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

When two light waves, both with wavelength \(\lambda\) and amplitude \(A\), interfere constructively, they produce a light wave of the same wavelength but with amplitude \(2 A .\) What will be the intensity of this light wave? a) same intensity as before b) double the intensity c) quadruple the intensity d) not enough information

Short Answer

Expert verified
Question: When two light waves of the same wavelength and amplitude interfere constructively, the intensity of the resulting wave is: a) Half the intensity b) Double the intensity c) Quadruple the intensity d) Equal to the intensity Answer: c) Quadruple the intensity

Step by step solution

01

Constructive interference of two waves

When two light waves of the same wavelength and amplitude interfere constructively, their amplitudes add up, which results in a wave with the same wavelength but twice the amplitude. So the amplitude of the resulting wave will be \(2 A\).
02

Relation between amplitude and intensity

The intensity of a wave is proportional to the square of its amplitude. Mathematically, this can be expressed as: \(I \propto A^2\) where \(I\) is the intensity and \(A\) is the amplitude of the wave.
03

Find the intensity of the resulting wave

As per the relationship, the intensity of the resulting wave will be proportional to the square of its amplitude \((2A)^2\). \(I_{result} \propto (2A)^2 = 4A^2\)
04

Compare the intensity of the resulting wave with the initial intensity

Since the intensity of the original waves is proportional to \(A^2\), let's compare the resulting intensity with the initial intensity: \(\frac{I_{result}}{I_{initial}} = \frac{4A^2}{A^2} = 4\)
05

Choose the correct option

The intensity of the resulting wave is 4 times the intensity of the original waves. So, the correct answer is option c) Quadruple the intensity.

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wave Amplitude
Wave amplitude is a measure of the size or height of a wave. It defines how much the medium is displaced from its resting position when the wave passes through. In simple terms, the amplitude is the peak value of wave displacement.
In the context of light waves, amplitude affects how bright or dim the light appears. The greater the amplitude, the brighter the light. When waves interfere constructively, as mentioned in the exercise, their amplitudes add up. This results in a wave with double the amplitude compared to individual waves.
  • Example: If one wave has an amplitude of 2 units and another wave with the same amplitude interferes constructively, the resulting wave has an amplitude of 4 units.
The concept of amplitude is crucial because it directly relates to the wave's energy and intensity, which gives insight into the wave's power and effect.
Wave Intensity
Wave intensity describes the power of a wave per unit area. When discussing light waves, intensity often relates to how much energy a wave carries over an area in a given time interval.
  • The formula to represent intensity is:
    \(I \propto A^2\)
    This means intensity is proportional to the square of the amplitude.
When thinking about constructive interference, remember that if the amplitude doubles, the intensity doesn't just double but increases by four times. That's because \((2A)^2 = 4A^2\).
This squared relationship indicates that even small changes in amplitude can cause significant changes in intensity. Therefore, understanding intensity helps in predicting how waves behave when they combine or interact with each other.
Wavelength
Wavelength refers to the distance between consecutive points of a wave that are in phase, such as crest to crest or trough to trough. Represented by the Greek letter \(\lambda\) (lambda), wavelength is critical in determining the energy and type of wave.
  • For light waves, it helps determine the color of the light in the visible spectrum.
  • In many wave phenomena, like sound or light, wavelength impacts how waves move through different mediums.
The concept of wavelength is essential in the exercise because constructive interference happens most effectively when the interfering waves have the same wavelength.
For two waves to interfere constructively, they need to be in phase, meaning the peaks and troughs line up which typically occurs with identical wavelengths. Understanding wavelength provides insights into how waves can combine constructively or destructively.

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 two-slit apparatus is covered with a red \((670 \mathrm{nm})\) filter. When white light is shone on the filter, on the screen beyond the two-slit apparatus, there are nine interference maxima within the 4.50 -cm-wide central diffraction maximum. When a blue \((450 \mathrm{nm})\) filter replaces the red, how many interference maxima will there be in the central diffraction maximum, and how wide will that diffraction maximum be?

A Young's interference experiment is performed with monochromatic green light \((\lambda=540 \mathrm{nm}) .\) The separation between the slits is \(0.100 \mathrm{~mm},\) and the interference pattern on a screen shows the first side maximum \(5.40 \mathrm{~mm}\) from the center of the pattern. How far away from the slits is the screen?

What minimum path difference is needed to cause a phase shift by \(\pi / 4\) in light of wavelength \(700 . \mathrm{nm} \)

You are making a diffraction grating that is required to separate the two spectral lines in the sodium \(D\) doublet, at wavelengths 588.9950 and \(589.5924 \mathrm{nm}\), by at least \(2.00 \mathrm{~mm}\) on a screen that is \(80.0 \mathrm{~cm}\) from the grating. The lines are to be ruled over a distance of \(1.50 \mathrm{~cm}\) on the grating. What is the minimum number of lines you should have on the grating?

Which close binary pair of stars will be more easily resolvable with a telescope - two red stars, or two blue ones? Assume the binary star systems are the same distance from Earth and are separated by the same angle.

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