The Michelson Interferometer A Michelson interferometer is adjusted so that a bright fringe appears on the screen. As one of the mirrors is moved \(25.8 \mu \mathrm{m}, 92\) bright fringes are counted on the screen. What is the wavelength of the light used in the interferometer?

Short Answer

Expert verified
Answer: The wavelength of the light used in the Michelson interferometer is approximately \(5.61 \times 10^{-7}\) meters, or \(561 \, \mathrm{nm}\).

Step by step solution

01

Write down the given information

We're given that the mirror is moved \(25.8 \mu \mathrm{m}\), and there are 92 bright fringes counted. This means \(\Delta L = 25.8 \times 10^{-6}\) meters, and \(m = 92\).
02

Rearrange the formula to solve for the wavelength

We need to find \(\lambda\). To do this, we can rearrange the formula for the Michelson interferometer fringe shift: $$\lambda = \frac{2\Delta L}{m}$$
03

Plug in the given values and solve for the wavelength

Now, we can plug in the values for \(\Delta L\) and \(m\): $$\lambda = \frac{2(25.8 \times 10^{-6})}{92}$$ $$\lambda = \frac{51.6 \times 10^{-6}}{92}$$ $$\lambda = 5.61 \times 10^{-7} \mathrm{m}$$
04

Write the final answer

So, the wavelength of the light used in the Michelson interferometer is approximately \(5.61 \times 10^{-7}\) meters, or \(561 \, \mathrm{nm}\).

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 Vaia!

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 grating is made of exactly 8000 slits; the slit spacing is $1.50 \mu \mathrm{m} .\( Light of wavelength \)0.600 \mu \mathrm{m}$ is incident normally on the grating. (a) How many maxima are seen in the pattern on the screen? (b) Sketch the pattern that would appear on a screen \(3.0 \mathrm{m}\) from the grating. Label distances from the central maximum to the other maxima.
In a double-slit interference experiment, the wavelength is \(475 \mathrm{nm}\), the slit separation is \(0.120 \mathrm{mm},\) and the screen is $36.8 \mathrm{cm}$ away from the slits. What is the linear distance between adjacent maxima on the screen? [Hint: Assume the small-angle approximation is justified and then check the validity of your assumption once you know the value of the separation between adjacent maxima.] (tutorial: double slit 1 ).
Parallel light of wavelength \(\lambda\) strikes a slit of width \(a\) at normal incidence. The light is viewed on a screen that is \(1.0 \mathrm{m}\) past the slits. In each case that follows, sketch the intensity on the screen as a function of \(x\), the distance from the center of the screen, for $0 \leq x \leq 10 \mathrm{cm}$ (a) \(\lambda=10 a\). (b) \(10 \lambda=a,\) (c) \(30 \lambda=a.\)
White light containing wavelengths from \(400 \mathrm{nm}\) to \(700 \mathrm{nm}\) is shone through a grating. Assuming that at least part of the third-order spectrum is present, show that the second-and third-order spectra always overlap, regardless of the slit separation of the grating.
The radio telescope at Arecibo, Puerto Rico, has a reflecting spherical bowl of \(305 \mathrm{m}(1000 \mathrm{ft})\) diameter. Radio signals can be received and emitted at various frequencies with appropriate antennae at the focal point of the reflecting bowl. At a frequency of \(300 \mathrm{MHz}\), what is the angle between two stars that can barely be resolved? (Tutorial:radio telescope).
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.

Sign-up for free