/*! 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 A flat observation screen is pla... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 27: Problem 5

A flat observation screen is placed at a distance of \(4.5 \mathrm{~m}\) from a pair of slits. The separation on the screen between the central bright fringe and the first-order bright fringe is \(0.037 \mathrm{~m}\). The light illuminating the slits has a wavelength of \(490 \mathrm{~nm}\). Determine the slit separation.

Short Answer

Expert verified
The slit separation is approximately \( 5.96 \mathrm{~\mu m} \).

Step by step solution

01

Understand the given information

We know a few important pieces of information: The distance from the slits to the screen, denoted as \( L \), is \(4.5 \mathrm{~m}\). The wavelength of the light \( \lambda \) is given as \(490 \mathrm{~nm} = 490 \times 10^{-9} \mathrm{~m}\). The separation between the central bright fringe and the first-order bright fringe on the screen, \( y_1 \), is \(0.037 \mathrm{~m}\). We need to find the slit separation \( d \).
02

Understand the formula for fringe separation

In a double-slit interference pattern, the separation between consecutive bright fringes (or dark fringes) can be calculated using the formula: \( y_m = \frac{m \lambda L}{d} \), where \( y_m \) is the fringe separation for the \( m \)-th order bright fringe, \( m \) is the order of the fringe, \( \lambda \) is the wavelength, \( L \) is the distance to the screen, and \( d \) is the slit separation. For the first-order bright fringe, \( m = 1 \).
03

Rearrange formula to solve for slit separation

Rearrange the formula from Step 2 to solve for \( d \):\[ d = \frac{m \lambda L}{y_m} \]In this case, \( m = 1 \), \( \lambda = 490 \times 10^{-9} \mathrm{~m} \), \( L = 4.5 \mathrm{~m} \), and \( y_1 = 0.037 \mathrm{~m} \).
04

Substitute the known values

Substitute the known values into the formula from Step 3:\[ d = \frac{1 \times 490 \times 10^{-9} \times 4.5}{0.037} \]
05

Calculate the slit separation

Calculate \( d \):\[ d = \frac{490 \times 10^{-9} \times 4.5}{0.037} \approx \frac{2.205 \times 10^{-7}}{0.037} \approx 5.959 \times 10^{-6} \mathrm{~m} = 5.96 \mathrm{~\mu m}\]Thus, the slit separation \( d \) is approximately \( 5.96 \mathrm{~\mu m} \).

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.

Fringe Separation
In the fascinating world of physics, the concept of fringe separation emerges prominently when studying light interference patterns, especially in the double slit experiment. Fringe separation refers to the distance between successive bright or dark bands, known as fringes. These fringes result from constructive or destructive interference of light waves after passing through two closely spaced slits.

To determine fringe separation mathematically, we use the formula:
  • \( y_m = \frac{m \lambda L}{d} \)
Here, \( y_m \) represents the separation of the \( m \)-th order fringe from the central one, \( \lambda \) is the light wavelength, \( L \) the distance from slits to the screen, and \( d \) the slit separation. The variable \( m \) signifies the order of the fringe, with \( m = 1 \) corresponding to the first-order bright fringe.

This formula allows you to predict where bright and dark areas will appear on the screen, providing valuable insights into the wave nature of light.
Wavelength of Light
The wavelength of light is a fundamental parameter in understanding wave behavior, particularly in interference phenomena. It denotes the distance over which a wave's shape repeats, and is crucial when examining light passing through slits.

In our exercise example, we have a wavelength of \( 490 \mathrm{~nm} \), which stands for nanometers, where 1 nanometer equals \( 10^{-9} \) meters. This small scale highlights how precise and subtle the measurements in interference experiments must be. The wavelength affects the spacing between fringes on an observation screen; shorter wavelengths result in closer fringes, while longer ones cause them to spread apart.

Light wavelength variations can indicate different colors when visible light is used. This connection between wavelength and color is why interference patterns often exhibit colorful displays, adding an aesthetic layer to their scientific significance.
Slit Separation
Slit separation is a vital factor in experiments involving wave interference. It refers to the distance between two tiny, closely spaced slits through which light waves pass. These slits enable the overlapping of light waves, crucial to observing interference patterns. The formula linking fringes, slit separation, and wavelength is:
  • \( d = \frac{m \lambda L}{y_m} \)
This rearrangement from the fringe separation formula shows how to compute slit separation if fringe spacing is known. By carefully measuring fringe separation and using the known wavelength and distance to the screen, you can determine how far apart the slits must be.Understanding slit separation is essential not only for performing accurate experiments but also in various technological applications, such as diffraction gratings in spectroscopy. By influencing fringe patterns, slit separation becomes a tool for exploring wave properties and applying them in scientific and industrial contexts.

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

The transmitting antenna for a radio station is \(7.00 \mathrm{~km}\) from your house. The frequency of the electromagnetic wave broad cast by this station is \(536 \mathrm{kHz}\). The station builds a second transmitting antenna that broadcasts an identical electromagnetic wave in phase with the original one. The new antenna is \(8.12 \mathrm{~km}\) from your house. Does constructive or destructive interference occur at the receiving antenna of your radio? Show your calculations.

Refer to Interactive Solution 27.7 at for help in solving this problem. In a Young's double-slit experiment the separation \(y\) between the second-order bright fringe and the central bright fringe on a flat screen is \(0.0180 \mathrm{~m}\) when the light has a wavelength of 425 nm. Assume that the angles that locate the fringes on the screen are small enough so that \(\sin \theta \approx \tan \theta .\) Find the separation \(y\) when the light has a wavelength of \(585 \mathrm{nm}\)

Late one night on a highway, a car speeds by you and fades into the distance. Under these conditions the pupils of your eyes have diameters of about \(7.0 \mathrm{~mm} .\) The taillights of this car are separated by a distance of \(1.2 \mathrm{~m}\) and emit red light (wavelength \(=660 \mathrm{nm}\) in vacuum). How far away from you is this car when its taillights appear to merge into a single spot of light because of the effects of diffraction?

At the surface of the moon, which is \(3.77 \times 10^{8} \mathrm{~m}\) away, the light strikes a reflector left there by astronauts. The reflected light returns to the earth, where it is detected. When it leaves the spotlight, the circular beam of light has a diameter of about \(0.20 \mathrm{~m}\), and diffraction causes the beam to spread as the light travels to the moon. In effect, the first circular dark fringe in the diffraction pattern defines the size of the central bright spot on the moon. Determine the diameter (not the radius) of the central bright spot on the moon.

The same diffraction grating is used with two different wavelengths of light, \(\lambda_{\mathrm{A}}\) and \(\lambda_{\mathrm{B}}\). The fourth-order principal maximum of light A exactly overlaps the third-order principal maximum of light \(\mathrm{B}\). Find the ratio \(\lambda_{\mathrm{A}} / \lambda_{\mathrm{B}}\).
See all solutions

Recommended explanations on Physics Textbooks

Thermodynamics

Read Explanation

Wave Optics

Read Explanation

Astrophysics

Read Explanation

Particle Model of Matter

Read Explanation

Circular Motion and Gravitation

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.