/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 60 When a converging lens is used i... [FREE SOLUTION] | 91影视

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Chapter 26: Problem 60

When a converging lens is used in a camera (as in Figure 26.26b), the film must be at a distance of 0.210 m from the lens to record an image of an object that is 4.00 m from the lens. The same lens and film are used in a projector (see Figure 26.27b), with the screen 0.500 m from the lens. How far from the projector lens should the film be placed?

Short Answer

Expert verified
The film should be placed 0.333 m from the projector lens.

Step by step solution

01

Understand the Lens Formula

The lens formula relates the object distance \( (d_o) \), image distance \( (d_i) \), and focal length \( (f) \) of a lens. It is given by the equation \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). Our goal is to find the image distance for the projector lens.
02

Calculate the Focal Length

First, calculate the focal length of the lens using the camera setup. The object distance \( d_o \) is 4.00 m and the image distance \( d_i \) is 0.210 m. Plug these into the lens formula: \[ \frac{1}{f} = \frac{1}{4.00} + \frac{1}{0.210} \]Calculate \( f \).
03

Perform Calculations

Calculate the reciprocals:\( \frac{1}{4.00} = 0.25 \) and \( \frac{1}{0.210} \approx 4.76 \).Thus, \( \frac{1}{f} = 0.25 + 4.76 = 5.01 \).So, \( f \approx \frac{1}{5.01} \approx 0.200 \) m.
04

Apply Lens Formula for Projector Set-up

With the projector setup, the screen's distance is the new image distance \( d_i = 0.500 \) m. Use the lens formula again:\[ \frac{1}{0.200} = \frac{1}{d_o} + \frac{1}{0.500} \].Rearrange to solve for \( \frac{1}{d_o} \).
05

Solve for Object Distance

Plug the values into the rearranged formula:\( \frac{1}{0.200} = \frac{1}{d_o} + 2 \).Thus, \( 5 = \frac{1}{d_o} + 2 \).Solve for \( \frac{1}{d_o} \):\( \frac{1}{d_o} = 5 - 2 = 3 \).Therefore, \( d_o = \frac{1}{3} \approx 0.333 \) m.

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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, focuses parallel rays of light to a point called the focal point. This kind of lens bulges out in the center more than at the edges. The primary characteristic of a converging lens is its ability to bend light rays inwards, allowing them to meet at a single point.

Some key features of a converging lens include:
  • Biconvex in shape
  • Used for focusing light
  • Capable of forming real and virtual images
These properties make converging lenses essential in devices like cameras and projectors. They help in gathering light from a point and projecting it onto film or a screen to form clear images.
Focal Length
The focal length of a lens is a critical measure. It is the distance from the center of the lens to its focal point. A shorter focal length means a lens can bend light rays more sharply, ensuring they converge more quickly, leading to a stronger magnifying effect.

When using the lens formula, the focal length is expressed in meters and often determined through the equation: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]
Understanding the focal length is fundamental for setting the correct distance between the lens and the film or sensor in devices like cameras and projectors. It influences both the magnification and clarity of images produced.
Image Distance
The image distance, denoted as \( d_i \), is the distance from the lens to the point where the image is formed. This can be on a screen, a sheet of film, or even virtually in space.

To find the image distance given the focal length and object distance, you can use the lens formula.
  • If \( d_i \) is positive, it indicates a real image formed on the opposite side of the object.
  • A negative \( d_i \) means a virtual image is produced on the same side as the object.
Calculating the image distance accurately ensures that the image is focused correctly, providing sharp and clear visuals.
Object Distance
The object distance, represented as \( d_o \), is the measure from the object to the lens along the principal axis. The object distance plays a vital role in determining how and where the image will be formed.

Here's how it typically functions in scenarios:
  • In cameras, the object distance is usually large, meaning the object is far from the lens.
  • In projectors, the object distance might be shorter as the film or digital sensor needs to be closely aligned to the lens for proper image projection.
Like other variables in the lens equation, knowing the object distance lets you manipulate other factors such as image size and clarity to achieve the optimal focus for various visual applications.

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Most popular questions from this chapter

A paperweight consists of a 9.00-cm-thick plastic cube. Within the plastic a thin sheet of paper is embedded, parallel to opposite faces of the cube. On each side of the paper is printed a different joke that can be read by looking perpendicularly straight into the cube. When read from one side (the top), the apparent depth of the paper in the plastic is 4.00 cm. When read from the opposite side (the bottom), the apparent depth of the paper in the plastic is 1.63 cm. What is the index of refraction of the plastic?

A dentist is examining a dental filling in a patient鈥檚 tooth. The diameter of the filling is 2.4 mm, and the dentist鈥檚 near point is 17.0 cm. To get a better look at the filling, the dentist dons safety goggles fitted with magnifying glasses \((f=6.0 \mathrm{cm}) .\) Find the greatest possible angular size (in radians) of the patient's filling when viewed by the dentist, both (a) without and \(\quad\) (b) with the magnifying glasses.

A ray of light is traveling in glass and strikes a glass鈥搇iquid interface. The angle of incidence is \(58.0^{\circ}\) , and the index of refraction of glass is \(n=1.50 .\) (a) What must be the index of refraction of the liquid so that the direction of the light entering the liquid is not changed? (b) What is the largest index of refraction that the liquid can have, so that none of the light is transmitted into the liquid and all of it is reflected back into the glass?

In a compound microscope, the objective has a focal length of 0.60 cm, while the eyepiece has a focal length of 2.0 cm. The separation between the objective and the eyepiece is L 12.0 cm. Another microscope that has the same angular magnification can be constructed by interchanging the two lenses, provided that the distance between the lenses is adjusted to a value \(L^{\prime}\) , Find \(L^{\prime}\)

A ray of sunlight is passing from diamond into crown glass; the angle of incidence is \(35.00^{\circ} .\) The indices of refraction for the blue and red components of the ray are: blue \(\left(n_{\text { diamood }}=2.444, n_{\text { cromgless }}=1.531\right)\) and red \(\left(n_{\text { dianood }}=2.410, n_{\text { crown glass }}=1.520\right) .\) Determine the angle between the refracted blue and red rays in the crown glass.
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