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The distance between the eyepiece and the objective lens in a certain compound microscope is \(20.0 \mathrm{~cm}\). The focal length of the objective is \(0.500 \mathrm{~cm}\), and that of the eyepiece is \(1.70 \mathrm{~cm}\). Find the overall magnification of the microscope.

Short Answer

Expert verified
The overall magnification of the microscope is the product of the magnification of the objective lens and the eyepiece. Apply the respective values in the formulas for each and then multiply these magnifications for the result.

Step by step solution

01

Calculate the magnification of the objective lens

Using the formula 'focal length of the objective lens / (distance between eyepiece and objective lens - focal length of the objective lens)', substitute the given values: Magnification of the objective lens = \(0.500cm / (20.0cm - 0.500cm)\)
02

Calculate the magnification of the eyepiece

Using the formula '25 cm / focal length of eyepiece', substitute the focal length of the eyepiece: Magnification of eyepiece = \(25cm / 1.70cm\)
03

Calculate the overall magnification

The overall magnification is the product of the magnification of the objective lens and the eyepiece. So, overall magnification = Magnification of Objective lens * Magnification of eyepiece

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Key Concepts

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

Compound Microscope
A compound microscope is an essential tool that allows us to see small details and tiny organisms that are invisible to the naked eye. It is specifically designed to provide high magnification, allowing researchers and students to observe minute structures.

A compound microscope uses multiple lenses to achieve this magnification:
  • The **objective lens** is close to the sample being observed and provides a real image of the sample.
  • The **eyepiece**, or ocular lens, is the lens you look through, which further enlarges this image.
By combining these lenses, compound microscopes can reach much higher levels of magnification than a simple handheld lens. This design is what differentiates them from single-lens microscopes, allowing for the detailed observation needed in fields like biology and mineralogy.

When calculating the total magnification provided by a compound microscope, it is the product of the magnification power of both the objective lens and the eyepiece that needs to be considered. This two-step magnification process is at the heart of how compound microscopes work.
Objective Lens
The objective lens in a compound microscope is crucial for providing the first step in magnification. Its role is to produce a magnified image of the sample that is then further enlarged by the eyepiece.

The objective lens is located close to the specimen and comes in various magnification strengths, often represented by numbers like 4x, 10x, 40x, indicating how many times it magnifies the specimen. The higher the number, the more it magnifies the sample.

Key features of the objective lens include:
  • Short focal length: This refers to the distance over which the lens can focus light. In the exercise, this was given as 0.500 cm.
  • High numerical aperture: This is a measure of the lens's ability to gather light and resolve fine specimen detail at a fixed object distance.
  • Different types, such as achromatic and plan lenses, designed to reduce optical aberrations and enhance image quality.
The amount of magnification an objective lens provides can be determined using the formula:\[\text{Magnification of objective lens} = \frac{\text{focal length of objective lens}}{\text{distance between eyepiece and objective lens} - \text{focal length of objective lens}}\] This equation helps in understanding how significant the spacing between the eyepiece and objective lens is in shaping the final magnification outcome.
Eyepiece
The eyepiece, also known as the ocular lens, is the second critical component in a compound microscope's lens system. Located at the top of the microscope, it is where you place your eye to view the magnified image produced by the objective lens.

The eyepiece generally provides an additional fixed magnification, typically ranging from 5x to 30x. This magnification is multiplied by the objective lens's magnification to give the total magnification.

Some key aspects of the eyepiece include:
  • Its simplicity of use – you simply look through it to see the enlarged image.
  • Fixed focal length, which in this exercise is 1.70 cm, a value used to calculate the eyepiece magnification.
  • Some eyepieces come with a built-in ruler, called a reticule, which can help measure small structures directly.
The formula used to determine the magnification provided by the eyepiece is:\[\text{Magnification of eyepiece} = \frac{\text{standard near point (usually 25 cm)}}{\text{focal length of eyepiece}}\]In this exercise, this formula allowed us to calculate how much the eyepiece magnifies the image it receives from the objective lens. Understanding this part of the system is crucial for utilizing a compound microscope effectively.

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Most popular questions from this chapter

A spy satellite circles Earth at an altitude of \(200 \mathrm{~km}\) and carries out surveillance with a special high-resolution telescopic camera having a lens diameter of \(35 \mathrm{~cm}\). If the angular resolution of this camera is limited by diffraction, estimate the separation of two small objects on Earth's surface that are just resolved in yellow-green light \((\lambda=550 \mathrm{~nm})\)

(a) Use conceptual arguments to show that the intensity of light (energy per unit area per unit time) reaching the film in a camera is proportional to the square of the reciprocal of the \(f\) -number as $$ I \propto \frac{1}{(f / D)^{2}} $$ (b) The correct exposure time for a camera set to \(f / 1.8\) is \((1 / 500)\) s. Calculate the correct exposure time if the \(f\) -number is changed to \(f / 4\) under the same lighting conditions. Note: "f/4," on a camera, means "an f-number of \(4 . "\)

A patient can't see objects closer than \(40.0 \mathrm{~cm}\) and wishes to clearly see objects that are \(20.0 \mathrm{~cm}\) from his eye. (a) Is the patient nearsighted or farsighted? (b) If the eye-lens distance is \(2.00 \mathrm{~cm}\), what is the minimum object distance \(p\) from the lens? (c) What image position with respect to the lens will allow the patient to see the object? (d) Is the image real or virtual? Is the image distance \(q\) positive or negative? (e) Calculate the required focal length. (f) Find the power of the lens in diopters. (g) If a contact lens is to be prescribed instead, find \(p, q\), and \(f\), and the power of the lens.

The Yerkes refracting telescope has a \(1.00-\mathrm{m}\) -diameter objective lens of focal length \(20.0 \mathrm{~m}\). Assume it is used with an eyepiece of focal length \(2.50 \mathrm{~cm}\). (a) Determine the magnification of the planet Mars as seen through the telescope. (b) Are the observed Martian polar caps right side up or upside down?

A laboratory (astronomical) telescope is used to view a scale that is \(300 \mathrm{~cm}\) from the objective, which has a focal length of \(20.0 \mathrm{~cm} ;\) the eyepiece has a focal length of \(2.00\) \(\mathrm{cm}\). Calculate the angular magnification when the telescope is adjusted for minimum eyestrain. Note: The object is not at infinity, so the simple expression \(m=f_{o} / f_{e}\) is not sufficiently accurate for this problem. Also, assume small angles, so that \(\tan \theta \approx \theta\).

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