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Two rectangular pieces of plane glass are laid one upon the other on a table. A thin strip of paper is placed between them at one edge so that a very thin wedge of air is formed. The plates are illuminated at normal incidence by 546-nm light from a mercury-vapor lamp. Interference fringes are formed, with 15.0 fringes per centimeter. Find the angle of the wedge

Short Answer

Expert verified
The angle of the wedge is approximately \( 4.095 \times 10^{-6} \) radians.

Step by step solution

01

Understand the Problem Setup

We have two glass plates forming a thin air wedge due to a strip of paper at one edge. The wedge is illuminated with light of wavelength \( \, \lambda = 546 \, \text{nm} \). We know the fringe density is 15.0 fringes per centimeter.
02

Relation between Fringe Spacing and the Wedge Angle

The number of fringes per centimeter is the inverse of the fringe spacing. We can use the equation for the path difference between two successive bright fringes to find the angle \( \theta \) of the wedge. The path difference \( \Delta x \) for constructive interference (bright fringes) is given by \( 2t = m\, \lambda \).
03

Expression for Wedge Angle

For a small angle approximation, the thickness \( t \) at a distance \( x \) from the edge where the paper is placed is \( t = x \tan \theta \approx x \theta \). The fringe spacing \( \Delta x \) is then derived from \( m \lambda = 2t \). Thus, \( x \theta = \frac{m \lambda}{2} \), giving us \( \theta = \frac{m \lambda}{2x} \).
04

Calculate Fringe Spacing

We know there are 15.0 fringes per centimeter, which means each fringe is \( \frac{1}{15.0} \) cm wide or \( x = \frac{1}{15.0} \times 10^{-2} \) meters.
05

Substitute Values to Find the Angle

We use \( m = 1 \) (for the first fringe), \( \lambda = 546 \times 10^{-9} \) meters, and \( x = \frac{1}{15.0} \times 10^{-2} \) meters in \( \theta = \frac{m \lambda}{2x} \) to calculate the angle. \[ \theta = \frac{1 \times 546 \times 10^{-9}}{2 \times \frac{1}{15.0} \times 10^{-2}} = \frac{546 \times 15.0}{2} \times 10^{-9} \approx 4.095 \times 10^{-6} \, \text{radians} \].

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Key Concepts

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

Air Wedge
An air wedge is a fascinating phenomenon created when two surfaces are slightly inclined relative to each other, with air filling the gap between them. This slight inclination forms a wedge-like shape. In many experiments, placing a thin strip of material, like paper, at one edge between two glass plates can achieve such a setup. The purpose of this setup is to explore optical interference effects.

The resulting thin wedge of air can cause interference patterns to appear when monochromatic light shines on it. This happens because the path length through which light must travel differs across the wedge. Consequently, an air wedge is useful in experiments to visualize and calculate effects related to light interference, such as measuring very small angles accurately.
Wavelength
Wavelength is a fundamental property of waves, defined as the distance between successive peaks (or troughs) of a wave. For light waves, wavelength plays a crucial role in determining the behavior and color of light. In optics, the unit of measurement for wavelength is typically nanometers (nm).

In the context of the air wedge experiment, understanding the wavelength of the illuminating light is essential. In this case, a mercury-vapor lamp with a wavelength of 546 nm is used. This specific wavelength helps dictate the conditions under which interference fringes appear. Light of different wavelengths will create different fringe patterns, allowing researchers to observe how light interacts in the air wedge.
Constructive Interference
Constructive interference occurs when two waves meet in such a way that their crests and troughs align perfectly. This alignment produces a new wave with a higher amplitude resulting in bright fringes in optical experiments. The concept of constructive interference is pivotal in experiments with air wedges, where the varied path difference causes light waves to either reinforce or cancel each other out.

For two waves to constructively interfere, the condition is that the path difference must be a multiple of the wavelength, i.e., \( 2t = m \lambda \), where \( t \) is the thickness of the air wedge and \( m \) is an integer. This relationship is what results in bright fringes appearing across the wedge.
Fringe Spacing
Fringe spacing refers to the distance between consecutive bright fringes in an interference pattern. It is an important measure because it allows one to deduce properties like angles and path differences in optical setups such as the air wedge experiment. For example, in our scenario, 15 fringes per centimeter leads to specific fringe spacing.

The expression for fringe spacing \( \Delta x \) can be derived under the condition \( m \lambda = 2t \), connecting it to the wedge geometry. With \( x \approx \frac{1}{15.0} \times 10^{-2} \) meters in this problem, we find the angle of the wedge, which correlates to the physical spacing of these interference fringes.
  • Fringe spacing inversely relates to the number of fringes per unit length.
  • Accurate calculation of fringe spacing helps in determining small angles precisely, which is often a goal in interference experiments.

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Most popular questions from this chapter

Two slits spaced \(0.0720 \mathrm{~mm}\) apart are \(0.800 \mathrm{~m}\) from a screen. Coherent light of wavelength \(\lambda\) passes through the two slits. In their interference pattern on the screen, the distance from the center of the central maximum to the first minimum is \(3.00 \mathrm{~mm} .\) If the intensity at the peak of the central maximum is \(0.0600 \mathrm{~W} / \mathrm{m}^{2},\) what is the intensity at points on the screen that are (a) \(2.00 \mathrm{~mm}\) and (b) \(1.50 \mathrm{~mm}\) from the center of the central maximum?

Suppose you illuminate two thin slits by monochromatic coherent light in air and find that they produce their first interference \(minima\) at \(\pm35.20^\circ\) on either side of the central bright spot. You then immerse these slits in a transparent liquid and illuminate them with the same light. Now you find that the first minima occur at \(\pm19.46^\circ\) instead. What is the index of refraction of this liquid?

Two speakers \(A\) and \(B\) are 3.50 m apart, and each one is emitting a frequency of 444 Hz. However, because of signal delays in the cables, speaker \(A\) is one-fourth of a period ahead of speaker \(B\). For points far from the speakers, find all the angles relative to the centerline (Fig. P35.44) at which the sound from these speakers cancels. Include angles on both sides of the centerline. The speed of sound is 340 m/s.

Coherent light with wavelength 600 nm passes through two very narrow slits and the interference pattern is observed on a screen 3.00 m from the slits. The first-order bright fringe is at 4.84 mm from the center of the central bright fringe. For what wavelength of light will the first-order dark fringe be observed at this same point on the screen?

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

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