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Chapter 35: Problem 54

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 \(\mathrm{nm}\) . What is the thickness of the plate if the next longer wavelength for which there is constructive interference is 540.6 \(\mathrm{nm}\) ?

Short Answer

Expert verified
The thickness of the glass plate is approximately 156.58 nm.

Step by step solution

01

Understand the Problem

We are asked to find the thickness of a glass plate when white light reflects off its surfaces. The condition for observing constructive interference is given for two wavelengths. The question involves thin film interference where the refractive index of the glass is 1.52.
02

Condition for Constructive Interference

For constructive interference in thin film, the condition is given by \(2nt = m\lambda\), where \(n\) is the refractive index, \(t\) is the thickness, \(\lambda\) is the wavelength in air, and \(m\) is the order of interference. The thickness \(t\) remains constant for both wavelengths given.
03

Determine the Order of Interference

Assume \(m_1\) is the order for 477.0 nm and \(m_2 = m_1 + 1\) for 540.6 nm, as 540.6 nm is the next longer wavelength for constructive interference.
04

Formulate Equations for Each Wavelength

For 477.0 nm, the condition becomes \(2 \times 1.52 \times t = m_1 \times 477.0\). Similarly, for 540.6 nm, it becomes \(2 \times 1.52 \times t = (m_1 + 1) \times 540.6\).
05

Substitute and Solve the Equations

We have two equations:1. \(2 \times 1.52 \times t = m_1 \times 477.0\)2. \(2 \times 1.52 \times t = (m_1 + 1) \times 540.6\).Subtract the first equation from the second: \((m_1 + 1) \times 540.6 - m_1 \times 477.0 = 2 \times 1.52 \times t - 2 \times 1.52 \times t\).Simplifying gives: \(540.6 = m_1 \times 540.6 - m_1 \times 477.0 + 540.6\).
06

Calculate Thickness of the Glass Plate

From the subtraction and rearrangement, solve for \(m_1\):\(63.6 = m_1 (540.6 - 477.0)\).\(m_1 = \frac{63.6}{63.6}\).Since \(m_1 = 1\), substitute back into either thickness equation. For 477.0 nm: \(2 \times 1.52 \times t = 1 \times 477.0\).\(t = \frac{477.0}{2 \times 1.52}\).\(t \approx 156.58 \ nm\).

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Key Concepts

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

Constructive Interference
In the context of thin film interference, constructive interference occurs when waves combine to produce a wave with a larger amplitude. This happens when the path difference between two reflecting waves is an integer multiple of the wavelength. In our problem, constructive interference is observed for specific wavelengths of light reflecting off the surfaces of a thin glass plate.

For constructive interference in thin films, the formula used is \(2nt = m\lambda\), where:
  • \(n\) is the refractive index of the film
  • \(t\) is the thickness of the film
  • \(m\) is the order of interference
  • \(\lambda\) is the wavelength of light in air
Constructive interference occurs only at certain wavelengths, as the path difference must align perfectly with these wavelength multiples. Thus, by knowing the wavelengths for which constructive interference occurs, we can calculate properties of the thin film such as its thickness.
Refractive Index
The refractive index (\(n\)) is a measure of how much the speed of light is reduced inside a medium compared to vacuum. It's a critical factor in calculating interference in thin films.

The formula \(2nt = m\lambda\) reveals its role in interference:
  • The greater the refractive index, the more the light slows down and bends when entering the medium.
  • In our exercise, the refractive index of the glass is \(1.52\). This indicates that light travels 1.52 times slower in the glass than in air.
The value of \(n\) affects the thickness required to achieve constructive interference for a specific wavelength. In our problem, knowing \(n\) allows us to relate observed wavelengths to the physical characteristics of the thin film.
Wavelength Calculation
Calculating the wavelengths at which constructive interference occurs helps us deduce the physical characteristics of a thin film, such as its thickness. In the given solution, we start with two known wavelengths: 477.0 nm and 540.6 nm.

The key steps in wavelength calculation are:
  • Use the formula \(2nt = m\lambda\) for each wavelength.
  • Assume orders of interference \(m_1\) and \(m_2\) for each wavelength.
  • Set \(m_2 = m_1 + 1\) since 540.6 nm is the next longer wavelength.
By solving these equations for the interference order (\(m\)), we determine that \(m_1 = 1\) fits the conditions. From this, substituting back into the thickness equation with the value for the refractive index, we calculate the thickness of the plate to be approximately 156.58 nm.

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Most popular questions from this chapter

Coherent light that contains two wavelengths, 660 \(\mathrm{nm}\) (red) and 470 nm (blue), passes through two narrow slits separated by \(0.300 \mathrm{mm},\) and the interference pattern is observed on a screen 5.00 \(\mathrm{m}\) from the slits. What is the distance on the screen between the first-order bright fringes for the two wavelengths?

Two speakers \(A\) and \(B\) are 3.50 \(\mathrm{m}\) apart, and each one is emitting a frequency of 444 \(\mathrm{Hz}\) . However, because of signal delays in the cables, speaker \(A\) is one-fourth of a period ahead of speaker \(B\) . For points far from the speakers, find all the angles relative to the centerline (Fig. \(\mathrm{P} 35.49\) ) at which the sound from these speakers cancels. Include angles on both sides of the centerline. The speed of sound is 340 \(\mathrm{m} / \mathrm{s}\) .

Coherent light with wavelength 400 nm passes through two very narrow slits that are separated by \(0.200 \mathrm{mm},\) and the interference pattern is observed on a screen 4.00 \(\mathrm{m}\) from the slits. (a) What is the width (in mm) of the central interference maximum? (b) What is the width of the first-order bright fringe?

Two slits spaced 0.260 \(\mathrm{mm}\) apart are placed 0.700 \(\mathrm{m}\) from a screen and illuminated by coherent light with a wavelength of 660 nm. The intensity at the center of the central maximum \(\left(\theta=0^{\circ}\right)\) is \(I_{0}\) . (a) What is the distance on the screen from the center of the central maximum to the first minimum? (b) What is the distance on the screen from the center of the central maximum to the point where the intensity has fallen to \(I_{0} / 2 ?\)

Nonglare Glass. 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. (a) If the glass has a refractive index of 1.62 and you use \(\mathrm{TiO}_{2}\) , 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 \(\mathrm{nm} ?\) (b) If this coating is too thin to stand up to wear, what other thickness would also work? Find only the three thinnest ones.
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