/*! 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 5 (II) If 720 -nm and 660 -nm ligh... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 34: Problem 5

(II) If 720 -nm and 660 -nm light passes through two slits \(0.68 \mathrm{~mm}\) apart, how far apart are the second-order fringes for these two wavelengths on a screen \(1.0 \mathrm{~m}\) away?

Short Answer

Expert verified
The second-order fringes are approximately 0.18 mm apart.

Step by step solution

01

Understanding the Problem Context

This problem involves a double-slit interference experiment, where light of different wavelengths creates interference patterns on a screen. We need to find the distance between the second-order fringes produced by two wavelengths鈥�720 nm and 660 nm.
02

Identify Relevant Formula

The position of fringes is given by the formula: \( y = \frac{m \cdot \lambda \cdot L}{d} \), where \( y \) is the fringe position, \( m \) is the order number, \( \lambda \) is the wavelength, \( L \) is the distance to the screen, and \( d \) is the distance between slits.
03

Plug in Values for 720 nm Fringe

For the 720 nm wavelength and the second-order fringe (\( m=2 \)), plug into the formula: \( y_{720} = \frac{2 \cdot 720 \times 10^{-9} \cdot 1.0}{0.68 \times 10^{-3}} \). Calculate \( y_{720} \).
04

Calculate 720 nm Fringe Position

Doing the calculations: \[ y_{720} = \frac{2 \cdot 720 \times 10^{-9} \cdot 1.0}{0.68 \times 10^{-3}} = \frac{1.44 \times 10^{-6}}{0.68 \times 10^{-3}} \approx 2.12 \times 10^{-3} \text{ m} \].
05

Plug in Values for 660 nm Fringe

For the 660 nm wavelength and the second-order fringe, use: \( y_{660} = \frac{2 \cdot 660 \times 10^{-9} \cdot 1.0}{0.68 \times 10^{-3}} \). Calculate \( y_{660} \).
06

Calculate 660 nm Fringe Position

Performing the calculations: \[ y_{660} = \frac{2 \cdot 660 \times 10^{-9} \cdot 1.0}{0.68 \times 10^{-3}} = \frac{1.32 \times 10^{-6}}{0.68 \times 10^{-3}} \approx 1.94 \times 10^{-3} \text{ m} \].
07

Find Distance Between Fringes

The distance between the second-order fringes is \( y_{720} - y_{660} \). This calculates to \( 2.12 \times 10^{-3} - 1.94 \times 10^{-3} \approx 0.18 \times 10^{-3} \text{ m} \) or 0.18 mm.

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.

Wavelength in Double-Slit Interference
In a double-slit interference experiment, light waves are characterized by their wavelength, which is the distance between consecutive peaks of the wave. Wavelength, denoted by \( \lambda \), plays a crucial role in determining the interference pattern formed on a screen. When light passes through the slits, it bends and overlaps with waves from the other slit, creating an interference pattern of bright and dark fringes. Key factors about wavelength include:
  • Wavelength is typically measured in nanometers (nm) when dealing with light, where 1 nm equals \( 10^{-9} \) meters.
  • The different wavelengths of light will lead to different positions of the interference fringes.
  • Longer wavelengths will cause the fringes to be spaced further apart on the screen.
In our exercise, 720 nm and 660 nm light are used, showing that even small differences in wavelength will affect where these fringes occur on the screen.
Fringe Position Calculation
The fringe position in a double-slit experiment refers to where these bright or dark lines appear on a viewing screen. This is calculated via the formula:\[ y = \frac{m \cdot \lambda \cdot L}{d} \]Here:
  • \( y \) is the fringe position, or how far the fringe is from the central bright line.
  • \( m \) is the order number, which tells us which bright or dark line we are measuring for.
  • \( \lambda \) is the wavelength of the light in question.
  • \( L \) is the distance from the slits to the screen.
  • \( d \) is the spacing between the slits.
The position of the fringe changes with each of these variables. By calculating the fringe positions for both 720 nm and 660 nm light, we determined the second-order fringes' positions and subsequently found their difference to identify how far apart they appear on the screen.
Understanding Order Number
The order number, represented by the symbol \( m \), is a critical variable in the context of wave interference patterns. It refers to the sequence of the fringe from the central bright fringe, which is known as the zeroth-order fringe. Here's more about the order number:
  • The central bright fringe is \( m = 0 \), and this is where the path difference between the two slits is zero.
  • First-order fringes (\( m = 1 \)) are those immediately adjacent to the central fringe, second-order (\( m = 2 \)) follow those, and so on.
  • In the double-slit experiment, each order represents a multiple of the wavelength overlap, leading to constructive interference that reinforces brightness at that fringe location.
In our exercise, we calculated the position of the second-order fringes (\( m = 2 \)) for both wavelengths, which helped us in determining the fringe separation for the respective light sources on the screen.

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

(III) The yellow sodium \(\mathrm{D}\) lines have wavelengths of 589.0 and 589.6 \(\mathrm{nm}\) . When they are used to illuminate a Michelson interferometer, it is noted that the interference fringes disappear and reappear periodically as the mirror \(\mathrm{M}_{1}\) is moved. Why does this happen? How far must the mirror move between one disappearance and the next?

(II) A very thin sheet of plastic \((n=1.60)\) covers one slit of a double-slit apparatus illuminated by 680 -nm light. The center point on the screen, instead of being a maximum, is dark. What is the (minimum) thickness of the plastic?

(II) Light of wavelength \(470 \mathrm{nm}\) in air falls on two slits \(6.00 \times 10^{-2} \mathrm{~mm}\) apart. The slits are immersed in water, as is a viewing screen \(50.0 \mathrm{~cm}\) away. How far apart are the fringes on the screen?

(I) The third-order bright fringe of \(610 \mathrm{nm}\) light is observed at an angle of \(28^{\circ}\) when the light falls on two narrow slits. How far apart are the slits?

(II) A total of 31 bright and 31 dark Newton's rings (not counting the dark spot at the center) are observed when 560-nm light falls normally on a planoconvex lens resting on a flat glass surface (Fig. \(34-18) .\) How much thicker is the center than the edges?
See all solutions

Recommended explanations on Physics Textbooks

Turning Points in Physics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Measurements

Read Explanation

Fields in Physics

Read Explanation

Dynamics

Read Explanation

Modern 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.