/*! 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 49 An interferometer is used to mea... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An interferometer is used to measure the length of a bacterium. The wavelength of the light used is \(650 \mathrm{~nm}\). As one arm of the interferometer is moved from one end of the cell to the other, 310 fringe shifts are counted. How long is the bacterium?

Short Answer

Expert verified
The length of the bacterium is calculated to be \(2.02 \times 10^{-4}\) meters, or approximately \(0.2\) millimeters.

Step by step solution

01

Convert units

Firstly, convert the wavelength of light from nanometers to meters. As 1 meter consists of \(1 \times 10^{9}\) nanometers, convert 650 nm to meters by multiplying by \(1 \times 10^{-9}\). So, \(650 \: nm = 650 \times 1 \times 10^{-9}\) m.
02

Apply the interference principle

Path length difference in interferometry corresponds to the number of fringes. In this case, 310 fringes are counted as the interferometer arm moves from one end of the bacterium to the other. This implies that the bacterium is 310 wavelengths long.
03

Calculate the length of the bacterium

The length of the bacterium can be calculated by multiplying the number of wavelengths (fringe shifts) by the length of each wavelength. So, using the interference principle, calculate the length as \(310 \times 650 \times 1 \times 10^{-9}\) m.

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.

Understanding Wavelength
In the context of interferometry, the wavelength is a vital concept. Put simply, the wavelength is the distance between two consecutive peaks of a wave. Light waves have very tiny wavelengths, which is why they are measured in nanometers (nm). Just to give you an idea—1 nanometer equals one billionth of a meter! When conducting experiments like the one described, it's crucial to convert the wavelength into meters. This helps in calculations and understanding the size of the subject. For example, if the wavelength is 650 nm, its conversion into meters is necessary as it becomes part of the path length that the light covers while moving through the interferometer.

Why use wavelength in interferometry? Because it gives us a measure to calculate distances with great precision. Knowing the precise wavelength of the light source helps in determining the interference pattern, leading to accurate measurement outcomes.
What Are Fringe Shifts?
Fringe shifts are a fascinating phenomenon observed in interferometry. They occur when two beams of light travel different distances and then overlap, creating a pattern of bright and dark lines, called fringes. As one arm of the interferometer moves, the change in distance causes these lines to shift. This shift can be counted, providing crucial information about the object causing it.

In our exercise, 310 fringe shifts were noted. But what does this mean? Essentially, it tells us how many full cycles (or wavelengths) of change have occurred due to the movement of the interferometer's arm across the bacterium. Each fringe shift corresponds to one full wavelength of length change. Therefore, by counting these shifts, we can determine the length of the object being measured—like the bacterium in this exercise. Fringe shifts can thus translate these tiny wavelengths into something tangible and measurable.
Path Length Difference Explained
In interferometry, the concept of path length difference is fundamental. It refers to the difference in the distance traveled by two light beams. This difference is what causes the interference pattern—the bright and dark fringes we discussed earlier. Here's how it works: imagine the two beams start off aligned perfectly. When one beam travels a longer path, it's out of sync with the other, creating interference.

This path length difference is directly linked to the fringe shifts we observed. So, in our case of measuring a bacterium, the path length difference results in 310 fringe shifts as one beam's path is altered by the length of the bacterium. This means the path length difference is exactly 310 times the wavelength. To find the length of the bacterium, you multiply the number of fringe shifts by the wavelength. This calculation gives you the exact length of the object being measured, illustrating the elegance and precision of interferometry. By understanding path length differences, we harness the unique behavior of light to measure minuscule objects with amazing accuracy.

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 converging lens with a diameter of \(30.0 \mathrm{~cm}\) forms an image of a satellite passing overhead. The satellite has two green lights (wavelength \(500 \mathrm{~nm}\) ) spaced \(1.00 \mathrm{~m}\) apart. If the lights can just be resolved according to the Rayleigh criterion, what is the altitude of the satellite?

A lens has a focal length of \(28 \mathrm{~cm}\) and a diameter of \(4.0 \mathrm{~cm}\). What is the \(f\) -number of the lens?

Galileo devised a simple terrestrial telescope that produces an upright image. It consists of a converging objective lens and a diverging eyepiece at opposite ends of the telescope tube. For distant objects, the tube length is the objective focal length less the absolute value of the eyepiece focal length. (a) Does the user of the telescope see a real or virtual image? (b) Where is the final image? (c) If a telescope is to be constructed with a tube of length \(10.0 \mathrm{~cm}\) and a magnification of \(3.00\), what are the focal lengths of the objective and eyepiece?

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 \(\tan \theta \approx \theta\).

Find an equation for the length \(L\) of a refracting telescope in terms of the focal length of the objective \(f_{0}\) and the magnification \(m .\) (b) \(A\) knob adjusts the eyepiece forward and backward. Suppose the telescope is in focus with an eyepiece giving a magnification of \(50.0 .\) By what distance must the eyepiece be adjusted when the eyepiece is replaced, with a resulting magnification of \(1.00 \times 10^{22}\) Must the eyepiece be adjusted backward or forward? Assume the objective lens has a focal length of \(2.00 \mathrm{~m}\).

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.