Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 24: Problem 70

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 \(25 \mathrm{cm} ?\)

Short Answer

Expert verified
The angular magnification when the final image is at infinity is 3, and when the final image is at the near point (25 cm), it is 2.

Step by step solution

01

Calculate the focal length of the lens

We are given the power of the lens (P) and we can find the focal length (f) using the formula: \( f = \frac{1}{P} \) In our case, P = +12 D. So the focal length of the lens is: \( f = \frac{1}{12} = 0.0833 \mathrm{m} \)
02

Find the object distance (u) when the image is at infinity

When the final image is at infinity, the object is placed at the focal point of the lens (f). In this case, the object distance u is equal to the focal length (0.0833 m): \( u = f = 0.0833 \mathrm{m} \)
03

Calculate the angular magnification (a)

We can now find the angular magnification using the formula: \( M = \frac{d}{u} \) Here, d is the least distance of distinct vision, which is 25 cm or 0.25 m. So the angular magnification for case (a) is: \( M = \frac{0.25}{0.0833} = 3 \) Thus, the angular magnification when the final image is at infinity is 3.
04

Find the object distance (u) when the image is at the near point

When the image is at the near point (25 cm), we need to rearrange the lens formula and solve for u: \( \frac{1}{u} = \frac{1}{f} - \frac{1}{v} \) We know the focal length (f) and the image distance (v) is equal to the near point distance, which is 25 cm or 0.25 m. Substituting these values, we get: \( \frac{1}{u} = \frac{1}{0.0833} - \frac{1}{0.25} = 8 \) Thus, the object distance (u) when the image is at the near point is: \( u = \frac{1}{8} = 0.125 \mathrm{m} \)
05

Calculate the angular magnification (b)

Now, we can find the angular magnification for case (b) using the formula: \( M = \frac{d}{u} \) Using the object distance (u) calculated in Step 4, we have: \( M = \frac{0.25}{0.125} = 2 \) Thus, the angular magnification when the final image is at the near point is 2. In conclusion, the angular magnification of the convex lens is 3 when the final image is at infinity and 2 when the final image is at the near point (25 cm).

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

The wing of an insect is \(1.0 \mathrm{mm}\) long. When viewed through a microscope, the image is \(1.0 \mathrm{m}\) long and is located \(5.0 \mathrm{m}\) away. Determine the angular magnification.
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 converging lens with focal length \(3.00 \mathrm{cm}\) is placed $4.00 \mathrm{cm}$ to the right of an object. A diverging lens with focal length \(-5.00 \mathrm{cm}\) is placed \(17.0 \mathrm{cm}\) to the right of the converging lens. (a) At what location(s), if any, can you place a screen in order to display an image? (b) Repeat part (a) for the case where the lenses are separated by \(10.0 \mathrm{cm}.\)
You plan to project an inverted image \(30.0 \mathrm{cm}\) to the right of an object. You have a diverging lens with focal length \(-4.00 \mathrm{cm}\) located \(6.00 \mathrm{cm}\) to the right of the object. Once you put a second lens at \(18.0 \mathrm{cm}\) 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?
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).
See all solutions

Recommended explanations on Physics Textbooks

Rotational Dynamics

Read Explanation

Classical Mechanics

Read Explanation

Scientific Method Physics

Read Explanation

Famous Physicists

Read Explanation

Physics of Motion

Read Explanation

Engineering Physics

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