/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 Consider the circular pattern of... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 29

Consider the circular pattern of Haidinger's fringes resulting from a film with a thickness of \(2 \mathrm{mm}\) and an index of refraction of 1.5. For monochromatic illumination of \(\lambda_{0}=600 \mathrm{nm}\), find the value of \(m\) for the central fringe \(\left(\theta_{t}=0\right) .\) Will it be bright or dark?

Short Answer

Expert verified
The central fringe is bright with \( m = 10 \).

Step by step solution

01

Understand the Problem

The problem involves finding the value of the interference order \( m \) for Haidinger's fringes, which occur due to thin film interference. We need to determine the condition for constructive (bright) or destructive (dark) interference at the center.
02

Write the Thin Film Interference Condition

For normal incidence \( \theta_t = 0 \), fringes are characterized by the optical path difference. The condition for constructive interference (bright fringe) is given by the equation:\[2nt = m \lambda,\]and for destructive interference (dark fringe), it's:\[2nt = (m + \frac{1}{2}) \lambda.\]Where \( n \) is the refractive index, \( t \) is the thickness, and \( \lambda \) is the wavelength of the light.
03

Substitute Known Values

We know:- \( n = 1.5 \)- \( t = 2 \text{ mm} = 2000 \text{ nm} \)- \( \lambda = 600 \text{ nm} \)Substituting these values into the equations:For constructive interference:\[2(1.5)(2000) = m(600)\]For destructive interference:\[2(1.5)(2000) = (m + \frac{1}{2})(600)\]
04

Solve for m (Constructive)

For constructive interference, simplify and solve:\[2(1.5)(2000) = m(600) \\Rightarrow 6000 = 600m \\Rightarrow m = 10\]
05

Solve for m (Destructive)

For destructive interference, substitute similarly:\[2(1.5)(2000) = (m + \frac{1}{2})(600) \\Rightarrow 6000 = 600m + 300 \\Rightarrow 5700 = 600m \\Rightarrow m = 9.5\]
06

Determine the Central Fringe

Since \( m \) must be an integer, the central fringe corresponds to \( m = 10 \). According to our calculation, \( m = 10 \) corresponds to constructive interference, indicating that the central fringe is bright.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Haidinger's Fringes
Haidinger's fringes are an intriguing optical phenomenon that occurs due to interference in thin films. Imagine looking at a soap bubble with different colors swirling on its surface. That's somewhat similar to observing Haidinger's fringes, but we are dealing with a much more fine-tuned scale. These circular patterns are formed because of the interference of light waves reflecting between the interfaces of a thin film.
The central fringe, in a circular pattern of Haidinger's fringes, can be either bright or dark depending on the interference conditions at the film's center. These conditions are dictated by the thickness, refractive index, and the wavelength of light used to illuminate the film.
Haidinger's fringes serve as a practical demonstration of thin film interference where light intensity changes due to varying thickness and other parameters, creating a beautiful set of concentric rings.
Constructive Interference
Constructive interference is one of the two types of interference that can affect the appearance of Haidinger's fringes. When light waves coincide in such a way that their peaks (or valleys) align, they add up to produce a higher amplitude wave鈥攁 bright or intense light spot. This is deemed constructive interference.
In the context of thin films, constructive interference occurs when the optical path difference between two light waves is an integer multiple of the wavelength. Mathematically, for normal incidence, this condition is represented by:
  • \[2nt = m \lambda\]
where \(n\) is the refractive index, \( t \) is the film's thickness, and \( \lambda \) is the wavelength of light. This formula helps determine where on the film constructive interference will create bright fringes.
Parameters of Film Thickness
The properties of a thin film greatly influence the interference patterns observed, such as Haidinger's fringes. One of these critical properties is the film's thickness.
The thickness of the film dictates how the light waves traveling through it will interfere with each other once reflected. If the film thickness is altered, it directly affects the optical path difference, which in turn affects whether we observe constructive or destructive interference at given locations.
When we calculate the order of interference, which results in either bright or dark fringes, knowing the exact thickness is crucial. In practical applications, thin films with varying thickness parameters are used for things like anti-reflective coatings, enhancing visibility in optical devices, and creating particular visual effects in artistic displays.
Monochromatic Illumination
Using monochromatic illumination in experiments like observing Haidinger's fringes simplifies the analysis of interference phenomena. Monochromatic light means light of a single wavelength or color, which allows precise control and predictions about where bright and dark fringes will appear. If multiple wavelengths were used, the interference patterns would overlap, making it difficult to distinguish individual fringes.
When monochromatic light passes through a thin film, due to consistent wavelength, we obtain clear interference patterns such as Haidinger's fringes. For instance, in the given problem, the use of light with a wavelength of 600 nm allows us to accurately calculate the interference order \( m \) and predict whether the central fringe is bright or dark based on the conditions of interference.
Thus, monochromatic illumination plays a vital role in the precision of studying and utilizing thin film interference.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A soap film of index 1.34 has a region where it is \(550.0 \mathrm{nm}\) thick. Determine the vacuum wavelengths of the radiation that is not reflected when the film is illuminated from above with sunlight.

Imagine that you have an opaque screen with three horizontal very narrow parallel slits in it. The second slit is a center-to-center distance \(a\) beneath the first, and the third is a distance \(5 a / 2\) beneath the first. (a) Write a complex exponential expression in terms of \(\delta\) for the amplitude of the electric field at some point \(P\) at an elevation \(\theta\) on a distant screen where \(\delta=k a \sin \theta .\) Prove that $$I(\theta)=\frac{I(0)}{3}+\frac{2 I(0)}{9}(\cos \delta+\cos 3 \delta / 2+\cos 5 \delta / 2)$$ Verify that at \(\theta=0, I(\theta)=I(O)\)

A Michelson Interferometer is illuminated with monochromatic light. One of its mirrors is then moved \(2.53 \times 10^{-5} \mathrm{m}\), and it is observed that 92 fringe-pairs, bright and dark, pass by in the process. Determine the wavelength of the incident beam.

The Fresnel biprism is used to obtain fringes from a point source that is placed \(2 \mathrm{m}\) from the screen, and the prism is midway between the source and the screen. Let the wavelength of the light be \(\lambda_{0}=500 \mathrm{nm}\) and the index of refraction of the glass be \(n=1.5\) What is the prism angle, if the separation of the fringes is \(0.5 \mathrm{mm} ?\)

Using Lloyd's mirror, X-ray fringes were observed, the spacing of which was found to be \(0.0025 \mathrm{cm}\). The wavelength used was \(8.33 \AA .\) If the source-screen distance was \(3 \mathrm{m}\), how high above the mirror plane was the point source of X-rays placed?
See all solutions

Recommended explanations on Physics Textbooks

Mechanics and Materials

Read Explanation

Famous Physicists

Read Explanation

Electric Charge Field and Potential

Read Explanation

Electricity and Magnetism

Read Explanation

Classical Mechanics

Read Explanation

Quantum Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.