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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

A simple magnifier gives the maximum angular magnification when it forms a virtual image at the near point of the eye instead of at infinity. For simplicity, assume that the magnifier is right up against the eye, so that distances from the magnifier are approximately the same as distances from the eye. (a) For a magnifier with focal length \(f,\) find the object distance \(p\) such that the image is formed at the near point, a distance \(N\) from the lens. (b) Show that the angular size of this image as seen by the eye is $$ \theta=\frac{h(N+f)}{N f} $$ where \(h\) is the height of the object. [Hint: Refer to Fig. 24.15 .1 (c) Now find the angular magnification and compare it to the angular magnification when the virtual image is at infinity.
A nearsighted man cannot clearly see objects more than \(2.0 \mathrm{m}\) away. The distance from the lens of the eye to the retina is \(2.0 \mathrm{cm},\) and the eye's power of accommodation is \(4.0 \mathrm{D}\) (the focal length of the cornea-lens system increases by a maximum of \(4.0 \mathrm{D}\) over its relaxed focal length when accommodating for nearby objects). (a) As an amateur optometrist, what corrective eyeglass lenses would you prescribe to enable him to clearly see distant objects? Assume the corrective lenses are $2.0 \mathrm{cm}$ from the eyes. (b) Find the nearest object he can see clearly with and without his glasses.
You would like to project an upright image at a position \(32.0 \mathrm{cm}\) to the right of an object. You have a converging lens with focal length $3.70 \mathrm{cm}\( located \)6.00 \mathrm{cm}$ to the right of the object. By placing a second lens at \(24.65 \mathrm{cm}\) to the right of the object, you obtain an image in the proper location. (a) What is the focal length of the second lens? (b) Is this lens converging or diverging? (c) What is the total magnification? (d) If the object is \(12.0 \mathrm{cm}\) high, what is the image height?
A convex lens of power +12 D is used as a magnifier to examine a wildflower. What is the angular magnification if the final image is at (a) infinity or (b) the near point of \(25 \mathrm{cm} ?\)
An object is placed \(12.0 \mathrm{cm}\) in front of a lens of focal length $5.0 \mathrm{cm} .\( Another lens of focal length \)4.0 \mathrm{cm}\( is placed \)2.0 \mathrm{cm}$ past the first lens. (a) Where is the final image? Is it real or virtual? (b) What is the overall magnification? (interactive: virtual optics lab).
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