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Chapter 33: Problem 39

A person with a near-point distance of \(24.0 \mathrm{~cm}\) finds that a magnifying glass gives an angular magnification that is 1.25 times larger when the image of the magnifier is at the near point than when the image is at infinity. What is the focal length of the magnifying glass?

Short Answer

Expert verified
Answer: The focal length of the magnifying glass lens is 24.0 cm.

Step by step solution

01

Understand the given information

The near-point distance \(d_N = 24.0 cm\). The angular magnification when the image is at the near-point is \(1.25\) times larger than when the image is at infinity.
02

Write the lens formula

The lens formula states: \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\), where \(f\) is the focal length, \(d_o\) is the object distance, and \(d_i\) is the image distance.
03

Write the magnification formula

The magnification formula for a magnifying glass is: \(M = \frac{d_i}{d_N} = 1 + \frac{d_N}{f}\), where \(M\) is the angular magnification, and \(d_N\) is the near-point distance.
04

Relate the magnification when the image is at the near-point and at infinity

Let \(M_1\) be the magnification when the image is at infinity and \(M_2\) be the magnification when the image is at the near-point. We are given that \(M_2 = 1.25M_1\). Using the magnification formula: \(M_1 = 1 + \frac{d_N}{f}\Rightarrow f = \frac{d_N}{M_1 - 1}\) \(M_2 = 1 + \frac{d_N}{f - d_N}\)
05

Substitute and solve for focal length

Substitute \(M_2 = 1.25M_1\) and \(f = \frac{d_N}{M_1 - 1}\) into the equation for \(M_2\): \(1.25M_1 = 1 + \frac{d_N}{\frac{d_N}{M_1 - 1} - d_N}\) Now, solve for \(M_1\): \(M_1 = \frac{\frac{5}{4}}{1 - \frac{1}{5}} = 2\) Now we can find the focal length using the equation \(f = \frac{d_N}{M_1 - 1}\): \(f = \frac{24.0cm}{2 - 1} = 24.0cm\) Therefore, the focal length of the magnifying glass is \(24.0cm\).

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Most popular questions from this chapter

An instructor wants to use a lens to project a real image of a light bulb onto a screen \(1.71 \mathrm{~m}\) from the bulb. In order to get the image to be twice as large as the bulb, what focal length lens will be needed?

An object is \(6.0 \mathrm{~cm}\) from a thin lens along the axis of the lens. If the lens has a focal length of \(9.0 \mathrm{~cm},\) determine the image distance.

Two refracting telescopes are used to look at craters on the Moon. The objective focal length of both telescopes is \(95.0 \mathrm{~cm}\) and the eyepiece focal length of both telescopes is \(3.80 \mathrm{~cm} .\) The telescopes are identical except for the diameter of the lenses. Telescope A has an objective diameter of \(10.0 \mathrm{~cm}\) while the lenses of telescope \(\mathrm{B}\) are scaled up by a factor of two, so that its objective diameter is \(20.0 \mathrm{~cm}\). a) What are the angular magnifications of telescopes \(A\) and \(B\) ? b) Do the images produced by the telescopes have the same brightness? If not, which is brighter and by how much?

You have found in the lab an old microscope, which has lost its eyepiece. It still has its objective lens, and markings indicate that its focal length is \(7.00 \mathrm{~mm}\). You can put in a new eyepiece, which goes in \(20.0 \mathrm{~cm}\) from the objective. You need a magnification of about 200. Assume you want the comfortable viewing distance for the final image to be \(25.0 \mathrm{~cm}\). You find in a drawer eyepieces marked \(2.00-, 4.00-,\) and \(8.00-\mathrm{cm}\) focal length. Which is your best choice?

As a high-power laser engineer you need to focus a 1.06-mm diameter laser beam to a 10.0 - \(\mu\) m diameter spot \(20.0 \mathrm{~cm}\) behind the lens. What focal length lens would you use?
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