Use the thin-lens equation to show that the transverse magnification due to the objective of a microscope is \(m_{o}=-U f_{o} .\) [Hints: The object is near the focal point of the objective: do not assume it is at the focal point. Eliminate \(p_{0}\) to find the magnification in terms of \(q_{0}\) and $\left.f_{0} . \text { How is } L \text { related to } q_{0} \text { and } f_{0} ?\right].$

Short Answer

Expert verified
Answer: The transverse magnification due to the objective of a microscope is given by \(m_o = -Uf_o\), where \(U = \frac{L-f_e}{f_o}\), \(L\) is the tube length, and \(f_o\) and \(f_e\) are the focal lengths of the objective and eyepiece, respectively.

Step by step solution

01

Define the thin-lens equation

The thin-lens equation is given by \(\frac{1}{f} = \frac{1}{p} + \frac{1}{q}\), where \(f\) is the focal length, \(p\) is the object distance, and \(q\) is the image distance.
02

Apply the lens equation to the microscope objective

For the microscope objective, we can write the lens equation as \(\frac{1}{f_o} = \frac{1}{p_o} + \frac{1}{q_o}\), where \(f_o\) is the objective focal length, \(p_o\) is the object distance, and \(q_o\) is the image distance.
03

Rearrange the equation to eliminate \(p_o\)

To eliminate \(p_o\), we can rearrange the equation as \(p_o = \frac{f_o q_o}{q_o - f_o}\).
04

Calculate the transverse magnification of the objective

The transverse magnification formula is \(m_o = \frac{-q_o}{p_o}\). Using the rearranged equation from Step 3, we get \(m_o = \frac{-q_o}{\frac{f_o q_o}{q_o - f_o}} = - \frac{q_o (q_o - f_o)}{f_o q_o}= -\frac{q_o - f_o}{f_o}\).
05

Find the relationship between \(L\), \(q_o\), and \(f_o\)

To find the relationship between \(L\), \(q_o\), and \(f_o\), we need to consider the geometry of the microscope. The tube length \(L\) is the distance between the objective and eyepiece focal points. As the object is near the focal point of the objective, we have \(L \approx q_o + f_e - f_o\). We can rearrange this expression for \(q_o\) as \(q_o = L + f_o - f_e\), where \(f_e\) is the focal length of the eyepiece.
06

Substitute the relationship into the magnification formula

Substituting the relationship from Step 5 into the magnification formula in Step 4, we have \(m_o = -\frac{(L + f_o - f_e) - f_o}{f_o} = -\frac{L-f_e}{f_o}\).
07

Show the final expression for the transverse magnification

To express the transverse magnification due to the objective of a microscope, we can now write \(m_o = -Uf_o\), where \(U = \frac{L-f_e}{f_o}\).

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

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 camera lens has a fixed focal length of magnitude \(50.0 \mathrm{mm} .\) The camera is focused on a 1.0 -m-tall child who is standing \(3.0 \mathrm{m}\) from the lens. (a) Should the image formed be real or virtual? Why? (b) Is the lens converging or diverging? Why? (c) What is the distance from the lens to the film? (d) How tall is the image on the film? (e) To focus the camera, the lens is moved away from or closer to the film. What is the total distance the lens must be able to move if the camera can take clear pictures of objects at distances anywhere from \(1.00 \mathrm{m}\) to infinity?
You have a set of converging lenses with focal lengths $1.00 \mathrm{cm}, 10.0 \mathrm{cm}, 50.0 \mathrm{cm},\( and \)80.0 \mathrm{cm} .$ (a) Which two lenses would you select to make a telescope with the largest magnifying power? What is the angular magnification of the telescope when viewing a distant object? (b) Which lens is used as objective and which as eyepiece? (c) What should be the distance between the objective and eyepiece?
A microscope has an eyepicce that gives an angular magnification of 5.00 for a final image at infinity and an objective lens of focal length $15.0 \mathrm{mm}\(. The tube length of the microscope is \)16.0 \mathrm{cm} .$ (a) What is the transverse magnification due to the objective lens alone? (b) What is the angular magnification due to the microscope? (c) How far from the objective should the object be placed?
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.
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