/*! 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 51 A flood light is covered with a ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 51

A flood light is covered with a filter that transmits red light. The electric field of the emerging beam is represented by a sinusoidal plane wave $$ E_{x}=36 \sin \left(1.20 \times 10^{7} z-3.6 \times 10^{15} t\right) \mathrm{V} / \mathrm{m} $$ The average intensity of the beam will be (A) \(0.86 \mathrm{~W} / \mathrm{m}^{2}\) (B) \(1.72 \mathrm{~W} / \mathrm{m}^{2}\) (C) \(3.44 \mathrm{~W} / \mathrm{m}^{2}\) (D) \(6.88 \mathrm{~W} / \mathrm{m}^{2}\)

Short Answer

Expert verified
The correct answer is option (B): The average intensity of the beam is \(1.72 \,\mathrm{W} / \mathrm{m}^{2}\).

Step by step solution

01

Given the electric field equation: $$ E_x = 36 \sin\left(1.20 \times 10^7 z - 3.6 \times 10^{15} t\right) \frac{\mathrm{V}}{\mathrm{m}} $$ Notice that the maximum value of \(\sin\) is 1, so the maximum electric field, \(E_\textit{max}\) is: $$ E_\textit{max} = 36 \,\mathrm{V} / \mathrm{m} $$ #Step 2: Calculate the average intensity#

Now we can use the equation for the intensity of a sinusoidal plane wave to calculate the average intensity of the beam: $$ \textit{Intensity} = \frac{1}{2} \epsilon_0 c E_\textit{max}^2 $$ using the values \(\epsilon_0 = 8.85 \times 10^{-12} \,\mathrm{F/m}\) (permittivity of free space) and \(c = 3 \times 10^8 \,\mathrm{m/s}\) (speed of light), and the maximum electric field obtained in step 1, \(E_\textit{max} = 36 \,\mathrm{V} / \mathrm{m}\): $$ \textit{Intensity} = \frac{1}{2} (8.85 \times 10^{-12}\,\mathrm{F/m})(3 \times 10^8 \,\mathrm{m/s}) (36 \,\mathrm{V/m})^2 $$ #Step 3: Calculate the numerical value of the intensity#
02

Calculate the numerical value of the intensity: $$ \textit{Intensity} = 1.72 \,\mathrm{W} / \mathrm{m}^{2} $$ #Step 4: Compare the result to the given choices#

Comparing the calculated average intensity to the given answer choices, we find that the average beam intensity matches the value in option (B), which is: $$ 1.72 \,\mathrm{W} / \mathrm{m}^{2} $$ Thus, the correct answer is option (B).

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!

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 \(100 \mathrm{pF}\) capacitor is connected to a \(230 \mathrm{~V}, 50 \mathrm{~Hz} \mathrm{AC}\) source. The rms value of conduction current will be (A) \(7.2 \times 10^{-6} \mathrm{~A}\) (B) \(3.6 \times 10^{-5} \mathrm{~A}\) (C) \(1.8 \times 10^{-4} \mathrm{~A}\) (D) \(0.9 \times 10^{-3} \mathrm{~A}\)

This question has Statement 1 and Statement 2. Of the four choices given after the statements, choose the one that best describes the two statements. Statement-1: Sky wave signals are used for long distance radio communication. These signals are in general, less stable than ground wave signals are in general, less stable than ground wave signals. Statement-2: The state of ionosphere varies from hour to hour, day to day and season to season. (A) Statement- 1 is true, Statement-2 is false. (B) Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation of Statement-1. (C) Statement 1 is true, Statement 2 is true, Statement 2 is not the correct explanation of Statement 1 . (D) Statement 1 is false, Statement 2 is true.

The ozone layer in earth's atmosphere is crucial for human survival because it (A) has ions. (B) reflects radio signals. (C) reflects ultraviolet ray. (D) reflects infra red rays.

Infrared radiation is detected by (A) Spectrometer (B) Pyrometer (C) Nanometer (D) Photometer

The AM range of radio waves have frequency (A) Less than \(30 \mathrm{MHz}\) (B) More than \(30 \mathrm{MHz}\) (C) Less than \(20000 \mathrm{~Hz}\) (D) More than \(20000 \mathrm{~Hz}\)
See all solutions

Recommended explanations on Physics Textbooks

Electrostatics

Read Explanation

Modern Physics

Read Explanation

Fundamentals of Physics

Read Explanation

Particle Model of Matter

Read Explanation

Atoms and Radioactivity

Read Explanation

Scientific Method 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.