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Chapter 35: Problem 43

An illuminated slide is held \(44 \mathrm{~cm}\) from a screen. How far from the slide must a lens of focal length \(11 \mathrm{~cm}\) be placed to form an image of the slide's picture on the screen?

Short Answer

Expert verified
The lens must be placed approximately 14.67 cm from the slide.

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, and \(d_i\) is the image distance.
02

- Identify Known Values

The focal length \(f\) is \(11 \text{ cm}\), and the distance between the slide and the screen (which is the object distance plus the image distance) is \(44 \text{ cm}\).
03

- Set Up Equations for Object and Image Distance

Let \(d_o\) be the distance from the lens to the slide (object distance). Then, the distance from the lens to the screen (image distance) will be \(d_i = 44 - d_o\).
04

- Substitute Values into the Lens Formula

Substitute the known values and expressions into the lens formula:\[\frac{1}{11} = \frac{1}{d_o} + \frac{1}{44 - d_o}\]
05

- Solve for Object Distance

Solve the equation:\[\frac{1}{11} = \frac{1}{d_o} + \frac{1}{44 - d_o}\]Multiply through by \(44d_o\) to clear the denominators:\[\frac{44d_o}{11} = 44 + d_o\]This simplifies to:\[4d_o = 44 + d_o\]Then:\[4d_o - d_o = 44\]So:\[3d_o = 44\]Therefore:\[d_o = \frac{44}{3} \approx 14.67 \text{ cm}\]
06

- Verify Image Distance

Now verify the image distance \(d_i\):\[d_i = 44 - d_o = 44 - 14.67 = 29.33 \text{ cm}\]Substitute these values back into the lens formula to check correctness:\[\frac{1}{11} = \frac{1}{14.67} + \frac{1}{29.33}\] which come out correctly.

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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, represented by \(f\) in the lens formula, is one of the most crucial elements in optics. It signifies the distance between the center of a lens and its focal point, where it converges or diverges light rays. In simpler terms, it's where the lens focuses light. For a converging or convex lens, the focal length is positive, while for a diverging or concave lens, it is negative.

In our exercise, the focal length is given as \(11 \text{ cm}\). This information is vital for applying the lens formula and solving the problem.
object distance
The object distance, noted as \(d_o\), is the distance from the object to the lens. In our context, this is the distance from the illuminated slide to the lens. Determining \(d_o\) often requires using other known distances and the lens formula.

From the problem, we know the total distance from the slide to the screen is \(44 \text{ cm}\). To find the correct placement of the lens, we set the total distance as \(d_o + d_i = 44 \text{ cm}\). By solving the lens formula with this equation, we determine the precise object distance.
image distance
Image distance, represented as \(d_i\), is the distance from the lens to the image formed by the lens. In this exercise, it pertains to how far the image will be on the screen from the lens.

The total distance given is \(44 \text{ cm}\), which equals the sum of \(d_o\) (object distance) and \(d_i\) (image distance). This relationship helps us use the lens formula to piece together the values and find \(d_i\). Finally, we verify this image distance calculation to ensure correctness by checking it with the lens formula.
optics problem solving
Solving optics problems often revolves around understanding and applying the lens formula correctly. Follow these steps for a structured approach:

  • Identify and note down the known values (focal length, distances).
  • Express unknown distances using a variable and form related equations.
  • Substitute values into the lens formula: \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\).
  • Solve the equation for the unknown distance.
  • Verify your values by substituting back into the lens formula.


In our exercise, we first identified \(f\) and total distance (\(d_o + d_i = 44 \text{ cm}\)), then set up the equation and solved for \(d_o\) and \(d_i\). This structured approach ensures accurate results!

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Most popular questions from this chapter

A solid glass cube, of edge length \(10 \mathrm{~mm}\) and index of refraction \(1.5\), has a small spot at its center. (a) What parts of each cube face must be covered to prevent the spot from being seen, no matter what the direction of viewing? (Neglect light that reflects inside the cube and then refracts out into the air.) (b) What fraction of the cube surface must be so covered?

The index of refraction of benzene is \(1.8\). What is the critical angle for a light ray traveling in benzene toward a plane layer of air above the benzene?

Address each part of this question in two ways: (1) by drawing and interpreting appropriate geometrical diagrams and (2) by appealing to the lens equation and the expression for lateral magnification and demonstrating your result mathematically. If your two approaches do not agree, explain which one is correct and why the other is wrong. (a) Suppose you are using a camera and wish to have a larger image of a distant object than you are obtaining with the lens currently in use. Would you change to a lens with a longer or a shorter focal length? Explain your reasoning. (Hint: Note that the object distance is essentially fixed.) (b) Suppose you are using a slide projector and wish to obtain a larger image on the screen. You cannot achieve this by moving the screen farther from the projector because you are already using the entire length of the room. Would you change to a lens with a longer or a shorter focal length than the one you are using? Explain your reasoning. (Hint: Note that the image distance is essentially fixed.)

Light traveling in water of refractive index \(1.33\) is incident on a plate of glass with index of refraction 1.53. At what angle of incidence is the reflected light fully polarized?

A concave shaving mirror has a radius of curvature of \(35.0 \mathrm{~cm} .\) It is positioned so that the (upright) image of a man's face is \(2.50\) times the size of the face. How far is the mirror from the face?
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