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Ch 35: Interference
Young & Freedman Calc - University Physics 15th Edition
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook
All textbooksYoung & Freedman Calc 15th EditionCh 35: InterferenceProblem 25a
Chapter 34, Problem 25a

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?

Verified step by step guidance
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Step 1: Understand the concept of thin-film interference. Reflected light cancels when destructive interference occurs. This happens when the optical path difference between the two reflected rays is an odd multiple of half the wavelength in the film.
Step 2: Calculate the wavelength of light inside the TiO2 film. The wavelength in the film is given by \( \lambda_{film} = \frac{\lambda_{air}}{n_{film}} \), where \( \lambda_{air} \) is the wavelength in air (520.0 nm) and \( n_{film} \) is the refractive index of TiO2 (2.62).
Step 3: Determine the condition for destructive interference. For destructive interference, the optical path difference must satisfy \( 2t = (m + \frac{1}{2}) \lambda_{film} \), where \( t \) is the thickness of the film, \( m \) is an integer (order of interference), and \( \lambda_{film} \) is the wavelength in the film.
Step 4: Calculate the additional thickness needed. The current thickness of the film is 1036 nm. To achieve destructive interference, find the minimum additional thickness \( \Delta t \) such that the total thickness \( t_{total} = t_{current} + \Delta t \) satisfies the interference condition.
Step 5: Solve for \( \Delta t \). Rearrange the interference equation to isolate \( \Delta t \), ensuring that the total thickness corresponds to the next odd multiple of half the wavelength in the film. Use \( \Delta t = \frac{(m + \frac{1}{2}) \lambda_{film}}{2} - t_{current} \) to find the minimum additional thickness.

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Key Concepts

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

Interference of Light

Interference occurs when two or more light waves overlap, resulting in a new wave pattern. In thin films, constructive or destructive interference can happen depending on the path difference between reflected waves. For destructive interference, the path difference must equal an odd multiple of half the wavelength, leading to cancellation of reflected light.
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Refractive Index

The refractive index is a dimensionless number that describes how light propagates through a medium. It is defined as the ratio of the speed of light in a vacuum to the speed of light in the medium. In this scenario, the refractive indices of TiO2 and crown glass are crucial for determining how light reflects and refracts at the boundaries, affecting the interference pattern.
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Thin Film Thickness

The thickness of a thin film plays a critical role in determining the conditions for interference. For a film to cause destructive interference, its thickness must be adjusted so that the additional path length of light reflected from the top and bottom surfaces leads to a phase shift that results in cancellation. The minimum thickness can be calculated using the wavelength of light and the refractive indices involved.
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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

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

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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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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