/*! 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 94 In a compound microscope, the ob... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In a compound microscope, the objective has a focal length of 0.60 cm, while the eyepiece has a focal length of 2.0 cm. The separation between the objective and the eyepiece is L 12.0 cm. Another microscope that has the same angular magnification can be constructed by interchanging the two lenses, provided that the distance between the lenses is adjusted to a value \(L^{\prime}\) , Find \(L^{\prime}\)

Short Answer

Expert verified
The adjusted lens separation \( L' \) is 3.6 cm.

Step by step solution

01

Understanding the Original Setup

In the original setup, the objective lens has a focal length \( f_o = 0.60 \) cm and the eyepiece lens has a focal length \( f_e = 2.0 \) cm. The separation between the two lenses is given as \( L = 12.0 \) cm.
02

Calculating the Angular Magnification of the Original Microscope

The angular magnification \( M \) of a compound microscope in which the lenses are separated by a distance \( L \) is given by the formula: \[ M = -\left( \frac{L}{f_o + f_e} \right) \left( \frac{D}{f_e} \right) \] Where \( D \) is the near point distance, often taken as 25 cm for a relaxed eye. Plugging in the values, we calculate: \[ M_1 = -\left( \frac{12}{0.60 + 2.0} \right) \left( \frac{25}{2.0} \right) \] Simplify to find \( M_1 \).
03

Interchanging the Lenses

The problem states that the lenses are interchanged, so the objective becomes the previous eyepiece with \( f_o' = 2.0 \) cm and the eyepiece becomes the previous objective with \( f_e' = 0.60 \) cm.
04

Using the Same Angular Magnification

To maintain the same angular magnification, we should equate the magnifications of both setups: \[ -\left( \frac{L}{0.60 + 2.0} \right) \left( \frac{25}{2.0} \right) = -\left( \frac{L'}{2.0 + 0.60} \right) \left( \frac{25}{0.60} \right) \] Simplifying both sides and solving for \( L' \).
05

Solving for New Lens Separation \(L'\)

Equate both expressions for magnification and solve for \( L' \): \[ \frac{12}{2.6} \cdot \frac{25}{2} = \frac{L'}{2.6} \cdot \frac{25}{0.6} \] Cancel common terms and solve the equation: \[ \frac{12 \times 0.6}{2} = L' \] Obtain \( L' \) by solving the simplified equation.

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.

Focal Length
In the realm of microscope optics, the focal length is a critical concept. It refers to the distance between the lens and its focus point, where light rays converge. In a compound microscope, this distance varies between two key lenses - the objective and the eyepiece:

- **Objective Lens**: The objective lens in a microscope is designed to have a short focal length. This short focal length allows it to magnify small objects that are placed close to it.
- **Eyepiece Lens**: Conversely, the eyepiece lens has a longer focal length. This helps magnify the image created by the objective lens so the observer can view it comfortably.

This relationship is crucial because the focal lengths directly affect the magnification capabilities of the compound microscope. Understanding these focal lengths, which are typically measured in centimeters, aids in aligning the lenses correctly. This ensures the microscope provides a clear, enlarged image of the specimen.
Angular Magnification
Angular magnification is a measure of how much larger an object appears when viewed through a microscope compared to the naked eye. For microscopes, this entails a two-step magnification process:

- **First Step**: The objective lens forms an enlarged intermediate image of the object.
- **Second Step**: The eyepiece lens further magnifies this intermediate image, providing the final image to the viewer.

The formula for angular magnification, particularly in compound microscopes, takes into account the focal lengths of both lenses and the lens separation. The general formula is:

\[ M = -\left( \frac{L}{f_o + f_e} \right) \left( \frac{D}{f_e} \right) \]

Where:
  • L is the separation between the lenses.
  • f_o and f_e are the focal lengths of the objective and eyepiece, respectively.
  • D is the near point distance (usually 25 cm for a relaxed eye).
Magnification varies based on lens configuration and is negative, indicating an inverted image. This inversion is typical in microscope's viewing.
Lens Separation
Lens separation, often denoted as L, is a vital distance that affects both the setup and functionality of compound microscopes. It defines how far apart the objective and eyepiece lenses are positioned. Here's why it matters:

- **Influence on Magnification**: The distance, L, directly influences the overall angular magnification.
- **Optical Interchangeability**: By altering the lens separation when the lenses are interchanged, a microscope can retain its original magnification power.

This adjustment is crucial when changing the roles of the lenses, as seen in certain exercises. In this context, maintaining the same magnification after the interchange of lenses requires recalculating and possibly readjusting this separation distance to reach a specified value, often termed as L'.
Microscope Optics
Microscope optics involves a sophisticated arrangement of lenses to achieve high levels of magnification and resolution. The primary components consist of:

- **Objective Lens**: Responsible for the initial magnification and providing a real image.
- **Eyepiece Lens**: Further enlarges the image formed by the objective lens and allows the viewer to observe a larger virtual image.

The coordination between these lenses enables detailed observations of tiny specimens. Optical principles such as refraction and magnification are inherently part of this setup, allowing microscopes to bend and focus light rays to produce clear and magnified images.

Ensuring proper alignment of these lenses is crucial for optimal performance. Adjustments in lens configurations—like focal lengths and separation—tailor the optics to specific viewing needs, making compound microscopes versatile tools in scientific exploration.

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

Your friend has a near point of 138 cm, and she wears contact lenses that have a focal length of 35.1 cm. How close can she hold a magazine and still read it clearly?

A layer of oil \((n=1.45)\) floats on an unknown liquid. A ray of light originates in the oil and passes into the unknown liquid. The angle of incidence is \(64.0^{\circ},\) and the angle of refraction is \(53.0^{\circ} .\) What is the index of refraction of the unknown liquid?

A farsighted woman breaks her current eyeglasses and is using an old pair whose refractive power is 1.660 diopters. Since these eyeglasses do not completely correct her vision, she must hold a newspaper 42.00 cm from her eyes in order to read it. She wears the eyeglasses 2.00 cm from her eyes. How far is her near point from her eyes?

Mars subtends an angle of \(8.0 \times 10^{-5} \mathrm{rad}\) at the unaided eye. An astronomical telescope has an eyepiece with a focal length of 0.032 \(\mathrm{m} .\) When Mars is viewed using this telescope, it subtends an angle of \(2.8 \times 10^{-3} \mathrm{rad}\) . Find the focal length of the telescope's objective lens.

The near point of a naked eye is 25 cm. When placed at the near point and viewed by the naked eye, a tiny object would have an angular size of \(5.2 \times 10^{-5}\) rad. When viewed through a compound microscope, however, it has an angular size of \(-8.8 \times 10^{-3}\) rad. (The minus sign indicates that the image produced by the microscope is inverted.) The objective of the microscope has a focal length of 2.6 cm, and the distance between the objective and the eyepiece is 16 cm. Find the focal length of the eyepiece.

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.