Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 31: Problem 55

An object is \(68 \mathrm{cm}\) from a plano-convex lens whose curved side has curvature radius \(26 \mathrm{cm} .\) The refractive index of the lens is 1.62. Where is the image, and what type is it?

Short Answer

Expert verified
The image is at a distance v cm from the lens and it's of type based on the magnificaiton value.

Step by step solution

01

Apply Lens Formula to Establish Focal Length

The lens is a plano-convex lens with its curved side turned towards the object, so r1=26cm and r2=∞. Use the formula 1/f = (n-1)(1/r1 - 1/r2) to calculate the focal length. Replace r1 and n to calculate the value of f. It comes as f = 26 cm/(1.62-1)
02

Apply Lens Formula to Establish Image Position

Now use the lens formula 1/v -1/u = 1/f . The object distance u is -68 cm (negative because it's opposite to the direction of light). We know f from step 1. Substitute u and f and solve for v (image distance). Remember to take the inverse to obtain the value of v. If the value of v is positive, the image is virtual, on the same side as the object and if it's negative, the image is real and on the opposite side.
03

Check Magnification to Determine Image Type

The magnification formula is m = -v/u. Substitute v and u in this formula to get the magnification. If m is positive, the image is erect and if it's negative, the image is inverted.

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.

Start your free trial

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

A converging lens has focal length \(4.0 \mathrm{cm} .\) A 1.0 -cm-high arrow is located \(7.0 \mathrm{cm}\) from the lens with its lowest point \(5.0 \mathrm{mm}\) above the lens axis. Make a full-scale ray-tracing diagram to locate both ends of the image. Confirm using the lens equation.

You're given two lenses with different diameters. Knowing nothing else, you can conclude that a. the larger lens is faster. b. the smaller lens has the shorter focal length. c. the smaller lens suffers less spherical aberration. d. none of the above

A lens has focal length \(f=35 \mathrm{cm} .\) Find the type and height of the image produced when a 2.2 -cm-high object is placed at distances (a) \(f+10 \mathrm{cm}\) and (b) \(f-10 \mathrm{cm} .\)

An object is placed two focal lengths from a diverging lens. (a) What type of image forms, (b) what's the magnification, and (c) where is the image?

If a lens suffers from spherical aberration, stopping down will a. worsen the focus. b. improve the focus. c. not affect the focus.
See all solutions

Recommended explanations on Physics Textbooks

Electric Charge Field and Potential

Read Explanation

Engineering Physics

Read Explanation

Modern Physics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Oscillations

Read Explanation

Linear Momentum

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.

Sign-up for free