Sketch a sinusoidal wave with an amplitude of \(2 \mathrm{cm}\) and a wavelength of \(6 \mathrm{cm} .\) This wave represents the electric field portion of a visible EM wave traveling to the right with intensity \(I_{0}\). (a) Sketch an identical wave beneath the first. What is the amplitude (in centimeters) of the sum of these waves? (b) What is the intensity of the new wave? (c) Sketch two more coherent waves beneath the others, one of amplitude \(3 \mathrm{cm}\) and one of amplitude \(1 \mathrm{cm},\) so all four are in phase. What is the amplitude of the four waves added together? (d) What intensity results from adding the four waves?

Short Answer

Expert verified
Question: Given four coherent sinusoidal waves with amplitudes of 2 cm, 2 cm, 3 cm, and 1 cm, and a wavelength of 6 cm, find the intensity of the resulting wave when all four waves are added together. Answer: The intensity of the resulting wave when all four coherent sinusoidal waves are added together is 16I₀.

Step by step solution

01

Understanding the relationship between amplitude and intensity

The intensity (\(I\)) of a wave is directly proportional to the square of its amplitude (\(A\)). Mathematically, it is given by \(I \propto A^{2}\).
02

Amplitude of the sum of two identical sinusoidal waves

In part (a), we are asked to sketch two identical sinusoidal waves with amplitude 2 cm and wavelength 6 cm and find the amplitude of the sum of these two waves. Since they are identical and in phase, their amplitudes simply add up: \(A_{sum} = A_1 + A_2 = 2 + 2 = 4 \mathrm{cm}\).
03

Intensity of the new wave

In part (b), we find the intensity of the new wave formed after summing the two identical sinusoidal waves. Since intensity is proportional to the square of amplitude, the new intensity is: \(I_{new} = I_0 \left( \frac{A_{sum}}{A_1} \right)^2 = I_0 \left( \frac{4}{2} \right)^2 = 4 I_0\).
04

Adding more coherent waves

In part (c), we are adding two more coherent sinusoidal waves beneath the previous two waves, one with an amplitude of 3 cm and one with an amplitude of 1 cm, so that all waves are in phase. The total amplitude can be found as: \(A_{total} = A_{sum} + A_3 + A_4 = 4 + 3 + 1 = 8 \mathrm{cm}\).
05

Intensity of the resulting wave

In part (d), we need to find the intensity of the resulting wave when all four waves are added together. The resulting intensity can be calculated as: \(I_{result} = I_0 \left( \frac{A_{total}}{A_1} \right)^2 = I_0 \left( \frac{8}{2} \right)^2 = 16 I_0\).

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

You are given a slide with two slits cut into it and asked how far apart the slits are. You shine white light on the slide and notice the first-order color spectrum that is created on a screen \(3.40 \mathrm{m}\) away. On the screen, the red light with a wavelength of 700 nm is separated from the violet light with a wavelength of 400 nm by 7.00 mm. What is the separation of the two slits?
A lens \((n=1.52)\) is coated with a magnesium fluoride film \((n=1.38) .\) (a) If the coating is to cause destructive interference in reflected light for \(\lambda=560 \mathrm{nm}\) (the peak of the solar spectrum), what should its minimum thickness be? (b) At what two wavelengths closest to 560 nm does the coating cause constructive interference in reflected light? (c) Is any visible light reflected? Explain.
A grating is made of exactly 8000 slits; the slit spacing is $1.50 \mu \mathrm{m} .\( Light of wavelength \)0.600 \mu \mathrm{m}$ is incident normally on the grating. (a) How many maxima are seen in the pattern on the screen? (b) Sketch the pattern that would appear on a screen \(3.0 \mathrm{m}\) from the grating. Label distances from the central maximum to the other maxima.
A double slit is illuminated with monochromatic light of wavelength $600.0 \mathrm{nm} .\( The \)m=0\( and \)m=1\( bright fringes are separated by \)3.0 \mathrm{mm}\( on a screen \)40.0 \mathrm{cm}$ away from the slits. What is the separation between the slits? [Hint: Is the small angle approximation justified?]
Parallel light of wavelength \(\lambda\) strikes a slit of width \(a\) at normal incidence. The light is viewed on a screen that is \(1.0 \mathrm{m}\) past the slits. In each case that follows, sketch the intensity on the screen as a function of \(x\), the distance from the center of the screen, for $0 \leq x \leq 10 \mathrm{cm}$ (a) \(\lambda=10 a\). (b) \(10 \lambda=a,\) (c) \(30 \lambda=a.\)
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