/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 60 The image formed by a microscope... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The image formed by a microscope objective with a focal length of 5.00 mm is 160 mm from its second focal point. The eyepiece has a focal length of 26.0 mm. (a) What is the angular magnification of the microscope? (b) The unaided eye can distinguish two points at its near point as separate if they are about 0.10 mm apart. What is the minimum separation between two points that can be observed (or resolved) through this microscope?

Short Answer

Expert verified
(a) 308.8; (b) 0.000324 mm.

Step by step solution

01

Calculate the Objective Magnification

First, we calculate the magnification of the objective lens. The magnification by the objective lens, \( M_o \), is given by:\[ M_o = \frac{d_i}{f_o} \]where \( d_i = 160 \, \text{mm} \) is the image distance from the focal point, and \( f_o = 5.00 \, \text{mm} \) is the focal length of the objective lens.Substituting the values:\[ M_o = \frac{160 \, \text{mm}}{5.00 \, \text{mm}} = 32 \]
02

Determine Eyepiece Magnification

Next, find the angular magnification of the eyepiece lens. The angular magnification, \( M_e \), is given by the formula:\[ M_e = \frac{25 \, \text{cm}}{f_e} \]where \( f_e = 26.0 \, \text{mm} = 2.60 \, \text{cm} \) is the focal length of the eyepiece lens (since 1 cm = 10 mm).Therefore, \[ M_e = \frac{25 \, \text{cm}}{2.60 \, \text{cm}} \approx 9.62 \]
03

Calculate Total Angular Magnification

The total angular magnification of the microscope, \( M_t \), is the product of the magnifications of the objective and eyepiece lenses:\[ M_t = M_o \times M_e \]Substituting the obtained values:\[ M_t = 32 \times 9.62 \approx 308.8 \]
04

Resolve Minimum Separation for Unaided Eye

The unaided eye can resolve a minimum separation of 0.10 mm. With magnification, the angular resolution improves. The equation for the observed separation \( d \) through the microscope is:\[ d = \frac{d_u}{M_t} \]where \( d_u = 0.10 \, \text{mm} \) is the unaided eye's minimum distinction.Therefore,\[ d = \frac{0.10 \, \text{mm}}{308.8} \approx 0.000324 \] mm.

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 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Objective Lens
In the world of microscopy, the objective lens is crucial as it is responsible for gathering light from the specimen and creating an enlarged image. This lens is positioned close to the object being examined. The main task of the objective lens is to magnify the specimen, and this is done by focusing on creating real, inverted images. - **Magnification:** The magnification power of the objective lens is determined by the ratio of the image distance (\( d_i \)) to its focal length (\( f_o \)). Therefore, in practical terms, the formula for magnification (\( M_o \)) is \( M_o = rac{d_i}{f_o} \).
- **High Magnification:** In our example, with a focal length of 5.00 mm and an image distance of 160 mm, the objective lens provides a significant magnification factor of 32. This high power magnification allows detailed observation of small objects, which is especially useful in biological and material science research.
Angular Magnification
Angular magnification is a critical measure in microscopes, as it represents how much larger the viewed angle through the microscope is compared to viewing with the naked eye at a normal distance. Essentially, angular magnification is about how big the image appears to our eyes. - **Formula:** The angular magnification of the microscope is calculated by multiplying the magnification of the objective lens (\( M_o \)) by that of the eyepiece lens (\( M_e \)). So the total angular magnification (\( M_t \)) is given by \( M_t = M_o imes M_e \).
- **Application:** In our scenario, the total angular magnification comes out to be 308.8, meaning the image appears over 300 times larger than what the naked eye would see under normal conditions. This extensive enlargement can reveal details that are invisible without such magnification.
Eyepiece Focal Length
The eyepiece of a microscope plays a substantial role in magnifying the image initially formed by the objective lens. The focal length of the eyepiece lens determines its magnification power and influences the comfort and ease with which the image can be viewed. - **Focal Length and Magnification:** The eyepiece focal length (\( f_e \)) is inversely related to its magnification power. Shorter focal lengths result in higher magnification. For example, an eyepiece with a focal length of 26.0 mm helps further enlarge the image due to its angular magnification (\( M_e \)), which is calculated as \( M_e = rac{25 ext{ cm}}{f_e} \).
- **Practical Importance:** In the exercise, the eyepiece contributing a magnification factor of approximately 9.62 enhances viewing significantly. This fine adjustment by the eyepiece ensures that the final image is not only magnified but also clear and comfortable to view.

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 microscope is focused on the upper surface of a glass plate. A second plate is then placed over the first. To focus on the bottom surface of the second plate, the microscope must be raised 0.780 mm. To focus on the upper surface, it must be raised another 2.10 mm. Find the index of refraction of the second plate.

The focal length of the eyepiece of a certain microscope is 18.0 mm. The focal length of the objective is 8.00 mm. The distance between objective and eyepiece is 19.7 cm. The final image formed by the eyepiece is at infinity. Treat all lenses as thin. (a) What is the distance from the objective to the object being viewed? (b) What is the magnitude of the linear magnification produced by the objective? (c) What is the overall angular magnification of the microscope?

What is the size of the smallest vertical plane mirror in which a woman of height \(h\) can see her full-length image?

A pencil that is 9.0 cm long is held perpendicular to the surface of a plane mirror with the tip of the pencil lead 12.0 cm from the mirror surface and the end of the eraser 21.0 cm from the mirror surface. What is the length of the image of the pencil that is formed by the mirror? Which end of the image is closer to the mirror surface: the tip of the lead or the end of the eraser?

A converging lens with a focal length of 12.0 cm forms a virtual image 8.00 mm tall, 17.0 cm to the right of the lens. Determine the position and size of the object. Is the image erect or inverted? Are the object and image on the same side or opposite sides of the lens? Draw a principal-ray diagram for this situation.

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.