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Chapter 19: Problem 51

A \(1.0-\mathrm{cm}\) -diameter microscope objective has a focal length of \(2.8 \mathrm{mm} .\) It is used with light of wavelength of \(550 \mathrm{nm}\). a. What is the objective's resolving power if used in air? b. What is the resolving power of the objective if it is used in an oil- immersion microscope with \(n_{\text {eal }}=1.45 ?\)

Short Answer

Expert verified
a. The resolving power of the objective when used in air is \(6.71 \times 10^{ -7} \, m\). b. The resolving power of the objective when used in an oil-immersion microscope with a refractive index of 1.45 is \(9.72 \times 10^{ -7} \, m\).

Step by step solution

01

Calculate Resolving Power in Air

In this step, you will use the formula for resolving power given as \(R = 1.22 \times (\lambda / D)\). Substitute the given values in this formula, e.g., \(D = 1.0 \, cm = 0.01 \, m\), and \(\lambda = 550 \, nm = 550 \times 10^{ -9}\, m \). Hence the formula becomes \(R = 1.22 \times (550 \times 10^{ -9} / 0.01)\) which gives \(R = 6.71 \times 10^{ -7} \, m\).
02

Calculate Resolving Power in Oil

In this step, you have to calculate the resolving power when the objective is used with an oil immersion. The refractive index of oil \(n_{\text {eal}}\) is given as 1.45. Therefore, use the formula for resolving power in a medium other than air, \(R = 1.22 \times n \times (\lambda / D)\). After substituting the values into this formula, \(R = 1.22 \times 1.45 \times (550 \times 10^{ -9} / 0.01)\), you get \(R = 9.72 \times 10^{ -7} \, m\).

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Key Concepts

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

Optics: Understanding the Basics
Optics is a branch of physics that focuses on the study of light and its interactions with different materials. It explores how light behaves in various environments, whether it is bouncing off a mirror or passing through a lens. Optics is essential in the development and understanding of many technologies, including cameras, glasses, and microscopes.
  • Light is considered both a wave and a particle; this dual nature is crucial for understanding optical phenomena.
  • Optics considers various properties of light such as speed, wavelength, and intensity.
  • The behavior of light can be altered by changing the medium through which it travels, like air, glass, or water.
In this context, optics plays a vital role in understanding concepts such as resolving power and how different mediums, such as air or oil, affect a microscope's ability to distinguish between closely spaced objects.
Microscope Objective: The Key Component
The microscope objective is one of the most critical components of a microscope. It is the lens closest to the specimen and is primarily responsible for magnification and resolution.
  • Microscope objectives are designed to collect light from the sample and focus it into an image.
  • They come in various magnification powers, typically marked by numbers on the lens itself.
  • The quality and kind of lens determine how well fine details in the sample can be observed.
In the given exercise, a microscope objective with a diameter of 1.0 cm and a focal length of 2.8 mm is used. Understanding these specifications helps determine the resolving power when different mediums are involved.
Focal Length: What Does It Mean?
In optics, focal length is the distance between the lens and the point where light rays converge to form a sharp image. It is a critical property of lenses, mirrors, and optical devices like microscopes.
  • Short focal lengths provide greater magnification, allowing you to see smaller details within a specimen.
  • Measuring the focal length can help determine how far the lenses can "see," essential for clarity and detail in microscopy.
  • In this exercise, the focal length is 2.8 mm, indicating that the microscope is designed for high magnification and detailed viewing.
The focal length contributes significantly to the objective's ability to resolve tiny features, making it an essential parameter along with wavelength and refractive index.
Wavelength: How It Connects to Resolution
Wavelength is the distance between two consecutive peaks of a wave. In optics, it determines the color of the light and plays a significant role in how well an optical instrument can resolve details.
  • In the context of microscopes, shorter wavelengths generally provide better resolution.
  • The exercise uses a wavelength of 550 nm, typical for visible light.
  • Resolution is directly linked with wavelength through the formula for resolving power, where shorter wavelengths allow microscopes to distinguish between closer objects.
Understanding how wavelength affects resolving power is crucial, as this concept explains why different light sources are used to enhance resolution in various scientific fields.
Refractive Index: Its Impact on Light
The refractive index is a measure of how much light bends when it enters a new medium. It is essential for understanding how lenses and prisms work in optic devices.
  • When light passes from one medium to another, its speed changes, leading to bending or refraction.
  • The refractive index determines the degree of bending; the higher the index, the more light bends.
  • In the exercise, a refractive index of 1.45 is used for oil, indicating light will bend more than in air, which has an index of 1.
This property influences the resolving power of the microscope objective in different environments, enhancing its ability to distinguish fine details when using an oil immersion technique.

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Most popular questions from this chapter

The Hubble Space Telescope has a mirror diameter of \(2.4 \mathrm{m}\). Suppose the telescope was used to photograph the surface of the moon from a distance of \(3.8 \times 10^{8} \mathrm{m} .\) What is the distance between two objects that the telescope can barely resolve? Assume the wavelength of light is \(600 \mathrm{nm}\).

Suppose you point a pinhole camera at a \(15-\mathrm{m}\) -tall tree that is \(75 \mathrm{m}\) away. If the detector is \(22 \mathrm{cm}\) behind the pinhole, what will be the size of the tree's image on the detector?

Makers of cameras often use pixel count as an indicator of quality, and add sensors with more and more pixels. But the number of pixels is only one determinant of image quality. If the width of the central maximum of a diffraction pattern is larger than twice the spacing between the pixels, then packing the pixels more densely won't change the image quality. It's hard to get a camera with good optical quality in a smartphone; the lens must be small, so diffraction will be an issue. One smartphone manufacturer includes a camera with \(20.7 \mathrm{MP}\) which sounds very good. But let's look at the numbers: The distance between the pixels is \(1.2 \mu \mathrm{m} ;\) the focal length of the lens is \(4.8 \mathrm{mm},\) with a \(2.2-\mathrm{mm}\) -diameter aperture. For \(600 \mathrm{nm}\) light, what is the width of the central maximum? Could a sensor with more widely spaced pixels provide similar image quality?

Two lightbulbs are \(1.0 \mathrm{m}\) apart. From what distance can these lightbulbs be marginally resolved by a small telescope with a 4.0-cm-diameter objective lens? Assume that the lens is limited only by diffraction and \(\lambda=600 \mathrm{nm}\).

Martha is viewing a distant mountain with a telescope that has a \(120-\mathrm{cm}\) -focal-length objective lens and an eyepiece with a \(2.0 \mathrm{cm}\) focal length. She sees a bird that's \(60 \mathrm{m}\) distant and wants to observe it. To do so, she has to refocus the telescope. By how far and in which direction (toward or away from the objective) must she move the eyepiece in order to focus on the bird?
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