/*! 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 51 Suppose the mirrors in a Michels... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose the mirrors in a Michelson interferometer are perfectly aligned and the path lengths to mirrors \(\mathrm{M}_{1}\) and \(\mathrm{M}_{2}\) are identical. With these initial conditions, an observer sees a bright maximum at the center of the viewing area. Now one of the mirrors is moved a distance \(x .\) Determine \(a\) formula for the intensity at the center of the viewing area as a function of \(x\), the distance the movable mirror is moved from the initial position.

Short Answer

Expert verified
The intensity formula is \( I(x) = I_0 \cos^2\left(\frac{2\pi x}{\lambda}\right) \).

Step by step solution

01

Understanding interference

In a Michelson interferometer, two beams of light travel to mirrors placed at different paths and combine to create an interference pattern. The phase difference between these beams determines whether the interference is constructive (bright fringe) or destructive (dark fringe).
02

Relate path difference to mirror movement

When one mirror is moved by a distance \(x\), the path length difference becomes \(2x\) because the beam travels to the mirror and back. This path difference changes the phase and, consequently, the observed intensity.
03

Phase difference and intensity

The phase difference \(\Delta\phi\) introduced by the path difference \(2x\) is given by \(\Delta\phi = \frac{4\pi x}{\lambda}\), where \(\lambda\) is the wavelength of light. The intensity \(I\) as a function of \(\Delta\phi\) is \(I = I_0 \cos^2\left(\frac{\Delta\phi}{2}\right)\), where \(I_0\) is the maximum intensity.
04

Substitute for \(\Delta\phi\)

Substitute \(\Delta\phi = \frac{4\pi x}{\lambda}\) into the intensity formula to get: \[ I(x) = I_0 \cos^2\left(\frac{2\pi x}{\lambda}\right) \]. This is the formula for the intensity at the center of the viewing area as a function of \(x\), the distance the mirror is moved.

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Key Concepts

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

Interference Pattern
In a Michelson interferometer, light is split into two beams that travel different paths and eventually recombine to create a visible pattern called an interference pattern. This pattern consists of alternating bright and dark areas. The bright spots occur when the light waves reinforce each other, while the dark spots appear when the waves cancel each other out. This effect is due to the wave nature of light, where the light waves can add up or subtract based on their relative phases.
One can describe the interference pattern using fringes, which are the bands of light and dark observed. The exact appearance of these fringes depends on the path difference between the two beams. This difference determines the relative phase shift between the beams. Small shifts in the mirror position, even by just a few wavelengths, can significantly change the observed pattern, a fundamental principle harnessed by interferometers for precise measurements.
Constructive Interference
Constructive interference occurs when the phases of the two light beams are aligned such that they amplify each other. This happens when the phase difference between the beams leads them to be "in phase," or perfectly synchronized. As a result, a bright fringe is observed in the interference pattern.
For constructive interference in a Michelson interferometer, the path difference between the beams must be an integer multiple of the wavelength of the light. Mathematically, this can be expressed as:
  • Phase difference = 0, 2Ï€, 4Ï€, ...
  • Path difference = mλ (where m is an integer)
Such alignment ensures that the peaks of the light waves add together, maximizing the light intensity at those points.
Path Difference
The path difference in a Michelson interferometer refers to the difference in distance traveled by the two light beams before they recombine. This difference is crucial as it directly influences the interference pattern formed.
When one of the mirrors in the interferometer is moved by a distance of \(x\), the path length of that beam increases by twice the distance moved. Hence, the path difference introduced is \(2x\). This additional distance alters the phase relationship between the two beams.
  • Path difference = 2x
  • This results in a phase difference, impacting interference type
This change in path length consequently affects the position and contrast of the fringes seen at the viewing area, allowing for the detection of even minute displacements.
Intensity Formula
The intensity of the light observed at the center of the viewing area in a Michelson interferometer depends on the path difference and, consequently, the phase difference between the two light beams. The intensity can be described using a mathematical formula:
  • Intensity \(I\) = Maximum Intensity \(I_0\) multiplied by the square of the cosine of half the phase difference
  • Mathematically, \(I = I_0 \cos^2\left(\frac{\Delta\phi}{2}\right)\)
By substituting the expression for phase difference \(\Delta\phi = \frac{4\pi x}{\lambda}\) into the formula, the intensity as a function of the mirror movement \(x\) becomes:
\[ I(x) = I_0 \cos^2\left(\frac{2\pi x}{\lambda}\right) \]
This formula demonstrates how the intensity, and thus the brightness of the interference pattern, varies with the displacement of the mirror, enabling precise control and measurement in interferometric applications.

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Most popular questions from this chapter

A highly reflective mirror can be made for a particular wavelength at normal incidence by using two thin layers of transparent materials of indices of refraction \(n_{1}\) and \(n_{2}\left(1n_{2}\right) .\) What should be the minimum thicknesses \(d_{1}\) and \(d_{2}\) in Fig. \(34-29\) in terms of the incident wavelength \(\lambda,\) to maximize reflection?

(III) (a) Consider three cqually spaced and equal-intensity coherent sources of light (such as adding a third slit to the two slits of Fig. 12 ). Use the phasor method to obtain the intensity as a function of the phase difference \(\delta\) \((\mathrm{Eq} .\) 4). \((b)\) Determine the positions of maxima and minima. $$ or \begin{array}{r}{\frac{\delta}{2 \pi}=\frac{d \sin \theta}{\lambda}} \\\ {\delta=\frac{2 \pi}{\lambda} d \sin \theta}\end{array} (4)$$

(II) A red laser from the physics lab is marked as producing 632.8-nm light. When light from this laser falls on two closely spaced slits, an interference pattern formed on a wall several meters away has bright fringes spaced \(5.00 \mathrm{~mm}\) apart near the center of the pattern. When the laser is replaced by a small laser pointer, the fringes are \(5.14 \mathrm{~mm}\) apart. What is the wavelength of light produced by the pointer?

(I) The third-order bright fringe of \(610 \mathrm{nm}\) light is observed at an angle of \(28^{\circ}\) when the light falls on two narrow slits. How far apart are the slits?

Light of wavelength \(\lambda\) strikes a screen containing two slits a distance \(d\) apart at an angle \(\theta_{i}\) to the normal. Determine the angle \(\theta_{m}\) at which the \(m^{\text {th }}\) -order maximum occurs.

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