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Chapter 27: Problem 18

A film of oil lies on wet pavement. The refractive index of the oil exceeds that of the water. The film has the minimum nonzero thickness such that it appears dark due to destructive interference when viewed in red light (wavelength \(=640.0 \mathrm{nm}\) in vacuum). Assuming that the visible spectrum extends from 380 to \(750 \mathrm{nm},\) for which visible wavelength(s) in vacuum will the film appear bright due to constructive interference?

Short Answer

Expert verified
The film appears bright at 640.0 nm due to constructive interference.

Step by step solution

01

Understand the Basics of Thin Film Interference

In thin film interference, light reflects off both the top and bottom surfaces of a film, creating two waves. Constructive interference (bright light) occurs when the path difference is a multiple of the wavelength, while destructive interference (dark light) occurs when the path difference is a half-multiple of the wavelength.
02

Condition for Minimum Thickness with Destructive Interference

For destructive interference in a thin film: \[ 2nt = (m+\frac{1}{2})\lambda \]where \(t\) is the thickness, \(n\) is the refractive index of the film, \(\lambda\) is the wavelength of light, and \(m\) is an integer. For minimum nonzero thickness, use \(m=0\):\[ t = \frac{\lambda}{4n} \]
03

Apply Destructive Interference Condition

Given: The oil film appears dark (destructive) for red light at \(640.0 \text{ nm}\). Thus, \[ t = \frac{640.0}{4n} \].
04

Condition for Constructive Interference

For constructive interference (bright), the condition is:\[ 2nt = m\lambda \].Here, \(m\) is an integer (0,1,2,...) and the same thickness \(t = \frac{640.0}{4n} \). The goal is to find \(\lambda\) for which:\[ m\lambda = 640.0 \].
05

Solve for Visible Wavelengths (\(\lambda\))

Set the equation \( m\lambda = 640.0 \) within the visible range \(380 \text{ nm} \) to \(750 \text{ nm}\):- For \(m=1\), \(\lambda = 640.0 \text{ nm}\)- For \(m=2\), \(\lambda = 320.0 \text{ nm}\) (not visible)- Verify each \(\lambda\) to ensure it is in the visible spectrum.
06

Identify Valid Constructive Wavelengths

The film shows bright interference at \(\lambda = 640.0 \text{ nm}\) because it satisfies the constructive condition and is visible within the given spectrum.

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Key Concepts

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

Destructive Interference
In thin film interference, destructive interference occurs when two light waves cancel each other out. This happens when the path difference between the waves is a half-multiple of the wavelength. It's like two kids on a seesaw perfectly balancing each other out. To achieve this balance, the formula is: \[ 2nt = (m+\frac{1}{2})\lambda \]where:
  • \( t \) is the thickness of the film,
  • \( n \) is the refractive index,
  • \( \lambda \) is the wavelength of light,
  • \( m \) is an integer representing the order of interference.
For minimum thickness where the interference is destructive (and the film looks dark), the smallest nonzero thickness is: \[ t = \frac{\lambda}{4n} \]This leaves the film appearing dark, as the red light at 640.0 nm is effectively cancelled out.
Constructive Interference
When it comes to constructive interference, things get bright! The light waves amplify each other when their path difference is a full multiple of the wavelength. The formula for constructive interference is:\[ 2nt = m\lambda \]Here, the path difference is such that the waves are in sync, increasing the overall light observed. This type of interference makes certain colors appear more strongly.In our oil film scenario, to find which wavelengths will appear bright we use:\[ m\lambda = 640.0 \]This indicates that whenever \( m \times \lambda = 640.0 \text{ nm} \), those specific \( \lambda \) values will be constructive. However, only those that lie within the visible spectrum of 380 nm to 750 nm are visible. Hence, the film will appear bright for 640.0 nm in red light.
Visible Spectrum
The visible spectrum is like a rainbow of colors our eyes can see. It ranges from 380 nm to 750 nm. Within this range, different light waves correspond to different visible colors.
  • Shorter wavelengths around 380 nm fall into the violet end of the spectrum.
  • Longer wavelengths around 750 nm are red.
In our problem, we focus on this visible spectrum to find which wavelengths the oil film will appear bright in. Only the parts of the spectrum that fall within our visible range will show up as colors. Since 640.0 nm is comfortably within this range, it aligns with constructive interference, showcasing a bright red hue when observed in the light of this wavelength.

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Most popular questions from this chapter

Two stars are \(3.7 \times 10^{11} \mathrm{m}\) apart and are equally distant from the earth. A telescope has an objective lens with a diameter of \(1.02 \mathrm{m}\) and just detects these stars as separate objects. Assume that light of wavelength 550 nm is being observed. Also assume that diffraction effects, rather than atmospheric turbulence, limit the resolving power of the telescope. Find the maximum distance that these stars could be from the earth.

A spotlight sends red light (wavelength \(=694.3 \mathrm{nm}\) ) to the moon. At the surface of the moon, which is \(3.77 \times 10^{8} \mathrm{m}\) away, the light strikes a reflector left there by astronauts. The reflected light returns to the earth, where it is detected. When it leaves the spotlight, the circular beam of light has a diameter of about \(0.20 \mathrm{m},\) and diffraction causes the beam to spread as the light travels to the moon. In effect, the first circular dark fringe in the diffraction pattern defines the size of the central bright spot on the moon. Determine the diameter (not the radius) of the central bright spot on the moon.

A diffraction grating has 2604 lines per centimeter, and it produces a principal maximum at \(\theta=30.0^{\circ} .\) The grating is used with light that contains all wavelengths between 410 and 660 nm. What is (are) the wavelength(s) of the incident light that could have produced this maximum?

Crude Thickness Monitor. You and your team need to determine the thicknesses of two extremely thin sheets of plastic and do not have a measurement instrument capable of the job. You find a white light source with color filters (i.e., tinted glass windows that only let certain colors pass through) including red \((\lambda=660 \mathrm{nm}),\) green \((\lambda=550 \mathrm{nm}),\) and blue \((\lambda=\) \(470 \mathrm{nm}) .\) You locate two glass microscope slides, each of length \(10.0 \mathrm{cm}\) stack one on top of the other, and tape their ends together on one side. On the very edge of the side opposite the tape, you sandwich a piece of plastic between the slides to form a wedge of air between the plates. (a) When you shine blue light perpendicular to the slides, 55 bright fringes form along the full length of the slide. How thick is the plastic? (b) When you replace the first sheet with the second sheet of plastic, the fringes that appear are too closely spaced to count. You switch to red light (a longer wavelength), and count 124 fringes. How thick is the plastic? (c) How many fringes would you expect in each case, (a) and (b), if you instead used green light?

At most, how many bright fringes can be formed on either side of the central bright fringe when light of wavelength \(625 \mathrm{nm}\) falls on a double slit whose slit separation is \(3.76 \times 10^{-6} \mathrm{m} ?\)
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