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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 24a
Chapter 35, Problem 24a

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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Step 1: Understand the problem. The goal is to determine the minimum thickness of a thin film (TiO2) that will cancel reflected light of wavelength 505 nm. This involves the concept of thin-film interference, where destructive interference occurs when the path difference between reflected light waves is equal to half the wavelength of the light in the film.
Step 2: Recall the condition for destructive interference. For a thin film with a higher refractive index than the surrounding medium, the condition for destructive interference is: \( 2t = \frac{\lambda}{n} \), where \( t \) is the thickness of the film, \( \lambda \) is the wavelength of light in vacuum, and \( n \) is the refractive index of the film. Rearrange this equation to solve for \( t \): \( t = \frac{\lambda}{2n} \).
Step 3: Adjust the wavelength for the refractive index of the film. The wavelength of light in the film is shorter than in vacuum. Use the formula \( \lambda_{film} = \frac{\lambda_{vacuum}}{n} \), where \( \lambda_{vacuum} \) is the given wavelength (505 nm) and \( n \) is the refractive index of the film (2.62). Substitute these values to find the wavelength of light in the film.
Step 4: Substitute the values into the thickness formula. Use the formula \( t = \frac{\lambda_{film}}{2} \) to calculate the minimum thickness of the film. This thickness corresponds to the first instance of destructive interference (minimum thickness).
Step 5: Verify the assumptions. Ensure that the refractive indices satisfy the conditions for thin-film interference (i.e., the refractive index of the film is greater than that of the glass and the surrounding medium). Also, confirm that the wavelength used is appropriate for the visible spectrum.

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Key Concepts

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

Refraction and Refractive Index

Refraction is the bending of light as it passes from one medium to another with a different refractive index. The refractive index quantifies how much light slows down in a medium compared to its speed in a vacuum. In this context, the refractive indices of glass and the coating material (TiO2) are crucial for determining how light behaves at their interfaces, affecting glare and visibility.
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Thin Film Interference

Thin film interference occurs when light waves reflect off the boundaries of a thin film, such as a coating on glass. The reflected waves can interfere constructively or destructively, depending on the film's thickness and the wavelength of light. To cancel glare, the film must be of a specific thickness that causes destructive interference for the wavelength of light in question, which is 505 nm in this scenario.
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Destructive Interference Condition

Destructive interference happens when two light waves combine to reduce or cancel each other out. For a thin film, this occurs when the path difference between the reflected waves is equal to an odd multiple of half the wavelength. The condition for minimum film thickness can be derived from the formula involving the refractive indices and the wavelength of light, ensuring that glare is minimized when viewing the art.
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Related Practice
Textbook Question

Two slits spaced 0.260 mm apart are 0.900 m from a screen and illuminated by coherent light of wavelength 660 nm. The intensity at the center of the central maximum (u = 0°) is I0. What is the distance on the screen from the center of the central maximum to the point where the intensity has fallen to I0/2?

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Textbook Question

Two slits spaced 0.260 mm apart are 0.900 m from a screen and illuminated by coherent light of wavelength 660 nm. The intensity at the center of the central maximum (u = 0°) is I0. What is the distance on the screen from the center of the central maximum to the first minimum

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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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Textbook Question

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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Textbook Question

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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