/*! 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} Q. 27 A200聽mWvertically polarized las... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 31: Q. 27 (page 902)

A200mWvertically polarized laser beam passes through a polarizing filter whose axis is 35掳 from horizontal. What is the power of the laser beam as it emerges from the filter?

Short Answer

Expert verified

The power of the laser beam is65.8mw.

Step by step solution

01

Formula for power of laser beam

When unpolarized light strikes a polarizer, the power of the polarised light that emerges is half that of the unpolarized light.

Incident light that isn't polarised.

So,

P=P0cos2胃

Where,

role="math" localid="1648929202807" P0-Vertically polarized laser beam

P-Power of laser beam

02

Figure for laser beam with axis of 35°

Figure is,

03

Calculation for power of laser beam

The beam angle is,

胃=90掳-35掳

=55掳

The beam's power is determined as follows:

P=P0cos2胃

=200mW脳cos255掳

P鈮�65.8mW

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

In FIGURE P31.33, a circular loop of radius rtravels with speed valong a charged wire having linear charge density I. The wire is at rest in the laboratory frame, and it passes through the center of the loop.

a. What are E鈫�and B鈫�at a point on the loop as measured by a scientist in the laboratory? Include both strength and direction.

b. What are the fields E鈫�and B鈫�at a point on the loop as measured by a scientist in the frame of the loop?

c. Show that an experimenter in the loop鈥檚 frame sees a currentI=位谓passing through the center of the loop.

d. What electric and magnetic fields would an experimenter in the loop鈥檚 frame calculate at distance r from the current of part c?

e. Show that your fields of parts b and d are the same.

Unpolarized light with intensity350W/m2 passes first through a polarizing filter with its axis vertical, then through a second polarizing filter. It emerges from the second filter with intensity 131W/m2. What is the angle from vertical of the axis of the second polarizing filter?

When the Voyager 2spacecraft passed Neptune in 1989, it was 4.5脳109kmfrom the earth. Its radio transmitter, with which it sent back data and images, broadcast with a mere 21Wof power. Assuming that the transmitter broadcast equally in all directions,

a. What signal intensity was received on the earth?

b. What electric field amplitude was detected?

The received signal was somewhat stronger than your result because the spacecraft used a directional antenna, but not by much.

A radio wave is traveling in the negativey-direction. What is the direction of E鈫�at a point whereB鈫�is in the positivex-direction?

FIGURE shows a vertically polarized radio wave of frequency 1.0*106Hztraveling into the page. The maximum electric field strength is 1000V/m. What are

a. The maximum magnetic field strength?

b. The magnetic field strength and direction at a point where Eu=1500V/m, down?
See all solutions

Recommended explanations on Physics Textbooks

Kinematics Physics

Read Explanation

Work Energy and Power

Read Explanation

Famous Physicists

Read Explanation

Energy Physics

Read Explanation

Quantum Physics

Read Explanation

Electrostatics

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.