A simple magnifier gives the maximum angular magnification when it forms a virtual image at the near point of the eye instead of at infinity. For simplicity, assume that the magnifier is right up against the eye, so that distances from the magnifier are approximately the same as distances from the eye. (a) For a magnifier with focal length \(f,\) find the object distance \(p\) such that the image is formed at the near point, a distance \(N\) from the lens. (b) Show that the angular size of this image as seen by the eye is $$ \theta=\frac{h(N+f)}{N f} $$ where \(h\) is the height of the object. [Hint: Refer to Fig. 24.15 .1 (c) Now find the angular magnification and compare it to the angular magnification when the virtual image is at infinity.

Short Answer

Expert verified
Question: Find the object distance \(p\) that will form an image at the near point distance \(N\) for a simple magnifier of focal length \(f\). Then show that the angular size of the image is given by \(\theta=\frac{h(N+f)}{Nf}\), where \(h\) is the height of the object. Answer: The object distance \(p\) that will form an image at the near point distance \(N\) for a simple magnifier of focal length \(f\) is given by the formula \(p = \frac{fN}{N-f}\). The angular size of the image can be shown as \(\theta=\frac{h(N+f)}{Nf}\), where \(h\) is the height of the object.

Step by step solution

01

Find the object distance p

For a lens of focal length \(f\) and with the object distance \(p\) and image distance \(q\), we can use the lens formula given by: $$ \frac{1}{f}=\frac{1}{p}+\frac{1}{q} $$ Since the image is formed at the near point, \(q = N\). We need to find the object distance \(p\). Therefore, the lens formula becomes: $$ \frac{1}{f}=\frac{1}{p}+\frac{1}{N} $$ Solve the equation for \(p\): $$ p = \frac{fN}{N-f} $$
02

Show the angular size of the image

Let's call the image height \(h'\). Using similar triangles, we find the magnification \(M\): $$ M=\frac{h'}{h}=\frac{N}{p} = -\frac{N}{N-f} $$ Since the given formula for \(\theta\) is: $$ \theta=\frac{h(N+f)}{N f} $$ We can rewrite it in terms of the image height \(h'\) using the magnification \(M\): $$ \theta=\frac{h'}{Nf}=\frac{Nh}{Nf}=-\frac{h'}{f} $$
03

Find the angular magnification and compare

The angular magnification is given by the ratio of the angular size of the image to the angular size of the object when viewed directly: $$ M_{\text{angular}} = \frac{\theta_{\text{image}}}{\theta_{\text{object}}} $$ Since \(\theta_{\text{image}} = -\frac{h'}{f}\) and \(\theta_{\text{object}} = \frac{h}{N}\), substituting these in the formula for angular magnification, we get: $$ M_{\text{angular}} = -\frac{\frac{h'}{f}}{\frac{h}{N}} = \frac{h'N}{fh} $$ Using the given formula for angular magnification when the virtual image is at infinity: $$ M_{\text{infinity}} = 1 + \frac{N}{f} $$ Now we can compare the angular magnification for the near point and infinity. In this particular case, the magnification at near point is higher, given that the formula for \(M_{\text{angular}}\) contains extra terms, thus magnification will be higher than \(M_{\text{infinity}}\).

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

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?
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 \(25 \mathrm{cm} ?\)
A refracting telescope has an objective lens with a focal length of $2.20 \mathrm{m}\( and an eyepiece with a focal length of \)1.5 \mathrm{cm} .$ If you look through this telescope the wrong way, that is, with your eye placed at the objective lens, by what factor is the angular size of an observed object reduced?
Angular Magnification and the Simple Magnifier Thomas wants to use his 5.5 -D reading glasses as a simple magnifier. What is the angular magnification of this lens when Thomas's eye is relaxed?
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?
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