Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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 Vaia!

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 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 .)
Veronique is nearsighted; she cannot sce clearly anything more than $6.00 \mathrm{m}$ away without her contacts. One day she doesn't wear her contacts; rather, she wears an old pair of glasses prescribed when she could see clearly up to \(8.00 \mathrm{m}\) away. Assume the glasses are \(2.0 \mathrm{cm}\) from her eyes. What is the greatest distance an object can be placed so that she can see it clearly with these glasses?
A cub scout makes a simple microscope by placing two converging lenses of +18 D at opposite ends of a \(28-\mathrm{cm}^{-}\) long tube. (a) What is the tube length of the microscope? (b) What is the angular magnification? (c) How far should an object be placed from the objective lens?
Two converging lenses, separated by a distance of \(50.0 \mathrm{cm},\) are used in combination. The first lens, located to the left, has a focal length of \(15.0 \mathrm{cm} .\) The second lens, located to the right, has a focal length of \(12.0 \mathrm{cm} .\) An object, \(3.00 \mathrm{cm}\) high, is placed at a distance of \(20.0 \mathrm{cm}\) in front of the first lens. (a) Find the intermediate and final image distances relative to the corresponding lenses. (b) What is the total magnification? (c) What is the height of the final image?

A converging lens with a focal length of \(15.0 \mathrm{cm}\) and a diverging lens are placed \(25.0 \mathrm{cm}\) apart, with the converging lens on the left. A 2.00 -cm-high object is placed \(22.0 \mathrm{cm}\) to the left of the converging lens. The final image is \(34.0 \mathrm{cm}\) to the left of the converging lens. (a) What is the focal length of the diverging lens? (b) What is the height of the final image? (c) Is the final image upright or inverted?

See all solutions

Recommended explanations on Physics Textbooks

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.

Sign-up for free