the moth, \(30 \mathrm{~cm}\) from the mirror. What is the distance between your eyes and the apparent position of the moth's image in the mirror?

Figure \(34-47 a\) shows the basic structure of a human eye. Light refracts into the eye through the cornea and is then further redirected by a lens whose shape (and thus ability to focus the light) is controlled by muscles. We can treat the cornea and eye lens as a single effective thin lens (Fig. \(34-47 b\) ). A "normal" eye can focus parallel light rays from a distant object \(O\) to a point on the retina at the back of the eye, where processing of the visual information begins. As an object is brought close to the eye, however, the muscles must change the shape of the lens so that rays form an inverted real image on the retina (Fig. \(34-47 c\) ). (a) Suppose that for the parallel rays of Figs. \(34.47 a\) and \(b\), the focal length \(f\) of the effective thin lens of the eye is \(2.50 \mathrm{~cm}\). For an object at distance \(p=\) \(40.0 \mathrm{~cm}\), what focal length \(f^{\prime}\) of the effective lens is required for the object to be seen clearly? (b) Must the eye muscles increase or decrease the radii of curvature of the eye lens to give focal length \(f^{\prime} ?\)

Two thin lenses of focal lengths \(f_{1}\) and \(f_{2}\) are in contact and share the same central axis. Show that, in image formation, they are equivalent to a single thin lens for which the focal length is \(f=f_{1} f_{2} /\left(f_{1}+f_{2}\right)\)

A simple magnifier of focal length \(f\) is placed near the eye of someone whose near point \(P_{n}\) is \(25 \mathrm{~cm}\). An object is positioned so that its image in the magnifier appears at \(P_{n^{-}}\) (a) What is the angular magnification of the magnifier? (b) What is the angular magnification if the object is moved so that its image appears at infinity? For \(f=10 \mathrm{~cm}\), evaluate the angular magnifications of (c) the situation in (a) and (d) the situation in (b). (Viewing an image at \(P_{n}\) requires effort by muscles in the eye, whereas viewing an image at infinity requires no such effort for many people.)
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