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Chapter 26: Problem 82

The contacts worn by a farsighted person allow her to see objects clearly that are as close as \(25.0 \mathrm{cm},\) even though her uncorrected near point is \(79.0 \mathrm{cm}\) from her eyes. When she is looking at a poster, the contacts form an image of the poster at a distance of \(217 \mathrm{cm}\) from her eyes. (a) How far away is the poster actually located? (b) If the poster is \(0.350 \mathrm{m}\) tall, how tall is the image formed by the contacts?

Short Answer

Expert verified
(a) Poster distance is found using the lens formula; (b) Image height uses magnification.

Step by step solution

01

Understand the Lens Formula

For lenses, the relationship between the object distance \(d_o\), the image distance \(d_i\), and the focal length \(f\) is given by the lens formula: \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\). We need to find the actual distance to the poster (object distance \(d_o\)) by using the formula, where \(d_i = 217 \text{ cm}\) and \(d_o\) is unknown.
02

Determine the Focal Length

First, let's find the focal length of the contact lenses. Given her corrected near point (image distance) is \(d_i = 217 \text{ cm}\), and when used normally (without accommodation), the objects should appear at \(25 \text{ cm}\) (her modified near point). Thus, for her contacts, we use these values to find \(f\).Rearrange: \(\frac{1}{f} = \frac{1}{25} + \frac{1}{217}\)Calculate \(f\).
03

Utilize the Lens Formula Again for Poster

Knowing the focal length \(f\) from the first calculation, use the lens formula again for the poster. Here, \(d_i = 217 \text{ cm}\), and \(f\) is known. Rearrange to find \(d_o\):\(\frac{1}{d_o} = \frac{1}{f} - \frac{1}{d_i}\)Solve for \(d_o\) to find the distance of the poster.
04

Calculate Image Height

The magnification \(m\) caused by the lens is given by \(m = -\frac{d_i}{d_o}\). The image height \(h_i\) is related to the object height \(h_o\) and magnification by the formula: \(h_i = m \times h_o\).Given \(h_o = 0.350 \text{ m}\), calculate \(h_i\) using the found \(d_o\) and \(d_i = 217 \text{ cm}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Object Distance
The object distance, usually denoted as \(d_o\), is the distance between the object you are observing and the lens. It is a critical variable in the lens formula, which helps us understand how lenses form images.

In the exercise, you are tasked with finding the object distance of the poster from the observer. The lens formula is used to calculate this, expressed as:
\[\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\]
  • Here, \(f\) is the focal length
  • \(d_o\) is the object distance (what we need to find)
  • \(d_i\) is the image distance.
By rearranging the formula, we can solve for \(d_o\). This is done by isolating \(d_o\) on one side, giving:
\[\frac{1}{d_o} = \frac{1}{f} - \frac{1}{d_i}.\]Calculating \(d_o\) helps us determine how far away the "real" object is from the lens.
Focal Length
The focal length, \(f\), is a key measurement in understanding how lenses work. It indicates the distance at which parallel rays of light converge after passing through a lens.

For this exercise, the focal length is used to determine the interaction between the lenses and the light from the poster. Using the lens formula, the focal length can be derived from any known object and image distances. In this problem:
  • The corrected near point (acting as image distance \(d_i\)) is given as 217 cm.
  • The near point (modified due to the contact lenses) is 25 cm.
To find \(f\), rearrange the lens formula:
\[\frac{1}{f} = \frac{1}{25} + \frac{1}{217}.\]Here, plugging in the values gives the focal length, which allows us to solve further for object distances.
Image Distance
The image distance, \(d_i\), is the distance from the lens to the image it creates. In this context, it's how far the image formed by the contact lenses is from the observer's eyes.

In the problem, the image distance provided is 217 cm. This is the calculated placement of the image formed when the observer looks at the poster.
  • This value is used alongside the lens formula in calculations to find the object distance.
  • Allows us to calculate how far away "in reality" the image appears once corrected by the lenses.
The image distance is a reflection of how the lens can adjust our perception and assists in achieving clear vision at various real-world distances.
Magnification
Magnification, denoted as \(m\), is a measure of how much larger or smaller the image is compared to the object itself. It describes the magnification power of the lenses.

It can be calculated using the formula:
\[m = -\frac{d_i}{d_o}\]
  • The negative sign indicates that the image is formed on the opposite side, often inverted.
The exercise also connects magnification to image height. The height of the image \(h_i\) is calculated from the object height \(h_o\) and the magnification using:
\[h_i = m \times h_o\]For the exercise:
  • Given the object height as 0.350 m, calculate the image height using the magnification.
This gives insight into both the size and orientation of the image seen through the lens.

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Most popular questions from this chapter

An object is placed \(20.0 \mathrm{cm}\) to the left of a diverging lens \((f=\) \(-8.00 \mathrm{cm}$$) .\) A concave mirror \((f=12.0 \mathrm{cm})\) is placed \(30.0 \mathrm{cm}\) to the right of the lens. (a) Find the final image distance, measured relative to the mirror. (b) Is the final image real or virtual? (c) Is the final image upright or inverted with respect to the original object?

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.

The distance between the lenses in a compound microscope is \(18 \mathrm{cm} .\) The focal length of the objective is \(1.5 \mathrm{cm} .\) If the microscope is to provide an angular magnification of -83 when used by a person with a normal near point \((25 \mathrm{cm}\) from the eye), what must be the focal length of the eyepiece?

The lengths of three telescopes are \(L_{\mathrm{A}}=455 \mathrm{mm}, L_{\mathrm{B}}=615 \mathrm{mm},\) and \(L_{C}=824 \mathrm{mm} .\) The focal length of the eyepiece for each telescope is 3.00 mm. Find the angular magnification of each telescope.

A small logo is embedded in a thick block of crown glass \((n=1.52)\), \(3.20 \mathrm{cm}\) beneath the top surface of the glass. The block is put under water, so there is \(1.50 \mathrm{cm}\) of water above the top surface of the block. The logo is viewed from directly above by an observer in air. How far beneath the top surface of the water does the logo appear to be?
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