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- https://phys.libretexts.org/Courses/Muhlenberg_College/Physics_122%3A_General_Physics_II_(Collett)/12%3A_Waves/12.05%3A_Energy_and_Power_of_a_WaveThe energy and power of a wave are proportional to the square of the amplitude of the wave and the square of the angular frequency of the wave. Intensity is defined as the power divided by the area. A...The energy and power of a wave are proportional to the square of the amplitude of the wave and the square of the angular frequency of the wave. Intensity is defined as the power divided by the area. As the wave moves out from a source, the energy is conserved, but the intensity decreases as the area increases.
- https://phys.libretexts.org/Courses/Coalinga_College/Physical_Science_for_Educators_(CID%3A_PHYS_14)/14%3A_Property_of_Sound_Doppler_Effect_and_Interferences/14.05%3A_Intensity_and_Loudness_of_SoundThe loudness of sound is determined, in turn, by the intensity of the sound waves. Intensity results from two factors: the amplitude of the sound waves and how far they have traveled from the source o...The loudness of sound is determined, in turn, by the intensity of the sound waves. Intensity results from two factors: the amplitude of the sound waves and how far they have traveled from the source of the sound. The same amount of energy is spread over a greater area, so the intensity and loudness of the sound is less. Intensity of sound results from two factors: the amplitude of the sound waves and how far they have traveled from the source of the sound.
- https://phys.libretexts.org/Bookshelves/University_Physics/University_Physics_I_-_Classical_Mechanics_(Gea-Banacloche)/12%3A_Waves_in_One_Dimension/12.01%3A_Traveling_WavesIf we think of the momentum of a volume element in the medium as being proportional to the product of the instantaneous density and velocity, we see that for this wave, which is traveling in the posit...If we think of the momentum of a volume element in the medium as being proportional to the product of the instantaneous density and velocity, we see that for this wave, which is traveling in the positive \(x\) direction, there is more “positive momentum” than “negative momentum” in the medium at any given time (of course, if the wave had been traveling in the opposite direction, the sign of \(v_{med}\) in Equation (\ref{eq:12.6}) would have been negative, and we would have found the opposite re…
- https://phys.libretexts.org/Courses/Muhlenberg_College/Physics_122%3A_General_Physics_II_(Collett)/14%3A_Diffraction/14.03%3A_Intensity_in_Single-Slit_DiffractionThe intensity pattern for diffraction due to a single slit can be calculated using phasors as \(I = I_0 \left(\frac{sin \space \beta}{\beta}\right)^2,\) where \(\beta = \frac{\phi}{2} = \frac{\pi D \...The intensity pattern for diffraction due to a single slit can be calculated using phasors as \(I = I_0 \left(\frac{sin \space \beta}{\beta}\right)^2,\) where \(\beta = \frac{\phi}{2} = \frac{\pi D \space sin \space \theta}{\lambda}\), D is the slit width, λλ is the wavelength, and θθ is the angle from the central peak.
- https://phys.libretexts.org/Courses/Bowdoin_College/Phys1140%3A_Introductory_Physics_II%3A_Part_2/04%3A_Diffraction/4.03%3A_Intensity_in_Single-Slit_DiffractionThe intensity pattern for diffraction due to a single slit can be calculated using phasors as \(I = I_0 \left(\frac{sin \space \beta}{\beta}\right)^2,\) where \(\beta = \frac{\phi}{2} = \frac{\pi D \...The intensity pattern for diffraction due to a single slit can be calculated using phasors as \(I = I_0 \left(\frac{sin \space \beta}{\beta}\right)^2,\) where \(\beta = \frac{\phi}{2} = \frac{\pi D \space sin \space \theta}{\lambda}\), D is the slit width, λλ is the wavelength, and θθ is the angle from the central peak.
