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Chapter 35: Problem 47

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 \(\mathrm{mm}\) apart, are made in the center of the black face. When laser light of wavelength 632.8 \(\mathrm{nm}\) shines through the slits perpendicular to the black face, you find that the central bright fringe on the opposite face is 5.82 \(\mathrm{mm}\) wide, measured between the dark fringes that border it on either side. What is the index of refraction of the plastic?

Short Answer

Expert verified
The index of refraction is approximately 1.486.

Step by step solution

01

Understand the Problem

We're asked to find the index of refraction of the cylindrical pipe material. The setup describes a double slit interference pattern where laser light of a given wavelength is shone through two slits at the surface. The resulting fringes on the opposite face give us a central bright fringe width to use in our calculations.
02

Recall Relevant Formula

The width (\(\Delta x\)) of the central maximum in a double slit interference pattern is given by:\(\Delta x = \frac{2 \lambda D}{d}\)where \(\lambda\) is the wavelength of light, \(D\) is the distance from the slits to the screen, and \(d\) is the distance between the slits. This needs to be adjusted for the refractive medium.
03

Apply Thin Slit Formula with Refractive Index

The wavelength in the medium becomes \(\lambda/n\), where n is the index of refraction. So the formula becomes:\(\Delta x = \frac{2 (\lambda/n) D}{d}\)Rearrange for \(n\):\(n = \frac{2 \lambda D}{d \Delta x}\)
04

Plug in Values

Using the given values:- \(\lambda = 632.8 \times 10^{-9} \text{ m}\) (wavelength of light)- \(D = 3.25 \text{ m}\) (distance through the pipe)- \(d = 0.225 \times 10^{-3} \text{ m}\) (distance between slits)- \(\Delta x = 5.82 \times 10^{-3} \text{ m}\) (width of the central maximum)Plug these into the equation for \(n\),\[n = \frac{2 \times 632.8 \times 10^{-9} \text{ m} \times 3.25 \text{ m}}{0.225 \times 10^{-3} \text{ m} \times 5.82 \times 10^{-3} \text{ m}}\]
05

Calculate the Result

Calculate the value of \(n\):\[n = \frac{2 \times 632.8 \times 10^{-9} \text{ m} \times 3.25 \text{ m}}{0.225 \times 10^{-3} \text{ m} \times 5.82 \times 10^{-3} \text{ m}} \approx 1.486\]
06

Conclusion

The calculated index of refraction for the plastic pipe material is approximately 1.486, which describes how much the material slows down and bends the light relative to a vacuum.

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Key Concepts

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

Double Slit Interference
In the fascinating world of optics, the double slit experiment is a perfect demonstration of the wave nature of light. When light passes through two closely spaced slits, it creates an interference pattern on a screen placed behind the slits. This occurs because the light waves from each slit overlap and create regions of constructive and destructive interference.
  • Constructive interference leads to bright fringes, where light waves are in phase and reinforce each other.
  • Destructive interference leads to dark fringes, where light waves are out of phase and cancel each other out.
This pattern is not random; it forms a series of alternating bright and dark bands. The central band is the brightest, known as the central bright fringe. The positions of these fringes can be calculated using specific formulae related to the wavelength of light, the distance between the slits, and the distance from the slits to the observation point.
Central Bright Fringe
The central bright fringe is the most prominent feature in a double slit interference pattern. It lies directly in line with the light source and the center of the slits. This is where the maximum constructive interference occurs.
The width of the central bright fringe, denoted as \(\Delta x\), can be determined using the formula:\[\Delta x = \frac{2 \lambda D}{d}\]where \(\lambda\) is the wavelength of the light, \(D\) is the distance from the slits to the screen, and \(d\) is the distance between the slits. By measuring this width, it is possible to gain insight into other physical properties such as the refractive index of the material containing the slits, as the formula needs to be adjusted for the medium through which the light passes.
Wavelength of Light
The wavelength of light, denoted by \(\lambda\), is a crucial parameter in understanding interference patterns. It represents the physical distance between consecutive peaks (or troughs) in a wave and is usually measured in nanometers (nm) for visible light.
  • A shorter wavelength results in more densely packed interference fringes.
  • A longer wavelength leads to more widely spaced fringes.
In the given problem, the wavelength of the laser light used is 632.8 nm. This length of the light wave determines how the interference pattern will look on the screen. As the waves pass through the slits and overlap, they create the distinct patterns associated with this type of optical phenomenon.
Refractive Medium
When light travels through different materials, its speed and wavelength change, but its frequency remains constant. The material it travels through is known as the refractive medium. The index of refraction, often denoted by \(n\), describes how much a material slows down the light relative to its speed in a vacuum.
  • A higher index of refraction means light travels slower in the material.
  • The wavelength within the material is given as \(\lambda/n\)\, where \(n\) is the index of refraction.This adjustment is essential for correctly predicting interference patterns in various media.
In the problem, the plastic pipe acts as a refractive medium, altering the fringe pattern. By analyzing the interference pattern and using formulas related to the refraction, the index of refraction of the plastic can be calculated, which was found to be approximately 1.486. This value helps in understanding how light behaves as it travels through different materials.

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Most popular questions from this chapter

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 \(\mathrm{nm}\) ?

Sensitive Eyes. 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?

Coherent light of frequency \(6.32 \times 10^{14} \mathrm{Hz}\) passes through two thin slits and falls on a screen 85.0 \(\mathrm{cm}\) away. You observe that the third bright fringe occurs at \(\pm 3.11 \mathrm{cm}\) on either side of the central bright fringe. (a) How far apart are the two slits? (b) At what distance from the central bright fringe will the third dark fringe occur?

Two identical audio speakers connected to the same amplifier produce in-phase sound waves with a single frequency that can be varied between 300 and 600 Hz. The speed of sound is 340 \(\mathrm{m} / \mathrm{s} .\) You find that where you are standing, you hear minimum- intensity sound. (a) Explain why you hear minimum-intensity sound. (b) If one of the speakers is moved 39.8 \(\mathrm{cm}\) toward you, the sound you hear has maximum intensity. What is the frequency of the sound? (c) How much closer to you from the position in part (b) must the speaker be moved to the next position where you hear maximum intensity?

Two light sources can be adjusted to emit monochromatic light of any visible wavelength. The two sources are coherent, 2.04\(\mu \mathrm{m}\) apart, and in line with an observer, so that one source is 2.04\(\mu \mathrm{m}\) farther from the observer than the other. (a) For what visible wavelengths \((380\) to 750 \(\mathrm{nm})\) will the observer see the brightest light, owing to constructive interference? (b) How would your answers to part (a) be affected if the two sources were not in line with the observer, but were still arranged so that one source is 2.04\(\mu \mathrm{m}\) farther away from the observer than the other? (c) For what visible wavelengths will there be destructive interference at the location of the observer?
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