/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 77 A microscope has an eyepiece of ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The uncorrected far point of Colin's eye is \(2.0 \mathrm{m} .\) What refractive power contact lens enables him to clearly distinguish objects at large distances?
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 located at \(x=0 .\) At \(x=2.50 \mathrm{cm}\) is a converging lens with a focal length of \(2.00 \mathrm{cm},\) at \(x=16.5 \mathrm{cm}\) is an unknown lens, and at \(x=19.8 \mathrm{cm}\) is another converging lens with focal length \(4.00 \mathrm{cm} .\) An upright image is formed at $x=39.8 \mathrm{cm} .$ For each lens, the object distance exceeds the focal length. The magnification of the system is \(6.84 .\) (a) Is the unknown lens diverging or converging? (b) What is the focal length of the unknown lens? (c) Draw a ray diagram to confirm your answer.
A photographer wishes to take a photograph of the Eiffel Tower $(300 \mathrm{m}\( tall ) from across the Seine River, a distance of \)300 \mathrm{m}$ from the tower. What focal length lens should she use to get an image that is 20 mm high on the film?
The Ortiz family is viewing slides from their summer vacation trip to the Grand Canyon. Their slide projector has a projection lens of \(10.0-\mathrm{cm}\) focal length and the screen is located \(2.5 \mathrm{m}\) from the projector. (a) What is the distance between the slide and the projection lens? (b) What is the magnification of the image? (c) How wide is the image of a slide of width \(36 \mathrm{mm}\) on the screen? (See the figure with Problem 16 .)
See all solutions

Recommended explanations on Physics Textbooks

Scientific Method Physics

Read Explanation

Fluids

Read Explanation

Turning Points in Physics

Read Explanation

Medical Physics

Read Explanation

Famous Physicists

Read Explanation

Astrophysics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.