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Chapter 26: Problem 117

A converging lens \((f=12.0 \mathrm{cm})\) is held \(8.00 \mathrm{cm}\) in front of a newspaper that has a print size with a height of \(2.00 \mathrm{mm}\). Find (a) the image distance (in \(\mathrm{cm}\) ) and (b) the height (in \(\mathrm{mm}\) ) of the magnified print.

Short Answer

Expert verified
Image distance is -24.0 cm (virtual), and the height of the print is 6.00 mm.

Step by step solution

01

Identify Given Values

The problem provides us with the following values: the focal length of the lens \( f = 12.0 \ \mathrm{cm} \), the object distance \( d_o = 8.00 \ \mathrm{cm} \), and the height of the object (print size) \( h_o = 2.00 \ \mathrm{mm} \).
02

Use the Lens Formula

The lens formula is \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \). We will rearrange this formula to solve for the image distance \( d_i \).
03

Solve for Image Distance

Rearrange the lens formula to \( \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} \), and substitute the known values: \( \frac{1}{d_i} = \frac{1}{12.0} - \frac{1}{8.00} \). Calculate this to find \( \frac{1}{d_i} = \frac{1}{12.0} - \frac{1}{8.00} = \frac{2}{24.0} - \frac{3}{24.0} = -\frac{1}{24.0} \), thus \( d_i = -24.0 \ \mathrm{cm} \).
04

Determine the Image Nature

Since \( d_i \) is negative, it suggests that the image is virtual and located on the same side as the object.
05

Calculate the Magnification

The magnification \( m \) is given by \( m = -\frac{d_i}{d_o} \). Substitute the known values to find \( m = -\frac{-24.0}{8.00} = 3.0 \).
06

Find the Image Height

Using the magnification equation which also relates to height, \( m = \frac{h_i}{h_o} \), we substitute \( m = 3.0 \) and \( h_o = 2.00 \ \mathrm{mm} \) to get \( h_i = 3.0 \times 2.00 = 6.00 \ \mathrm{mm} \).

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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, often referred to as a convex lens, is a lens that brings light rays together at a point. This type of lens is thicker in the middle than at the edges.
A simple way to describe its shape is that it bulges outward. Converging lenses have several key concepts:
  • They have a focal point, which is the point where parallel light rays meet after passing through the lens.
  • The distance from the lens to the focal point is called the focal length, often denoted by \( f \).
  • These lenses can produce both real and virtual images depending on the object's position relative to the lens.
In our given exercise, the converging lens has a focal length of \(12.0 \, \text{cm}\) and helps in forming an image of the newspaper print.
Image Distance
The image distance, denoted by \(d_i\), is the distance between the lens and the image it forms. This is essential to determine what type of image is produced - whether it is real or virtual.
The sign and magnitude of \(d_i\) inform us about the image nature:
  • If \(d_i\) is positive, the image is real and inverted, formed on the opposite side of the object.
  • If \(d_i\) is negative, the image is virtual and upright, appearing on the same side as the object.
In this particular exercise, the calculation gives an image distance \(d_i = -24.0 \, \text{cm}\). The negative sign indicates the image is virtual, which occurs on the same side as the object.
Lens Formula
The lens formula is an equation that relates the focal length \( f \), the object distance \( d_o \), and the image distance \( d_i \). It is given by:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]This formula is fundamental in optics as it helps in determining the image's location and properties based on the given object and focal lengths.
In our exercise:
  • The object's distance \(d_o\) is \(8.00 \, \text{cm}\).
  • The focal length \(f\) is \(12.0 \, \text{cm}\).
Rearranging the formula allows us to find the image distance \(d_i\) by solving:\[ \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} \]Using this, the negative image distance indicates a virtual image.
Magnification
Magnification is a measure of how much larger or smaller the image is compared to the object. It is given by the formula:\[ m = -\frac{d_i}{d_o} \]This equation also relates the image height \( h_i \) to the object height \( h_o \):\[ m = \frac{h_i}{h_o} \]In our problem:
  • The calculated magnification is \( m = 3.0 \), meaning the image is three times larger than the object.
  • The object height \(h_o\) is \(2.00 \, \text{mm}\).
  • The image height \(h_i\) is then calculated as \(6.00 \, \text{mm}\) using the magnification relationship.
This positive magnification signifies that the image is upright and virtual.

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Most popular questions from this chapter

A spectator, seated in the left-field stands, is watching a baseball player who is \(1.9 \mathrm{m}\) tall and is \(75 \mathrm{m}\) away. On a \(\mathrm{TV}\) screen, located \(3.0 \mathrm{m}\) from a person watching the game at home, the image of this same player is \(0.12 \mathrm{m}\) tall. Find the angular size of the player as seen by (a) the spectator watching the game live and (b) the TV viewer. (c) To whom does the player appear to be larger?

A converging lens \(\left(f_{1}=24.0 \mathrm{cm}\right)\) is located \(56.0 \mathrm{cm}\) to the left of a diverging lens \(\left(f_{2}=-28.0 \mathrm{cm}\right) .\) An object is placed to the left of the converging lens, and the final image produced by the two-lens combination lies \(20.7 \mathrm{cm}\) to the left of the diverging lens. How far is the object from the converging lens?

A camper is trying to start a fire by focusing sunlight onto a piece of paper. The diameter of the sun is \(1.39 \times 10^{9} \mathrm{m},\) and its mean distance from the earth is \(1.50 \times 10^{11} \mathrm{m} .\) The camper is using a converging lens whose focal length is \(10.0 \mathrm{cm}\). (a) What is the area of the sun's image on the paper? (b) If \(0.530 \mathrm{W}\) of sunlight passes through the lens, what is the intensity of the sunlight at the paper?

The drawing shows a ray of light traveling through three materials whose surfaces are parallel to each other. The refracted rays (but not the reflected rays) are shown as the light passes through each material. A ray of light strikes the \(a-b\) interface at a \(50.0^{\circ}\) angle of incidence. The index of refraction of material \(a\) is \(n_{a}=1.20 .\) The angles of refraction in materials \(b\) and \(c\) are, respectively, \(45.0^{\circ}\) and \(56.7^{\circ} .\) Find the indices of refraction in these two media.

In an ultra-low-temperature experiment, a collection of sodium atoms enter a special state called a Bose-Einstein condensate in which the index of refraction is \(1.57 \times 10^{7} .\) What is the speed of light in this condensate?
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