Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 31

In a double-slit experiment, the fifth maximum is 2.8 \(\mathrm{cm}\) from the central maximum on a screen that is \(1.5 \mathrm{m}\) away from the slits. If the slits are \(0.15 \mathrm{mm}\) apart, what is the wavelength of the light being used?

Short Answer

Expert verified
The wavelength of the light being used in the double-slit experiment is approximately 560 nm.

Step by step solution

01

List all given information

We have the following information: - The fifth maximum (\(m=5\)) - Distance from the central maximum to the fifth maximum, y = 2.8 cm = 0.028 m (convert to meters) - Distance between the screen and slits, L = 1.5 m - Distance between the slits, d = 0.15 mm = 1.5E-4 m (convert to meters)
02

Plug in the given values into the equation

Now, we can plug in the given values into the equation: \(5 \lambda = 1.5E-4 \frac{0.028}{1.5} \)
03

Calculate the wavelength

To find the wavelength \(\lambda\), we need to solve the equation: \(\lambda = \frac{1.5E-4 \cdot 0.028}{5 \cdot 1.5} \) \(\lambda = \frac{4.2E-6}{7.5}\) \(\lambda = 5.6E-7 \mathrm{m}\)
04

Convert the wavelength into a readable format

As the wavelength value may be too small to read in meters, let's convert it into nanometers: \(\lambda = 5.6E-7 \mathrm{m} \cdot \frac{1E9 \mathrm{nm}}{1 \mathrm{m}}\) \(\lambda = 560 \mathrm{nm}\) So, the wavelength of the light being used is approximately 560 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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 Michelson interferometer has two equal arms. A mercury light of wavelength \(546 \mathrm{nm}\) is used for the interferometer and stable fringes are found. One of the arms is moved by \(1.5 \mu \mathrm{m}\). How many fringes will cross the observing field?

Is there a phase change in the light reflected from either surface of a contact lens floating on a person's tear layer? The index of refraction of the lens is about \(1.5,\) and its top surface is dry.

In a thermally stabilized lab, a Michelson interferometer is used to monitor the temperature to ensure it stays constant. The movable mirror is mounted on the end of a 1.00 -m-long aluminum rod, held fixed at the other end. The light source is a He Ne laser, \(\lambda=632.8 \mathrm{nm}\). The resolution of this apparatus corresponds to the temperature difference when a change of just one fringe is observed. What is this temperature difference?

What effect does increasing the wedge angle have on the spacing of interference fringes? If the wedge angle is too large, fringes are not observed. Why?

A thin film with \(n=1.32\) is surrounded by air. What is the minimum thickness of this film such that the reflection of normally incident light with \(\lambda=500 \mathrm{nm}\) is minimized?
See all solutions

Recommended explanations on Physics Textbooks

Fields in Physics

Read Explanation

Electricity

Read Explanation

Electromagnetism

Read Explanation

Particle Model of Matter

Read Explanation

Linear Momentum

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.

Sign-up for free