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A plate of glass 9.00 cm long is placed in contact with a second plate and is held at a small angle with it by a metal strip 0.0800 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 nm. How many interference fringes are observed per centimeter in the reflected light?

Short Answer

Expert verified
There are approximately 44 interference fringes per centimeter.

Step by step solution

01

Understand the Problem

The problem involves interference of light waves due to a thin film created by a small air gap between two glass plates. One plate is tilted at a small angle to the other, causing the air gap to vary linearly from one side to the other. Light is reflected from each surface of this air gap, leading to interference fringes.
02

Identify the Interference Condition

For constructive interference, the path difference should be equal to an integer multiple of the wavelength. The condition for constructive interference in reflected light in a thin film (the air gap in this case) can be given by:\[ 2d \cos \theta = m \lambda \]where \( d \) is the thickness of the air wedge, \( \lambda \) is the wavelength of the light in air, and \( m \) is an integer representing the order of the fringe.
03

Determine the Thickness Variation

The thickness of the air gap, \( d \), varies linearly from one end of the 9.00 cm plate to the other end. At the thicker end, \( d = 0.0800 \) mm, and at the thinner end, \( d = 0 \). You need to relate this thickness to the distance along the plate, \( x \).
04

Relate Thickness to Distance

Assuming a linear relationship, the thickness can be expressed as:\[ d(x) = \frac{0.0800 \text{ mm}}{9.00 \text{ cm}} \cdot x \]where \( x \) is the distance from the thinner end of the glass plate.
05

Calculate Fringes per Unit Distance

Substitute the expression for \( d(x) \) into the interference condition for a bright fringe:\[ 2 \left( \frac{0.0800 \times 10^{-3} \text{ m}}{0.090 \text{ m}} \cdot x \right) = m \times 656 \times 10^{-9} \text{ m} \]Solve the equation to find the number of fringes, \( m \), per unit length (1 cm).
06

Solve for Fringes per Centimeter

The number of fringes in one centimeter, \( m \), is given by:\[ m = \frac{2 \times 0.0800 \times 10^{-3} \text{ m}}{0.090 \text{ m} \times 656 \times 10^{-9} \text{ m}} \approx 44.27 \text{ fringes per centimeter} \]Round this to the nearest whole number since you cannot have a fraction of a fringe.

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Key Concepts

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

Constructive Interference
Constructive interference occurs when two or more light waves meet and overlap, amplifying the wave's amplitude. For this to happen, the waves must be in phase, meaning their crests and troughs align perfectly. In our exercise concerning a thin film of air between glass plates, constructive interference manifests as bright fringes in the reflected light.

In the context of thin film interference, like the air gap between the tilted glass plates, the condition for constructive interference is given by the formula:
  • \(2d \cos \theta = m \lambda\)
    This means that the path difference, which is twice the thickness of the air gap(\(2d\)), should be equal to an integer (\(m\)) multiple of the wavelength of light in the air (\(\lambda\)).
By satisfying the above condition, bright or intense light fringes are observed due to the alignment of wave patterns. This is why you see bands of light on the glass, known as interference fringes.
Wavelength in Air
The wavelength of light is a critical factor in understanding how interference patterns are formed. Light waves have different speeds in different media, which affects their wavelength. In air, the wavelength remains close to its value in a vacuum.

In our exercise, the given wavelength of light in air is 656 nm (nanometers), which is a typical wavelength for visible light. When applying it to the constructive interference formula, it helps determine how many interference fringes will appear.
  • The formula used is \(2d \cos \theta = m \lambda\), where \(\lambda\) is the wavelength in air.
Accurate knowledge of this wavelength is crucial in calculating the number of fringes because it directly impacts their occurrence by interacting with the varying thickness of the air gap.
Linear Thickness Variation
The unique aspect of thin film interference in this exercise is the linear thickness variation of the air gap between the glass plates. One end of the plate is thicker due to the metal strip, while the other end has no gap (thickness zero). This linear variation plays a significant role in creating interference patterns.

To understand how thickness varies along the plate, the thickness at any point, say distance \(x\) from the thinner edge, can be expressed linearly:
  • \(d(x) = \frac{0.0800 \text{ mm}}{9.00 \text{ cm}} \cdot x\)
This equation allows us to know precisely how thick the air gap is at any given point on the plate.

With this thickness variation, the path difference of the reflected light also changes, affecting constructive interference conditions across the plate. As \(x\) changes, so too does the amount of light meeting the constructive interference condition, resulting in a predictable pattern of bright and dark fringes.

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Most popular questions from this chapter

After a laser beam passes through two thin parallel slits, the first completely dark fringes occur at \(\pm\)19.0\(^\circ\) with the original direction of the beam, as viewed on a screen far from the slits. (a) What is the ratio of the distance between the slits to the wavelength of the light illuminating the slits? (b) What is the smallest angle, relative to the original direction of the laser beam, at which the intensity of the light is \(1 \over 10\) the maximum intensity on the screen?

Two flat plates of glass with parallel faces are on a table, one plate on the other. Each plate is 11.0 cm long and has a refractive index of 1.55. A very thin sheet of metal foil is inserted under the end of the upper plate to raise it slightly at that end, in a manner similar to that discussed in Example 35.4. When you view the glass plates from above with reflected white light, you observe that, at 1.15 mm from the line where the sheets are in contact, the violet light of wavelength 400.0 nm is enhanced in this reflected light, but no visible light is enhanced closer to the line of contact. (a) How far from the line of contact will green light (of wavelength 550.0 nm) and orange light (of wavelength 600.0 nm) first be enhanced? (b) How far from the line of contact will the violet, green, and orange light again be enhanced in the reflected light? (c) How thick is the metal foil holding the ends of the plates apart?

Two speakers that are 15.0 m apart produce in-phase sound waves of frequency 250.0 Hz in a room where the speed of sound is 340.0 m/s. A woman starts out at the midpoint between the two speakers. The room's walls and ceiling are covered with absorbers to eliminate reflections, and she listens with only one ear for best precision. (a) What does she hear: constructive or destructive interference? Why? (b) She now walks slowly toward one of the speakers. How far from the center must she walk before she first hears the sound reach a minimum intensity? (c) How far from the center must she walk before she first hears the sound maximally enhanced?

Two small stereo speakers \(A\) and \(B\) that are 1.40 m apart are sending out sound of wavelength 34 cm in all directions and all in phase. A person at point \(P\) starts out equidistant from both speakers and walks so that he is always 1.50 m from speaker \(B\) (Fig. E35.1). For what values of x will the sound this person hears be (a) maximally reinforced, (b) cancelled? Limit your solution to the cases where x \(\leq\) 1.50 m.

Coherent light that contains two wavelengths, 660 nm (red) and 470 nm (blue), passes through two narrow slits that are separated by 0.300 mm. Their interference pattern is observed on a screen 4.00 m from the slits. What is the distance on the screen between the first-order bright fringes for the two wavelengths?

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