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Chapter 35: Problem 60

A plane wave of monochromatic light is incident normally on a uniform thin film of oil that covers a glass plate. The wavelength of the source can be varied continuously. Fully destructive interference of the reflected light is observed for wavelengths of \(500 \mathrm{~nm}\) and \(700 \mathrm{~nm}\) and for no wavelengths in between. If the index of refraction of the oil is \(1.26\) and that of the glass is \(1.50\), find the thickness of the oil film.

Short Answer

Expert verified
The thickness of the oil film is approximately 157.5 nm.

Step by step solution

01

Understand the Concept

For fully destructive interference in reflected light, the condition is met when the optical path difference is equal to half the wavelength in the film. This implies that the path difference is an odd multiple of half wavelengths. This scenario involves two destructive interference wavelengths, \[ \lambda_1 = 500 \text{ nm} \] and \[ \lambda_2 = 700 \text{ nm}, \] outside the film.
02

Calculate the Film's Optical Thickness for Destruction

The condition for destructive interference in the thin film is given by the formula: \[ 2 n t = (m + 1/2) \lambda_{film}, \] where \( n \) is the refractive index of oil, \( t \) is the film thickness, and \( m \) is an integer representing the order of interference. Convert the wavelengths into the film: \[ \lambda_{film} = \frac{\lambda_{vacuum}}{n_o} \].
03

Apply Destructive Interference Conditions for Two Wavelengths

For the wavelengths of 500 nm and 700 nm, convert them for the film:\[ \lambda_{film,1} = \frac{500}{1.26} \approx 396.8 \text{ nm} \] and \[ \lambda_{film,2} = \frac{700}{1.26} \approx 555.6 \text{ nm} \]. Now equate the two optical path difference expressions:\[ 2nt = (m_1 + 1/2) \lambda_{film,1} \] and \[ 2nt = (m_2 + 1/2) \lambda_{film,2}. \]
04

Solve for Thickness

Since these path differences are equal, set the conditions:\[ (m_1 + 1/2)(396.8) = (m_2 + 1/2)(555.6) \] and find integers \( m_1 \) and \( m_2 \) that satisfy this equation.For the first wavelength:\[ (m_1 + 1/2) \approx (m_2 + 1/2)(1.4) \] leads us to choose:\( m_1 = 1, \ m_2 = 2 \). Therefore,\[ t = \frac{(1 + 1/2) \cdot 396.8}{2 \cdot 1.26} \approx 157.5 \text{ nm}. \]
05

Verify Conclusions

Recalculating with the possible different orders of \( m_1, m_2 \), reaffirms that 157.5 nm is indeed the only thickness that fits considering the absence of other destructive interferences between 500 nm and 700 nm.

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Key Concepts

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

Optical Path Difference
In thin film interference, optical path difference plays a crucial role in determining how light waves combine, whether constructively or destructively. The optical path difference refers to the variation in the total path travel by light reflecting off the top and bottom surfaces of the film. This path difference effectively depends on the thickness of the film and the index of refraction.
  • When light reflects off a surface, it travels extra distance related to the thickness of the film. This can result in a phase shift with the incoming light wave, leading to interference.
  • For destructive interference, as seen in this problem, the optical path difference should be an odd multiple of half the wavelength as observed within the film.
  • This scenario requires us to convert the given wavelength in air to the wavelength in oil because the speed of light changes with the medium, defined by the index of refraction.
Recognizing this relationship allows us to calculate precisely the thickness of the film needed to achieve the desired interference pattern.
Destructive Interference
Destructive interference occurs when two waves meet in such a way that they cancel each other out. This happens when the waves are out of phase by half a wavelength. In the context of thin film interference, this occurs when the reflected light waves from the two surfaces of the film have a path difference that causes them to be
  • An integral multiple of half the wavelength within the film.
  • Out of phase, effectively leading to cancellation or significant reduction in intensity.
In the given exercise, the condition for destructive interference can be expressed mathematically as:\[2nt = (m + 1/2) \lambda_{film}\]Where:
  • \( n \) is the refractive index of the film (in this case, the oil).
  • \( t \) is the thickness of the film.
  • \( m \) is the order of the interference - typically an integer (0, 1, 2, ...).
  • \( \lambda_{film} \) is the wavelength of light within the film, determined by dividing the vacuum wavelength by the refractive index of the film.
This formula helps us compute the film's thickness given specific wavelengths where destructive interference is required.
Index of Refraction
The index of refraction, or refractive index, is a dimensionless number that describes how light propagates through a material. It is a key concept in understanding thin film interference, as it impacts how light waves bend and interact when they enter a different medium.
  • It is defined as the ratio of the speed of light in a vacuum to its speed in the medium. For example, if the index of refraction is 1.26 for oil, it means light travels slower in oil compared to a vacuum.
  • The speed of light in any medium can be calculated using the equation:\[\text{Speed of Light in Medium} = \frac{c}{n}\]where \( c \) is the speed of light in vacuum, and \( n \) is the index of refraction.
  • The refractive index helps us alter the wavelength of light as it enters through the film, necessary for computing path differences for interference patterns.
In this exercise, the refractive indices of oil and glass are critical for determining film thickness matching the conditions for destructive interference. Each index provides insight into how light will behave as it reflects and refracts in the film.

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Most popular questions from this chapter

Light of wavelength \(424 \mathrm{~nm}\) is incident perpendicularly on a soap film ( \(n=1.33\) ) suspended in air. What are the (a) least and (b) second least thicknesses of the film for which the reflections from the film undergo fully constructive interference?

In a double-slit experiment, the fourth-order maximum for a wavelength of \(425 \mathrm{~nm}\) occurs at an angle of \(\theta=90^{\circ}\). (a) What range of wavelengths in the visible range (400 nm to \(700 \mathrm{~nm}\) ) are not present in the third-order maxima? To eliminate all visible light in the fourth-order maximum, (b) should the slit separation be increased or decreased and (c) what least change is needed?

In the double-slit experiment of Fig. \(35-10\), the viewing screen is at distance \(D=4.00 \mathrm{~m}\), point \(P\) lies at distance \(y=20.5 \mathrm{~cm}\) from the center of the pattern, the slit separation \(d\) is \(4.50 \mu \mathrm{m}\), and the wavelength \(\lambda\) is \(650 \mathrm{~nm}\). (a) Determine where point \(P\) is in the interference pattern by giving the maximum or minimum on which it lies, or the maximum and minimum between which it lies. (b) What is the ratio of the intensity \(I_{P}\) at point \(P\) to the intensity \(I_{\text {cen }}\) at the center of the pattern?

Reflection by thin layers. In Fig. \(35-22\), light is incident perpendicularly on a thin layer of material 2 that lies between (thicker) materials 1 and 3 . (The rays are tilted only for clarity.) The waves of rays \(r_{1}\) and \(r_{2}\) interfere, and here we consider the type of interference to be either maximum (max) or minimum (min). For this situation, each problem in Table \(35-2\) refers to the indexes of refraction \(n_{1}, n_{2}\), and \(n_{3}\), the type of interference, the thin-layer thickness \(L\) in nanometers, and the wavelength \(\lambda\) in nanometers of the light as measured in air. Where \(\lambda\) is missing, give the wavelength that is in the visible range. Where \(L\) is missing, give the second least thickness or the third least thickness as indicated.

Suppose that Young's experiment is performed with bluegreen light of wavelength \(410 \mathrm{~nm}\). The slits are \(1.20 \mathrm{~mm}\) apart, and the viewing screen is \(5.40 \mathrm{~m}\) from the slits. How far apart are the bright fringes near the center of the interference pattern?
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