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Ch 35: Interference
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
All textbooksYoung & Freedman Calc 14th EditionCh 35: InterferenceProblem 33
Chapter 35, Problem 33

How far must the mirror M2 (see Fig. 35.19) of the Michelson interferometer be moved so that 1800 fringes of He-Ne laser light (λ = 633 nm) move across a line in the field of view?
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Understand the problem: The Michelson interferometer produces interference fringes when the path length difference between the two arms changes. The number of fringes observed corresponds to the number of wavelengths of light that fit into the change in path length. Here, we are tasked with finding the distance the mirror M2 must be moved to produce 1800 fringes for He-Ne laser light with a wavelength of 633 nm.
Recall the relationship between the number of fringes (N), the wavelength of light (λ), and the change in path length (ΔL). The formula is: ΔL = N × λ. However, in a Michelson interferometer, moving one mirror by a distance d changes the path length by 2d because the light travels to the mirror and back. Thus, the formula becomes: 2d = N × λ.
Rearrange the formula to solve for the distance the mirror must be moved (d): d = (N × λ) / 2.
Substitute the given values into the formula: N = 1800 fringes and λ = 633 nm (convert this to meters: 633 nm = 633 × 10⁻⁹ m). The formula becomes: d = (1800 × 633 × 10⁻⁹) / 2.
Perform the calculation to find the value of d. Ensure the units are consistent and the result is expressed in meters. This will give the distance the mirror M2 must be moved to produce 1800 fringes.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Interference

Interference is a phenomenon that occurs when two or more waves overlap, resulting in a new wave pattern. In the context of the Michelson interferometer, light waves from a coherent source, like a He-Ne laser, combine to produce bright and dark fringes due to constructive and destructive interference. The position and number of these fringes are crucial for measuring changes in optical path length.
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Fringe Count

Fringe count refers to the number of bright or dark bands observed in an interference pattern. In the Michelson interferometer, moving one of the mirrors changes the optical path length, causing the fringes to shift. The total number of fringes that move across a line in the field of view is directly related to the distance the mirror is moved and the wavelength of the light used.
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Wavelength

Wavelength is the distance between successive peaks of a wave, typically measured in nanometers for light. In this problem, the wavelength of the He-Ne laser light is given as 633 nm. The relationship between the movement of the mirror and the number of fringes observed is determined by the wavelength, as each complete cycle of fringe movement corresponds to a change in optical path length equal to one wavelength.
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Related Practice
Textbook Question
What is the thinnest soap film (excluding the case of zero thickness) that appears black when illuminated with light with wavelength 480 nm? The index of refraction of the film is 1.33, and there is air on both sides of the film.
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Textbook Question

Two rectangular pieces of plane glass are laid one upon the other on a table. A thin strip of paper is placed between them at one edge so that a very thin wedge of air is formed. The plates are illuminated at normal incidence by 546-nm light from a mercury-vapor lamp. Interference fringes are formed, with 15.0 fringes per centimeter. Find the angle of the wedge.

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Textbook Question

A uniform film of TiO2, 1036 nm thick and having index of refraction 2.62, is spread uniformly over the surface of crown glass of refractive index 1.52. Light of wavelength 520.0 nm falls at normal incidence onto the film from air. You want to increase the thickness of this film so that the reflected light cancels. What is the minimum thickness of TiO2 that you must add so the reflected light cancels as desired?

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When viewing a piece of art that is behind glass, one often is affected by the light that is reflected off the front of the glass (called glare), which can make it difficult to see the art clearly. One solution is to coat the outer surface of the glass with a film to cancel part of the glare. If the glass has a refractive index of 1.62 and you use TiO2, which has an index of refraction of 2.62, as the coating, what is the minimum film thickness that will cancel light of wavelength 505 nm?

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What is the thinnest film of a coating with n = 1.42 on glass (n = 1.52) for which destructive interference of the red component (650 nm) of an incident white light beam in air can take place by reflection?

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