/*! 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 40 (II) What is the wavelength of t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(II) What is the wavelength of the light entering an interferometer if 384 bright fringes are counted when the movable mirror moves 0.125 \(\mathrm{mm}\) ?

Short Answer

Expert verified
The wavelength of the light is approximately 651 nm.

Step by step solution

01

Understand the Problem

We need to find the wavelength of light in an interferometer where 384 bright fringes are observed with a mirror movement of 0.125 mm. Bright fringes occur due to constructive interference where path difference is a multiple of the wavelength.
02

Formula Involvement

The number of bright fringes, often referred to as m, can be used to find the wavelength with the formula: \( m \times \lambda = 2 \times d \), where \( \lambda \) is the wavelength and \( d \) is the distance the mirror moved.
03

Substitute Known Values

Substitute the values given in the problem into the formula: \[384 \times \lambda = 2 \times 0.125 \text{ mm}\]Convert the distance to meters: \[0.125 \text{ mm} = 0.000125 \text{ meters}\]Thus, the equation becomes: \[384 \times \lambda = 2 \times 0.000125\]
04

Solve for Wavelength

Rearrange the equation to solve for \( \lambda \):\[\lambda = \frac{2 \times 0.000125}{384}\]Calculate the value: \[\lambda = \frac{0.00025}{384}\approx 6.51 \times 10^{-7} \text{ meters}\]
05

Convert Result to Nanometers

Convert the wavelength calculated into nanometers for a more conventional unit of measurement in the context of light:\[\lambda \approx 6.51 \times 10^{-7} \text{ meters} = 651 \text{ nm}\]

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 91Ó°ÊÓ!

Key Concepts

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

Interferometer
To comprehend the solution to the problem, we first delve into the concept of the interferometer. This device is used to measure the wavelength of light through interference patterns. By splitting a beam of light into two paths, it allows these beams to travel different distances and then recombine.
The key mechanism here is interference. When the light beams recombine, they can interfere with each other. This interference can either be constructive or destructive, depending on their relative phase. If the peaks of one fit into the troughs of the other, it's destructive, but if their peaks align, the interference is constructive.

Interferometers are widely used in various scientific investigations, including astronomical observations and optical testing, because they provide precise measurements of wavelengths and other properties. In this task, the interferometer helps measure the number of bright fringes created, giving insights into the light's wavelength.
Constructive Interference
Constructive interference occurs when two light waves meet in phase, meaning the peaks coincide. This results in brighter light, known as bright fringes in the context of an interferometer. Understanding this concept is essential when working with interference patterns, as it helps identify how differences in path length translate to visible changes in brightness.

When light waves are perfectly aligned in phase, their energies add up, leading to maximum brightness. This is calculated by considering the path difference between the waves: the difference must be a whole number multiple of the wavelength. This multiple, often referred to as m in the interferometry equation, directly impacts how we compute wavelength based on observed fringes.

In these scenarios, the formula used \[m \times \lambda = 2 \times d\] connects this relationship. By analyzing the number of bright fringes, we can work backward to find the wavelength, illustrating constructive interference's critical role in determining light properties.
Bright Fringes Counting
Counting bright fringes is a straightforward yet pivotal aspect of working with interferometry to determine the wavelength of light. This is because the appearance of these bright areas, or fringes, is directly tied to the constructive interference occurring as light waves overlap in phase.
  • The more fringes counted, the smaller the wavelength for the same mirror movement.
  • The positioning of the mirrors in the interferometer alters the path length difference, generating an interference pattern visible as alternating bright and dark bands.
In the given problem, 384 bright fringes were observed, which is critical information. Using these bright fringes, we can apply the constructive interference formula to calculate the wavelength.

By dividing twice the mirror movement by the number of fringes, we derive the wavelength \(\lambda = \frac{2 \times d}{m} \) allowing students to connect the physical observation of light patterns with mathematical equations and reinforcing the practical application of theoretical principles.

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

(II) Let's explore why only "thin" layers exhibit thin-film interference. Assume a layer of water, sitting atop a flat glass surface, is illuminated from the air above by white light (all wavelengths from \(400 \mathrm{nm}\) to \(700 \mathrm{nm}\) ). Further, assume that the water layer's thickness \(t\) is much greater than a micron \((=1000 \mathrm{nm})\); in particular, let \(t=200 \mu \mathrm{m}\). Take the index of refraction for water to be \(n=1.33\) for all visible wavelengths. (a) Show that a visible color will be reflected from the water layer if its wavelength is \(\lambda=2 n t / m,\) where \(m\) is an integer. (b) Show that the two extremes in wavelengths \((400 \mathrm{nm}\) and \(700 \mathrm{nm}\) ) of the incident light are both reflected from the water layer and determine the \(m\) -value associated with each. (c) How many other visible wavelengths, besides \(\lambda=400 \mathrm{nm}\) and \(700 \mathrm{nm},\) are reflected from the "thick" layer of water? (d) How does this explain why such a thick layer does not reflect colorfully, but is white or grey?

(II) A planoconvex lucite lens 3.4 \(\mathrm{cm}\) in diameter is placed on a flat piece of glass as in Fig. 18. When 580 -nm light is incident normally, 44 bright rings are observed, the last one right at the edge. What is the radius of curvature of the lens surface, and the focal length of the lens? [Hint: see Problem \(33 .\)]

(II) A physics professor wants to perform a lecture demonstration of Young's double-slit experiment for her class using the 633-nm light from a He-Ne laser. Because the lecture hall is very large, the interference pattern will be projected on a wall that is \(5.0 \mathrm{~m}\) from the slits. For easy viewing by all students in the class, the professor wants the distance between the \(m=0\) and \(m=1\) maxima to be \(25 \mathrm{~cm} .\) What slit separation is required in order to produce the desired interference pattern?

Light of wavelength \(690 \mathrm{nm}\) passes through two narrow slits \(0.66 \mathrm{~mm}\) apart. The screen is \(1.60 \mathrm{~m}\) away. A second source of unknown wavelength produces its second-order fringe \(1.23 \mathrm{~mm}\) closer to the central maximum than the 690 -nm light. What is the wavelength of the unknown light?

(I) Monochromatic light falling on two slits \(0.018 \mathrm{~mm}\) apart produces the fifth-order bright fringe at a \(9.8^{\circ}\) angle. What is the wavelength of the light used?

See all solutions

Recommended explanations on Physics Textbooks

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.