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Chapter 37: Problem 49

A flat piece of glass is held stationary and horizontal above the flat top end of a 10.0 -cm-long vertical metal rod that has its lower end rigidly fixed. The thin film of air between the rod and glass is observed to be bright by reflected light when it is illuminated by light of wavelength \(500 \mathrm{nm} .\) As the temperature is slowly increased by \(25.0^{\circ} \mathrm{C}\) the film changes from bright to dark and back to bright 200 times. What is the coefficient of linear expansion of the metal?

Short Answer

Expert verified
The coefficient of linear expansion of the metal is \(4 \times 10^{-6} \, \mathrm{C}^{-1}\).

Step by step solution

01

Understand the Interference Condition

The film changes from bright to dark and back to bright 200 times for a temperature increase of \(25.0^{\circ} \mathrm{C}\). This means the rod expands by 200 wavelengths. Therefore, the change in length of the rod is \(\Delta L = 200 \times 500 \times 10^{-9} m = 0.0001 m\).
02

Use the Formula for Linear Expansion

The formula for linear thermal expansion is \(\Delta L = L \alpha \Delta T\), where \(\alpha\) is the linear expansion coefficient, \(L\) is the initial length of the object, and \(\Delta T\) is the change in temperature. The goal is to isolate \(\alpha\) to solve for the coefficient of linear expansion.
03

Substitute the Known Values and Solve for \(\alpha\)

Substitute the values into the formula. We get \(\alpha = \frac{\Delta L}{L\Delta T} = \frac{0.0001 m}{(0.1 m)(25 ^{\circ}C)} = 4 \times 10^{-6} \, \mathrm{C}^{-1}\).

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Key Concepts

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

Linear Expansion
Linear expansion describes how the length of an object changes with temperature. This concept is crucial in understanding how substances behave when heated or cooled.

When an object is heated, the particles inside it move more vigorously. This causes the object to expand. For linear objects like rods or wires, this expansion is along their length.

The amount of expansion depends on several factors:
  • Initial length of the object
  • Change in temperature
  • Material鈥檚 coefficient of linear expansion
In our problem, the metal rod expands linearly, causing the air film's thickness to change.
Interference Condition
Interference condition refers to the patterns of light brightness due to constructive or destructive interference. When light waves overlap, they can enhance or cancel each other out.

For the air film between the metal rod and glass, it appears bright or dark depending on its thickness.
  • Bright: Constructive interference occurs
  • Dark: Destructive interference occurs
In the exercise, the film cycles 200 times between these conditions. This tells us the rod expands enough to change the film's thickness by multiples of the light's wavelength.
Thermal Expansion Formula
The thermal expansion formula calculates how much an object expands or contracts with temperature change. It's expressed as: \[ \Delta L = L \alpha \Delta T \] - \( \Delta L \): Change in length - \( L \): Original length of the object - \( \alpha \): Coefficient of linear expansion - \( \Delta T \): Change in temperature
In solving the problem, you substitute the change in length and temperature to find \( \alpha \). This gives the metal's linear expansion coefficient.
Wavelength Interference
Wavelength interference plays a key role in explaining the visual change in brightness for the air film. It's all about how light waves combine.
When two waves of light meet:
  • If they line up perfectly (constructively), the result is brightness.
  • If they are out of sync (destructively), it leads to darkness.
In the problem, the metal rod's expansion changes the air film's thickness by 200 wavelengths of the light. Each complete cycle (bright-dark-bright) corresponds to one full wavelength, which allows us to measure the extent of the rod's expansion.

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Most popular questions from this chapter

Astronomers observe the chromosphere of the Sun with a filter that passes the red hydrogen spectral line of wavelength \(656.3 \mathrm{nm},\) called the \(\mathrm{H}_{\alpha}\) line. The filter consists of a transparent dielectric of thickness \(d\) held between two partially aluminized glass plates. The filter is held at a constant temperature. (a) Find the minimum value of \(d\) that produces maximum transmission of perpendicular \(\mathrm{H}_{\alpha}\) light, if the dielectric has an index of refraction of 1.378 (b) What If? If the temperature of the filter increases above the normal value, what happens to the transmitted wavelength? (Its index of refraction does not change significantly.) (c) The dielectric will also pass what nearvisible wavelength? One of the glass plates is colored red to absorb this light.

Two coherent waves are described by $$E_{1}=E_{0} \sin \left(\frac{2 \pi x_{1}}{\lambda}-2 \pi f t+\frac{\pi}{6}\right)$$ $$E_{2}=E_{0} \sin \left(\frac{2 \pi x_{2}}{\lambda}-2 \pi f t+\frac{\pi}{8}\right)$$ Determine the relationship between \(x_{1}\) and \(x_{2}\) that produces constructive interference when the two waves are superposed.

Our discussion of the techniques for determining constructive and destructive interference by reflection from a thin film in air has been confined to rays striking the film at nearly normal incidence. What If? Assume that a ray is incident at an angle of \(30.0^{\circ}\) (relative to the normal) on a film with index of refraction \(1.38 .\) Calculate the minimum thickness for constructive interference of sodium light with a wavelength of \(590 \mathrm{nm}\)

A student holds a laser that emits light of wavelength \(633 \mathrm{nm} .\) The beam passes though a pair of slits separated by \(0.300 \mathrm{mm},\) in a glass plate attached to the front of the laser. The beam then falls perpendicularly on a screen,

The intensity on the screen at a certain point in a doubleslit interference pattern is \(64.0 \%\) of the maximum value. (a) What minimum phase difference (in radians) between sources produces this result? (b) Express this phase difference as a path difference for 486.1 -nm light.
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