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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 \(546 \mathrm{nm}\) 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 \(0.0234\) degrees.

Step by step solution

01

Understanding the Problem

We have two glass plates forming a thin wedge of air. When light reflects off the upper and lower surfaces of this air wedge, it creates interference fringes. The wavelength of light used is 546 nm and there are 15.0 fringes per centimeter. We need to find the angle of the wedge.
02

Interference Fringe Formula

The formula for interference fringes in a wedge of air is: \( m \lambda = 2t \sin(\theta) \), where \( m \) is the fringe order (integer), \( \lambda \) is the wavelength of light, \( t \) is the thickness of the wedge at the point of the \( m \, th \) fringe, and \( \theta \) is the angle of the wedge.
03

Relate Fringe Spacing to Wedge Angle

Given are 15 fringes per centimeter, meaning each fringe is \( \frac{1}{15} \, \text{cm} \) apart. This distance corresponds to the thickness increase of \( t \) by \( \lambda/2 \) per fringe since every half-wavelength corresponds to a fringe.
04

Express Angle in Terms of Fringe Data

The fringe spacing, which represents thickness change \( \Delta t \), in terms of angle can be expressed as: \( \Delta t = \frac{\lambda}{2} \). From wedge geometry, \( \Delta t = d \sin(\theta) \), where \( d \) is spacing between successive fringes. Hence, \( \sin(\theta) = \frac{\lambda}{2 \cdot d} \).
05

Plug in the Numbers

Using \( \lambda = 546 \, \text{nm} = 546 \times 10^{-7} \text{cm} \), and \( d = \frac{1}{15} \text{cm} \), we have: \[ \sin(\theta) = \frac{546 \times 10^{-7} \text{cm}}{2 \times \left(\frac{1}{15} \right) \text{cm}} \].
06

Calculate the Angle

Calculate \( \sin(\theta) \):\[ \sin(\theta) = \frac{546 \times 10^{-7}}{2 \times \frac{1}{15}} \approx 0.0004095 \].Thus, \( \theta = \sin^{-1}(0.0004095) \) which can be approximated further based on small angle trigonometry.
07

Determine the Angle

Approximating \( \theta \) using small angle approximation \[\theta \approx \sin(\theta)\], we find \( \theta \approx 0.0004095 \, \text{radians}\). Converting this into degrees: \( \theta \approx 0.0004095 \times \frac{180}{\pi} \approx 0.0234 \, \text{degrees} \).

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Key Concepts

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

Understanding the Wedge of Air
A wedge of air is formed when two pieces of glass are placed on top of each other, with a thin strip inserted at one edge, creating a tapering air gap. This simple arrangement is key in creating certain optical effects. Light entering the wedge from a source reflects off both the top and bottom surfaces of the air layer.

As these beams of light reflect, they can interfere with each other, producing patterns of light and dark bands known as interference fringes. This phenomenon is a direct consequence of the different path lengths the light travels due to the varying thickness across the wedge.

The formation of these fringes helps in precisely measuring the wedge angle and other optical properties. This is a fascinating demonstration of how minute differences in geometry can lead to significant observable effects.
Formation of Interference Fringes
Interference fringes are the bands of light and dark observed when waves, such as light waves, overlap and combine. The key to understanding these patterns lies in the concept of constructive and destructive interference.

Constructive interference occurs when waves add up to strengthen each other, resulting in bright fringes. On the other hand, destructive interference occurs when waves cancel each other out, leading to dark fringes.

In the context of the wedge of air between glass plates, the changing thickness of the air layer causes light waves reflecting from the top and bottom surfaces of the wedge to have different path lengths, which results in interference. Since the thickness changes uniformly across the wedge, the interference fringes appear as evenly spaced lines.
  • Bright fringes occur at positions where reflected waves are in phase.
  • Dark fringes occur where reflected waves are out of phase.
This predictable pattern allows for precise calculations of various physical quantities.
Role of Wavelength in Interference
The wavelength of light plays a crucial role in determining the spacing and appearance of interference fringes. In the given problem, light with a wavelength of 546 nm from a mercury-vapor lamp is used.

Wavelength is the distance between two successive crests or troughs in a wave. It is a fundamental property of waves and directly affects how they interfere with each other.

When light of this particular wavelength enters the wedge of air, each half-wavelength difference in path length between the reflected rays creates another fringe. Therefore, the specific wavelength directly correlates with the number of fringes per given length, in this case, 15 fringes per centimeter.
  • Wavelength influences the number of observable fringes.
  • Wavelength determines the color of fringes due to chromatic effects.
Understanding the relationship between wavelength and interference aids in calculating the angle of the wedge and other related metrics.
Calculating the Wedge Angle
Calculating the angle of the wedge involves using the interference pattern data, which includes the number of fringes per unit length and the wavelength of the light used. This angle is generally small, especially in precision experimental setups.

To find the angle, we relate the wedge's geometry to the interference pattern using the formula:\[ m \lambda = 2t \sin(\theta) \]

Here's how it's done:
  • Determine the spacing between fringes, which is the reciprocal of fringes per centimeter.
  • Understand that each fringe corresponds to a half-wavelength increase in thickness \(\Delta t = \lambda/2\).
  • Use the small angle approximation \(\theta \approx \sin(\theta)\) given the usually tiny wedge angles.
Inserting these values into the relationship derived in the solution, we can find \(\sin(\theta)\), and subsequently \(\theta\), in radians or degrees.

This method effectively combines geometric optics principles and trigonometry to reveal very accurate measurements of the tiny angle of the wedge.

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Most popular questions from this chapter

Two slits spaced \(0.450 \mathrm{~mm}\) apart are placed \(75.0 \mathrm{~cm}\) from a screen. What is the distance between the second and third dark lines of the interference pattern on the screen when the slits are illuminated with coherent light with a wavelength of \(500 \mathrm{nm} ?\)

Two radio antennas \(A\) and \(B\) radiate in phase at a frequency of \(15 \mathrm{MHz}\). Antenna \(B\) is \(120 \mathrm{~m}\) to the right of antenna \(A .\) Consider point \(Q\) along the line connecting the antennas, a horizontal distance \(x\) to the right of antenna \(A\). (a) What is the smallest nonzero value of \(x\) for which there will be constructive interference at point \(Q ?\) (b) What is the smallest nonzero value of \(x\) for which there will be destructive interference at point \(Q ?\)

Two satellites at an altitude of \(1200 \mathrm{~km}\) are separated by \(28 \mathrm{~km}\). If they broadcast \(3.6-\mathrm{cm}\) microwaves, what minimum receivingdish diameter is needed to resolve (by Rayleigh's criterion) the two transmissions?

Monochromatic light from a distant source is incident on a slit \(0.750 \mathrm{~mm}\) wide. On a screen \(2.00 \mathrm{~m}\) away, the distance from the central maximum of the diffraction pattern to the first minimum is measured to be \(1.35 \mathrm{~mm}\). Calculate the wavelength of the light.

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 \(600 \mathrm{~Hz}\). 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?

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