- https://phys.libretexts.org/Bookshelves/Astronomy__Cosmology/Planetary_Photometry_(Tatum_and_Fairbairn)/02%3A_Albedo/2.08%3A_IntensityThe intensity of an element of area is the product of its radiance and its projected area., and the intensity of a surface in a given direction is the integral of the radiance over the projected area ...The intensity of an element of area is the product of its radiance and its projected area., and the intensity of a surface in a given direction is the integral of the radiance over the projected area of the surface. If we consider a sphere of radius α centred in an Oxyz frame with directional spherical coordinates (Θ, Φ) irradiated from the x-direction with flux density F, an element of surface area is α 2 sin ΘdΘdΦ and its projected area in the direction μ is μα 2 sin ΘdΘdΦ.
- https://phys.libretexts.org/Courses/Merrimack_College/Conservation_Laws_Newton's_Laws_and_Kinematics_version_2.0/24%3A_Waves_in_One_Dimension/24.01%3A_Traveling_WavesIf we think of the momentum of a volume element in the medium as being proportional to the product of the instantaneous density and velocity, we see that for this wave, which is traveling in the posit...If we think of the momentum of a volume element in the medium as being proportional to the product of the instantaneous density and velocity, we see that for this wave, which is traveling in the positive \(x\) direction, there is more “positive momentum” than “negative momentum” in the medium at any given time (of course, if the wave had been traveling in the opposite direction, the sign of \(v_{med}\) in Equation (\ref{eq:12.6}) would have been negative, and we would have found the opposite re…
- https://phys.libretexts.org/Workbench/PH_245_Textbook_V2/06%3A_Module_5_-_Oscillations_Waves_and_Sound/6.03%3A_Objective_5.c./6.3.01%3A_Energy_and_Power_of_a_WaveThe energy and power of a wave are proportional to the square of the amplitude of the wave and the square of the angular frequency of the wave. Intensity is defined as the power divided by the area. A...The energy and power of a wave are proportional to the square of the amplitude of the wave and the square of the angular frequency of the wave. Intensity is defined as the power divided by the area. As the wave moves out from a source, the energy is conserved, but the intensity decreases as the area increases.
- https://phys.libretexts.org/Bookshelves/College_Physics/College_Physics_1e_(OpenStax)/17%3A_Physics_of_Hearing/17.03%3A_Sound_Intensity_and_Sound_LevelIntensity is defined to be the power per unit area carried by a wave. Power is the rate at which energy is transferred by the wave.
- https://phys.libretexts.org/Courses/Grand_Rapids_Community_College/PH246_Calculus_Physics_II_(2025)/13%3A_Diffraction/13.03%3A_Intensity_in_Single-Slit_DiffractionThe intensity pattern for diffraction due to a single slit can be calculated using phasors as \(I = I_0 \left(\frac{sin \space \beta}{\beta}\right)^2,\) where \(\beta = \frac{\phi}{2} = \frac{\pi D \...The intensity pattern for diffraction due to a single slit can be calculated using phasors as \(I = I_0 \left(\frac{sin \space \beta}{\beta}\right)^2,\) where \(\beta = \frac{\phi}{2} = \frac{\pi D \space sin \space \theta}{\lambda}\), D is the slit width, λλ is the wavelength, and θθ is the angle from the central peak.
- https://phys.libretexts.org/Bookshelves/University_Physics/University_Physics_(OpenStax)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/04%3A_Diffraction/4.03%3A_Intensity_in_Single-Slit_DiffractionThe intensity pattern for diffraction due to a single slit can be calculated using phasors as \(I = I_0 \left(\frac{sin \space \beta}{\beta}\right)^2,\) where \(\beta = \frac{\phi}{2} = \frac{\pi D \...The intensity pattern for diffraction due to a single slit can be calculated using phasors as \(I = I_0 \left(\frac{sin \space \beta}{\beta}\right)^2,\) where \(\beta = \frac{\phi}{2} = \frac{\pi D \space sin \space \theta}{\lambda}\), D is the slit width, λλ is the wavelength, and θθ is the angle from the central peak.
