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Chapter 26: Problem 22

\(\bullet\) 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 \( 4.1 \times 10^{-4} \) radians.

Step by step solution

01

Understand the Fringe Formation

Interference fringes are formed due to the difference in path lengths of light waves reflected from the top and bottom surfaces of the air wedge. The number of fringes per unit length of the wedge is related to the wavelength of the light and the geometry (i.e., the angle) of the wedge.
02

Write the Fringe Pattern Formula

The condition for bright fringes (constructive interference) in terms of path difference is given by: \( 2t \sin(\theta) = m\lambda \) for any integer \( m \), where \( t \) is the thickness of the air wedge at a point, \( \theta \) is the angle of the wedge, and \( \lambda \) is the wavelength of the light. Given that there are \( 15 \) fringes per centimeter, we can use this to find \( \theta \).
03

Relate Fringes to Wedge Geometry

The fringe spacing \( \Delta x \) is the inverse of the number of fringes per centimeter: \( \Delta x = \frac{1}{15\, \text{cm}^{-1}} = \frac{1}{15\cdot 100}\, \text{m}^{-1} \). The geometric relationship gives \( t = x \tan(\theta) \approx x\theta \) for small \( \theta \). Thus, \( \Delta x \approx \frac{\lambda}{2\theta} \).
04

Solve for the Wedge Angle

Use the equation \( \Delta x = \frac{\lambda}{2\theta} \). Rearrange to solve for \( \theta \): \[ \theta = \frac{\lambda}{2\Delta x} = \frac{546\times 10^{-9}\, \text{m}}{2\times \frac{1}{1500}\, \text{m}}. \] Calculating this gives the angle of the wedge.
05

Calculate the Angle

Substitute values into the equation from Step 4: \( \theta = \frac{546 \times 10^{-9}}{2 \times \frac{1}{1500}} \). Simplifying the numbers: \( \theta = \frac{546 \times 10^{-9} \times 1500}{2} \). Finally, \( \theta \approx 4.095 \times 10^{-4} \) radians.

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Key Concepts

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

Wavelength of Light
Light is a form of energy that travels in waves. Each type of light, such as the 546 nm mercury-vapor lamp light used in the exercise, has a specific wavelength. Wavelength is a critical parameter as it determines many properties of light, including its color and how it interacts with objects. For our problem, the wavelength of 546 nm is important because it determines the interference patterns produced when the light waves reflect off surfaces.
  • Wavelength (\( \lambda \)) is the distance between consecutive peaks (or troughs) in a wave.
  • It is typically measured in nanometers (nm) for light.
  • A specific wavelength can produce different kinds of interference depending on the surface it hits.

Understanding the wavelength helps us grasp how interference patterns such as fringes form, influencing how we perceive and calculate phenomena in experiments involving light.
Constructive Interference
Constructive interference occurs when two or more waves align with matching crests. This alignment strengthens the combined wave, resulting in what's known as bright fringes. In the context of the wedge of air formed in the exercise, constructive interference happens when the light waves reflecting off the top and bottom of the air wedge align.
  • Constructive interference results in enhanced light intensity.
  • Occurs when the path difference between two waves is a multiple of the wavelength.
  • Bright fringes indicate points of constructive interference.

This concept is fundamental in understanding how multiple reflections in films or layers, like the air wedge, create visible patterns that can be measured and analyzed to provide data about dimensions and angles.
Angle of the Wedge
The angle of the wedge is a crucial element in calculating how light waves interfere when passing through and reflecting off the surfaces in our experiment. It is the small angle formed by the wedge-shaped air layer trapped between the two glass plates.
  • The angle determines the changing thickness of the air wedge with distance from the starting point.
  • It influences the spacing and appearance of interference fringes.
  • Calculated using the formula: \[ \theta = \frac{\lambda}{2\Delta x} \]

A small angle means more widely spaced fringes, while a large angle means fringes are closer together. Understanding this helps us deduce not just the angle but the physical relationship between the glass plates.
Thin Film Interference
Thin film interference is a phenomenon where light waves reflecting off the two surfaces of a thin layer, like the air wedge, combine and interact. This interaction can result in interference patterns that provide information about the film's characteristics.
  • Occurs in layers close to the wavelength of the light used.
  • Causes colorful patterns seen on soap bubbles or oil slicks.
  • Used to measure thickness and surface quality in optical and manufacturing applications.

In our exercise, the thin film is the air between the glass, leading to the observed interference fringes. Understanding how thin film interference works allows us to use it as a powerful tool for measuring and engineering microscopic surfaces.

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Most popular questions from this chapter

\(\bullet\) Radio interference. Two radio antennas \(A\) and \(B\) radiate in phase. Antenna \(B\) is 120 \(\mathrm{m}\) to the right of antenna \(A .\) Consider point \(Q\) along the extension of the line connecting the anten- nas, a horizontal distance of 40 \(\mathrm{m}\) to the right of antenna \(B\) . The frequency, and hence the wavelength, of the emitted waves can be varied. (a) What is the longest wavelength for which there will be destructive interference at point \(Q ?\) (b) What is the longest wavelength for which there will be constructive interference at point \(Q ?\)

\(\bullet$$\bullet\) Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.350 \(\mathrm{mm}\) wide. The diffraction pattern is observed on a screen 3.00 \(\mathrm{m}\) away. Define the width of a bright fringe as the distance between the minima on either side. (a) What is the width of the central bright fringe? (b) What is the width of the first bright fringe on either side of the central one?

\(\bullet\) A laser beam of wavelength 600.0 \(\mathrm{nm}\) is incident normally on a transmission grating having 400.0 lines/mm. Find the angles of deviation in the first, second, and third orders of bright spots.

\(\bullet\) A plate of glass 9.00 \(\mathrm{cm}\) long is placed in contact with a sec- ond plate and is held at a small angle with it by a metal strip 0.0800 \(\mathrm{mm}\) thick placed under one end. The space between the plates is filled with air. The glass is illuminated from above with light having a wavelength in air of 656 \(\mathrm{nm}\) . How many interfer- ence fringes are observed per centimeter in the reflected light?

\(\bullet$$\bullet\) 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 \(\mathrm{nm}\) 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?
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