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Chapter 24: Problem 77

A microscope has an eyepiece of focal length \(2.00 \mathrm{cm}\) and an objective of focal length \(3.00 \mathrm{cm} .\) The eyepiece produces a virtual image at the viewer's near point \((25.0 \mathrm{cm}\) from the eye). (a) How far from the eyepiece is the image formed by the objective? (b) If the lenses are \(20.0 \mathrm{cm}\) apart, what is the distance from the objective lens to the object being viewed? (c) What is the angular magnification?

Short Answer

Expert verified
Question: Calculate a) the distance from the eyepiece to the image formed by the objective, b) the distance from the objective lens to the object being viewed, and c) the angular magnification for a compound microscope with given focal lengths and lens distances. Given: Focal length of eyepiece (f_e) = 2.00 cm, focal length of objective lens (f_o) = 3.00 cm, distance between lenses (d) = 20.0 cm, and final image distance (v_e) = 25.0 cm. Answer: a) The distance from the eyepiece to the image formed by the objective (u_e) is 3.33 cm. b) The distance from the objective lens to the object being viewed (u_o) is 4.76 cm. c) The angular magnification of the microscope is 3.5.

Step by step solution

01

Calculate the distance from the eyepiece to the image formed by the objective (part a)

Using the lens formula: \(1/f = 1/v - 1/u\) where f is the focal length of the eyepiece, v is the image distance, and u is the object distance. We are given the focal length of the eyepiece (f_e = 2.00 cm) and the final image distance (v_e = 25.0 cm). Plugging these values into the lens formula, we get: \(1/2.00 = 1/25.0 - 1/u_e\) Solving for \(u_e\), we get the distance from the eyepiece to the image formed by the objective: \(u_e = 3.33 \mathrm{cm}\)
02

Calculate the distance from the objective lens to the object being viewed (part b)

We can use the lens formula for the objective lens: \(1/f_o = 1/v_o - 1/u_o\) We are given the focal length of the objective lens (f_o = 3.00 cm) and the distance between the lenses (d = 20.0 cm). Since we know the distance from the eyepiece to the image formed by the objective (\(u_e = 3.33 \mathrm{cm}\)), we can find the image distance (v_o) using the given distance between the lenses: \(v_o = d - u_e = 20.0 - 3.33 = 16.67 \mathrm{cm}\) Now, we plug the values into the lens formula for the objective lens: \(1/3.00 = 1/16.67 - 1/u_o\) Solving for \(u_o\), we find the distance from the objective lens to the object being viewed: \(u_o = 4.76 \mathrm{cm}\)
03

Calculate the angular magnification (part c)

The angular magnification can be found using the formula: Angular Magnification = \(v_o / u_o\) Plugging our values of \(v_o\) and \(u_o\), we get: Angular Magnification = \(16.67 / 4.76 = 3.5\) The angular magnification of the microscope is 3.5.

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Most popular questions from this chapter

Jordan is building a compound microscope using an eyepiece with a focal length of \(7.50 \mathrm{cm}\) and an objective with a focal length of $1.500 \mathrm{cm} .\( He will place the specimen a distance of \)1.600 \mathrm{cm}$ from the objective. (a) How far apart should Jordan place the lenses? (b) What will be the angular magnification of this microscope?
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?
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