Problem 26 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

There are approximately 44 interference fringes per centimeter.

Step 1: 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.

Step 2: 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.

Step 3: 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, \( x \).

Step 4: 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.

Step 5: 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).

Step 6: 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.

Key Concepts

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.