Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A 60.0 -m \(\mathrm{W}\) pulsed laser produces a pulse of EM radia tion with wavelength \(1060 \mathrm{nm}\) (in air) that lasts for 20.0 ps (picoseconds). (a) In what part of the EM spectrum is this pulse? (b) How long (in centimeters) is : single pulse in air? (c) How long is it in water \((n=1.33) \equiv\) (d) How many wavelengths fit in one pulse? (e) What is the total electromagnetic energy in one pulse?

Short Answer

Expert verified
a) The 1060 nm wavelength pulse lies in the near-infrared range of the electromagnetic spectrum. b) The length of a single pulse in air is 6.0 mm. c) The length of a single pulse in water is 4.52 mm. d) There are approximately 5660 wavelengths in one pulse. e) The total electromagnetic energy in one pulse is approximately \(1.2 \times 10^{-12}\ \mathrm{J}\).

Step by step solution

01

a) Identify the part of the EM spectrum

Given the wavelength of 1060 nm, we can identify the part of the electromagnetic spectrum to which it belongs. After converting the wavelength to meters, we find that it lies in the near-infrared range, specifically, the range of 700 nm to 1400 nm.
02

b) Length of a single pulse in air

To find the length of a single pulse in air, we need the speed of light in air and the duration of the pulse. The speed of light in air is approximately the same as in vacuum, which is \(c \approx 3.0 \times 10^8\ \mathrm{m/s}\). The pulse duration is given as 20.0 ps, which is \(20.0 \times 10^{-12}\ \mathrm{s}\). To find the length of a single pulse, we multiply the speed of light by the pulse duration: Length in air = \(c \times t\) Length in air = \((3.0 \times 10^8\ \mathrm{m/s})(20.0 \times 10^{-12}\ \mathrm{s}) = 6.0 \times 10^{-3}\ \mathrm{m}\) or 6.0 mm
03

c) Length of a single pulse in water

To find the length of a single pulse in water, we need the speed of light in water. We are given the refractive index of water, \(n=1.33\). The speed of light in water, \(v\), can be found using the refractive index formula: \(v = \frac{c}{n}\) \(v = \frac{3.0 \times 10^8\ \mathrm{m/s}}{1.33}= 2.26 \times 10^8\ \mathrm{m/s}\) Now, we can find the length of a single pulse in water by multiplying the speed of light in water by the pulse duration: Length in water = \(v \times t\) Length in water = \((2.26 \times 10^8\ \mathrm{m/s})(20.0 \times 10^{-12}\ \mathrm{s}) = 4.52 \times 10^{-3}\ \mathrm{m}\) or 4.52 mm
04

d) Number of wavelengths in one pulse

To find the number of wavelengths in one pulse, we need to divide the length of the pulse in air by the given wavelength: Number of wavelengths = \(\frac{\text{Length in air}}{\lambda}\) Number of wavelengths = \(\frac{6.0 \times 10^{-3}\ \mathrm{m}}{1060 \times 10^{-9}\ \mathrm{m}} \approx 5660\) There are approximately 5660 wavelengths in one pulse.
05

e) Total electromagnetic energy in one pulse

To find the total electromagnetic energy in one pulse, we first need the frequency of the radiation using the speed of light equation: \(f = \frac{c}{\lambda}\) \(f = \frac{3.0 \times 10^8\ \mathrm{m/s}}{1060 \times 10^{-9}\ \mathrm{m}}= 2.83 \times 10^{14}\ \mathrm{Hz}\) Now we can find the total electromagnetic energy using the energy formula \(E = hf\). We are given the laser's power as 60.0 mW, and the pulse duration is 20.0 ps. First, convert the power to watts and the duration to seconds: Power = \(60.0 \times 10^{-3}\ \mathrm{W}\) Duration = \(20.0 \times 10^{-12}\ \mathrm{s}\) Next, find the total energy: Total energy = Power × Duration Total energy = \((60.0 \times 10^{-3}\ \mathrm{W})(20.0 \times 10^{-12}\ \mathrm{s}) = 1.2 \times 10^{-12}\ \mathrm{J}\) The total electromagnetic energy in one pulse is approximately \(1.2 \times 10^{-12}\ \mathrm{J}\).

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 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

Using Faraday's law, show that if a magnetic dipole antenna's axis makes an angle \(\theta\) with the magnetic field of an EM wave, the induced emf in the antenna is reduced from its maximum possible value by a factor of cos $\theta .$ [Hint: Assume that, at any instant, the magnetic field everywhere inside the loop is uniform. \(]\)
The electric field in a radio wave traveling through vacuum has amplitude $2.5 \times 10^{-4} \mathrm{V} / \mathrm{m}\( and frequency \)1.47 \mathrm{MHz}$. (a) Find the amplitude and frequency of the magnetic field. (b) The wave is traveling in the \(+x\) -direction. At \(x=0\) and \(t=0,\) the electric field is \(1.5 \times 10^{-4} \mathrm{V} / \mathrm{m}\) in the \- y-direction. What are the magnitude and direction of the magnetic field at \(x=0\) and \(t=0 ?\).
Unpolarized light is incident on a system of three polarizers. The second polarizer is oriented at an angle of \(30.0^{\circ}\) with respect to the first and the third is oriented at an angle of \(45.0^{\circ}\) with respect to the first. If the light that emerges from the system has an intensity of $23.0 \mathrm{W} / \mathrm{m}^{2},$ what is the intensity of the incident light?
Just after sunrise, you look straight up at the sky. Is the light you see polarized? If so, in what direction?
A polarized beam of light has intensity \(I_{0} .\) We want to rotate the direction of polarization by \(90.0^{\circ}\) using polarizing sheets. (a) Explain why we must use at least two sheets. (b) What is the transmitted intensity if we use two sheets, each of which rotates the direction of polarization by \(45.0^{\circ} ?\) (c) What is the transmitted intensity if we use four sheets, each of which rotates the direction of polarization by \(22.5^{\circ} ?\)
See all solutions

Recommended explanations on Physics Textbooks

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