/*! 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 57 Airtight Chamber In Fig. 36-28, ... [FREE SOLUTION] | 91Ó°ÊÓ

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Airtight Chamber In Fig. 36-28, an airtight chamber \(5.0 \mathrm{~cm}\) long with glass windows is placed in one arm of a Michelson interferometer. Light of wavelength \(\lambda=500 \mathrm{~nm}\) is used. Evacuating the air from the chamber causes a shift of 60 fringes. From these data, find the index of refraction of air at atmospheric pressure.

Short Answer

Expert verified
The index of refraction of air at atmospheric pressure is 1.0003.

Step by step solution

01

Identify Given Values

The given values are: chamber length, L = 5.0 cm = 0.05 m; wavelength of light, \( \lambda = 500 \text{nm} = 500 \times 10^{-9} \text{m} \); fringe shift, \( \Delta N = 60 \text{fringes} \).
02

Understand Fringe Shift Calculation

Fringe shift is determined by the formula: \[ \Delta N = 2L( \Delta n)/\lambda \. \] Here, \( \Delta n \) represents the change in the index of refraction of air when the chamber is evacuated, which is \( n_{air} - 1 \).
03

Isolate the Index of Refraction Change ( \( \Delta n \) )

Reorganizing the formula: \[ \Delta n = \Delta N \times \lambda / 2L \].
04

Calculate the Index of Refraction of Air

Substitute given values into the formula: \[ n_{air} - 1 = 60 \times 500 \times 10^{-9} / 2 \times 0.05 \]. This simplifies to: \[ n_{air} - 1 = 3 \times 10^{-4} \]. Finally, solve for \( n_{air} \): \[ n_{air} = 1 + 3 \times 10^{-4} = 1.0003 \].

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Key Concepts

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

Index of Refraction
The index of refraction (n) is a measure of how much light slows down when it enters a medium from a vacuum. It’s a fundamental property of materials, describing how light bends or refracts. For example, the index of refraction of air is close to 1, meaning light travels almost as fast in air as it does in a vacuum.
When light enters a medium like water or glass, it slows down, and its wavelength changes, causing the light to bend. The index of refraction is crucial in various optical devices, including lenses, microscopes, and interferometers.
The Michelson interferometer uses the principle of index of refraction to measure the shifts in light waves. By understanding how light behaves in different mediums, scientists can calculate precise measurements in experiments.
Fringe Shift Calculation
Fringe shift is an important concept in interferometry. It refers to the movement of dark or bright fringes observed when two beams of light combine and interfere. In the context of the Michelson interferometer, fringe shifts occur when there are changes in the distance that light travels.
The calculation of fringe shift in the example can be represented by the formula: \[ \Delta N = \frac{2L \Delta n}{\lambda} \], where \( \Delta N \) is the fringe shift, \( L \) is the length of the chamber, \( \Delta n \) is the change in the index of refraction, and \( \lambda \) is the wavelength of the light used.
By rearranging the formula to solve for \( \Delta n \), we use: \[\Delta n = \frac{\Delta N \cdot \lambda}{2L}\].
In the problem, substituting the given values, we find the change in the index of refraction when air is evacuated from the chamber. This shift tells us how much the refractive index has changed, enabling us to determine the index of refraction of air.
Wavelength of Light
The wavelength of light is the distance between successive crests of a light wave. It is usually measured in nanometers (nm). In the provided exercise, the wavelength used is \(500\ nm\). The wavelength is a key factor in calculating the fringe shifts in a Michelson interferometer.
Light's wavelength determines its color and how it interacts with different materials. In experiments like these, it's essential to use a known and consistent wavelength to ensure accurate results.
The wavelength affects the interference pattern that forms inside an interferometer. Shorter wavelengths lead to more closely spaced fringes, while longer wavelengths produce more widely spaced fringes. Therefore, accurately knowing the wavelength is critical for precise measurements in optical experiments.

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Most popular questions from this chapter

Laser Light Laser light of wavelength \(632.8 \mathrm{~nm}\) passes through a double-slit arrangement at the front of a lecture room, reflects off a mirror \(20.0 \mathrm{~m}\) away at the back of the room, and then produces an interference pattern on a screen at the front of the room. The distance between adjacent bright fringes is \(10.0 \mathrm{~cm} .\) (a) What is the slit separation? (b) What happens to the pattern when the lecturer places a thin cellophane sheet over one slit, thereby increasing by \(2.50\) the number of wavelengths along the path that includes the cellophane?

Coat Glass We wish to coat flat glass \((n=1.50)\) with a transparent material \((n=1.25)\) so that reflection of light at wavelength \(600 \mathrm{~nm}\) is eliminated by interference. What minimum thickness can the coating have to do this?

Directly Downward In Fig. \(36-25\), white light is sent directly downward through the top plate of a pair of glass plates. The plates touch at the left end and are separated by a wire of diameter \(0.048\) \(\mathrm{mm}\) at the right end; the air between the plates acts as a thin film. An observer looking down through the top plate sees bright and dark fringes due to that film. (a) Is a dark fringe or a bright fringe seen at the left end? (b) To the right of that end, fully destructive interference occurs at different locations for different wavelengths of the light. Does it occur first for the red end or the blue end of the visible spectrum?

Rhinestones The rhinestones in costume jewelry are glass with index of refraction \(1.50 .\) To make them more reflective, they are often coated with a layer of silicon monoxide of index of refraction 2.00. What is the minimum coating thickness needed to ensure that light of wavelength \(560 \mathrm{~nm}\) and of perpendicular incidence will be reflected from the two surfaces of the coating with fully constructive interference?

Two Glass Plates Two glass plates are held together at one end to form a wedge of air that acts as a thin film. A broad beam of light of wavelength \(480 \mathrm{~nm}\) is directed through the plates, perpendicular to the first plate. An observer intercepting light reflected from the plates sees on the plates an interference pattern that is due to the wedge of air. How much thicker is the wedge at the sixteenth bright fringe than it is at the sixth bright fringe, counting from where the plates touch?

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