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Chapter 24: Problem 41

Repeat Problem \(40(\mathrm{c})\) using a different eyepiece that gives an angular magnification of 5.00 for a final image at the viewer's near point \((25.0 \mathrm{cm})\) instead of at infinity.

Short Answer

Expert verified
Answer: The length of the microscope is 5.47 cm.

Step by step solution

01

Find linear magnification of eyepiece

We will find the linear magnification of the eyepiece, using the total angular magnification and the linear magnification of the objective lens. Formula: \(M_{total} = M_{objective} \times M_{eyepiece}\) Solve for \(M_{eyepiece}\): \(M_{eyepiece} = \frac{M_{total}}{M_{objective}}\)
02

Calculate linear magnification of eyepiece

Now substitute the values for \(M_{total}\) and \(M_{objective}\) in the above equation to find \(M_{eyepiece}\): \(M_{eyepiece} = \frac{5.00}{40} = 0.125\)
03

Find the image distance of the eyepiece

Since we are given that angular magnification is for a final image at viewer's near point, we can find the distance of an object placed \(25\text{ cm}\) away from the eyepiece and is magnified by \(0.125\) times. Formula: \(d_{image} = d_{near} \times M_{eyepiece}\)
04

Calculate the image distance of the eyepiece

Substitute the values for \(d_{near}\) and \(M_{eyepiece}\) in the above equation to find \(d_{image}\): \(d_{image} = 25\text{ cm} \times 0.125 = 3.125\text{ cm}\)
05

Find the object distance of the eyepiece

Now, we will use the lens formula on the eyepiece to determine the object distance of the eyepiece. Lens formula: \(\frac{1}{f_{eyepiece}} = \frac{1}{d_{object}} + \frac{1}{d_{image}}\) Solve for \(d_{object}\): \(d_{object} = \frac{f_{eyepiece} \times d_{image}}{f_{eyepiece} - d_{image}}\)
06

Calculate the object distance of the eyepiece

Substitute the values for \(f_{eyepiece}\) and \(d_{image}\) in the above equation to find \(d_{object}\): \(d_{object} = \frac{10\text{ cm} \times 3.125\text{ cm}}{10\text{ cm} - 3.125\text{ cm}} = 4.47\text{ cm}\)
07

Find the length of the microscope

Now that we have the object distance for the eyepiece and the focal length of the objective lens, we can add them to find the length of the microscope. Formula: \(L = f_{objective} + d_{object}\)
08

Calculate the length of the microscope

Substitute the values for \(f_{objective}\) and \(d_{object}\) in the above equation to find \(L\): \(L = 1.00\text{ cm} + 4.47\text{ cm} = 5.47\text{ cm}\) The length of the microscope is \(5.47\text{ cm}\).

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Most popular questions from this chapter

The area occupied by one frame on 35 -mm film is \(24 \mathrm{mm}\) by $36 \mathrm{mm}-\( see the figure with Problem \)16 .$ The focal length of the camera lens is \(50.0 \mathrm{mm}\). A picture is taken of a person $182 \mathrm{cm}$ tall. What is the minimum distance from the camera for the person to stand so that the image fits on the film? Give two answers; one for each orientation of the camera.
You have two lenses of focal length \(25.0 \mathrm{cm}\) (lens 1 ) and $5.0 \mathrm{cm}$ (lens 2 ). (a) To build an astronomical telescope that gives an angular magnification of \(5.0,\) how should you use the lenses (which for objective and which for eyepiece)? Explain. (b) How far apart should they be?
A man requires reading glasses with \(+2.0 \mathrm{D}\) power to read a book held \(40.0 \mathrm{cm}\) away with a relaxed eye. Assume the glasses are $2.0 \mathrm{cm}$ from his eyes. (a) What is his uncorrected far point? (b) What refractive power lenses should he use for distance vision? (c) His uncorrected near point is \(1.0 \mathrm{m} .\) What should the refractive powers of the two lenses in his bifocals be to give him clear vision from \(25 \mathrm{cm}\) to infinity?
Two lenses, of focal lengths \(3.0 \mathrm{cm}\) and \(30.0 \mathrm{cm},\) are used to build a small telescope. (a) Which lens should be the objective? (b) What is the angular magnification? (c) How far apart are the two lenses in the telescope?
Suppose that the lens system (cornea + lens) in a particular eye has a focal length that can vary between \(1.85 \mathrm{cm}\) and \(2.00 \mathrm{cm},\) but the distance from the lens system to the retina is only \(1.90 \mathrm{cm} .\) (a) Is this eye nearsighted or farsighted? Explain. (b) What range of distances can the eye see clearly without corrective lenses?
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