/*! 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 45 A thin uniform film of refractiv... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 35: Problem 45

A thin uniform film of refractive index 1.750 is placed on a sheet of glass of refractive index 1.50. At room temperature (20.0\(^\circ\)C), this film is just thick enough for light with wavelength 582.4 nm reflected off the top of the film to be cancelled by light reflected from the top of the glass. After the glass is placed in an oven and slowly heated to 170\(^\circ\)C, you find that the film cancels reflected light with wavelength 588.5 nm. What is the coefficient of linear expansion of the film? (Ignore any changes in the refractive index of the film due to the temperature change.)

Short Answer

Expert verified
The coefficient of linear expansion of the film is approximately \( 6.87 \times 10^{-6} \text{/掳C} \).

Step by step solution

01

Understand the Condition for Destructive Interference

To achieve destructive interference, the optical path difference must be an odd multiple of half the wavelength: \( 2n_{film}d = (m+\frac{1}{2}) \lambda \). Here \( n_{film} = 1.750 \), and \( \lambda \) represents the wavelengths (582.4 nm and 588.5 nm) at temperatures 20.0掳 C and 170掳 C. "\( m \)" is an integer that remains constant between temperatures since only the film length changes between those conditions, and thus the interference frequency remains the same.
02

Relate Film Thickness to Temperature-Dependent Wavelengths

Calculate the thickness condition at 20.0掳 C: \( d = \frac{(m+\frac{1}{2}) \times 582.4 \text{ nm}}{2 \times 1.750} \). Similarly, at 170掳 C: \( d_{170} = \frac{(m+\frac{1}{2}) \times 588.5 \text{ nm}}{2 \times 1.750} \). Since m is constant for both equations, their ratio will give the dilation due to thermal expansion.
03

Calculate Change in Thickness

By relating the two thicknesses, we have:\[ \frac{d_{170}}{d} = \frac{588.5 \text{ nm}}{582.4 \text{ nm}} = 1 + \alpha(T_{170} - T_{20}) \] where \( \alpha \) is the coefficient of linear expansion, \( T_{170} = 170 \) and \( T_{20} = 20 \) (both in degrees Celsius).
04

Solve for the Coefficient of Linear Expansion \( \alpha \)

We know:\[ 1 + \alpha(170 - 20) = \frac{588.5}{582.4} \]Thus,\[ \alpha = \frac{\frac{588.5}{582.4} - 1}{150} \] Calculate \( \alpha \) using the change in thickness ratio.

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 91影视!

Key Concepts

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

Refractive Index
When light travels through different media, its speed and direction change. This change is measured by a property called the refractive index (or index of refraction). The refractive index, denoted as \( n \), is a measure of how much the speed of light is reduced inside a medium compared to its speed in a vacuum. In this exercise, we are dealing with a thin film of refractive index 1.750 placed on glass with a refractive index of 1.50.
  • A higher refractive index means that light moves slower through the medium.
  • The speed reduction influences how light waves interfere with each other.
It is also crucial to understand that refractive indices can change with temperature, but in this problem, any such changes are ignored.
Wavelength Dependency
The interference pattern of light is highly dependent on its wavelength. Wavelength, which is the distance between successive peaks of a light wave, is a fundamental characteristic of light that affects how it interacts with materials. In this exercise, we observe that different wavelengths (582.4 nm and 588.5 nm) are involved at different temperatures.
  • At room temperature (20.0掳C), the wavelength involved in the destructive interference is 582.4 nm.
  • Upon heating to 170掳C, the wavelength at which destructive interference occurs changes to 588.5 nm.
The wavelength dependency is due to the fact that the optical path difference must match specific conditions for interference, which varies with the wavelength of light used.
Thermal Expansion
Materials expand when heated; this is known as thermal expansion. It's a common response to increasing temperature and is quantified in terms of the coefficient of linear expansion, \( \alpha \). This coefficient gives us a measure of how much a material's dimensions increase with temperature.
  • In this exercise, the film expands as the temperature increases from 20掳C to 170掳C.
  • Due to expansion, the film's thickness changes, altering the conditions for destructive interference.
The calculation involves comparing the film thickness at two temperatures to determine \( \alpha \), using the known change in interference wavelengths.
Destructive Interference
Destructive interference occurs when two light waves meet in such a way that they cancel each other out. For destructive interference to happen in reflected light, the optical path difference must be an odd multiple of half the wavelength. This condition ensures that the crests of one wave align with the troughs of another, leading to cancellation.
  • In the context of the film, when light reflects off the top and bottom surfaces, the corresponding waves interfere.
  • At room temperature, 582.4 nm light experiences destructive interference. After heating, this shifts to 588.5 nm.
This shift in wavelength is a critical indicator used to measure the thermal expansion of the film, as the change signifies a modification in the optical path contributing to the interference.

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

One round face of a 3.25-m, solid, cylindrical plastic pipe is covered with a thin black coating that completely blocks light. The opposite face is covered with a fluorescent coating that glows when it is struck by light. Two straight, thin, parallel scratches, 0.225 mm apart, are made in the center of the black face. When laser light of wavelength 632.8 nm shines through the slits perpendicular to the black face, you find that the central bright fringe on the opposite face is 5.82 mm wide, measured between the dark fringes that border it on either side. What is the index of refraction of the plastic?

A plate of glass 9.00 cm long is placed in contact with a second plate and is held at a small angle with it by a metal strip 0.0800 mm thick placed under one end. The space between the plates is filled with air. The glass is illuminated from above with light having a wavelength in air of 656 nm. How many interference fringes are observed per centimeter in the reflected light?

White light reflects at normal incidence from the top and bottom surfaces of a glass plate (\(n\) = 1.52). There is air above and below the plate. Constructive interference is observed for light whose wavelength in air is 477.0 nm. What is the thickness of the plate if the next longer wavelength for which there is constructive interference is 540.6 nm?

After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of refraction of 1.38, while the eyedrops have a refractive index of 1.45. After you put in the drops, your friends notice that your eyes look red, because red light of wavelength 600 nm has been reinforced in the reflected light. (a) What is the minimum thickness of the film of eyedrops on your cornea? (b) Will any other wavelengths of visible light be reinforced in the reflected light? Will any be cancelled? (c) Suppose you had contact lenses, so that the eyedrops went on them instead of on your corneas. If the refractive index of the lens material is 1.50 and the layer of eyedrops has the same thickness as in part (a), what wavelengths of visible light will be reinforced? What wavelengths will be cancelled?

Two small stereo speakers \(A\) and \(B\) that are 1.40 m apart are sending out sound of wavelength 34 cm in all directions and all in phase. A person at point \(P\) starts out equidistant from both speakers and walks so that he is always 1.50 m from speaker \(B\) (Fig. E35.1). For what values of x will the sound this person hears be (a) maximally reinforced, (b) cancelled? Limit your solution to the cases where x \(\leq\) 1.50 m.
See all solutions

Recommended explanations on Physics Textbooks

Physics of Motion

Read Explanation

Modern Physics

Read Explanation

Energy Physics

Read Explanation

Electricity

Read Explanation

Fluids

Read Explanation

Engineering Physics

Read Explanation
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.