![]() ![]() This means that the energy per photon of the light depends on its wavelength and velocity. The expression turns out to be I 4 Io cos2 (phi/2). Let's calculate the expression for the intensity of interfering waves due to coherent sources. Where $\lambda $ is the wavelength of the light and v is its velocity. Intensity in YDSE (Visual method-phasors) I 4Io cos2 (phi/2) Google Classroom. The frequency and wavelength of the light are related as $f\lambda =v$, Where h is Planck’s constant and f is the frequency of the light. The energy per photon is given as $E=hf$ …. Intensity-Field Relationship To see how intensity is related to the E-field and B-field, we multiply the numerator and denominator of Eq. The frequency of the light influences the energy per photon. Recall that intensity is proportional to amplitude squared. Since light is also a wave, its intensity will depend on the amplitude of the wave. Because a wave’s intensity is proportional to amplitude squared, the intensity of the resulting wave is four times as great as in the individual waves. This means that the intensity of a wave depends on the amplitude of the wave. $I\propto $, where I is intensity and A is amplitude of the wave. And intensity of a wave is directly proportional to the amplitude of the wave. ![]() Then the intensity of the light at the surface of the spherical source is equal to the energy emitted by it per unit time through a unit area of its surface.įrom electromagnetic theory we know that light is a wave. Suppose a source of light is in the form of a spherical shape and it is emitting a light of some wavelength. Intensity of light is defined as the energy transmitted per unit area in one unit of time. Let us first understand what intensity of light is. The quantity known as the wave’s frequency refers to the number of full wavelengths that pass by a given point in space every second the SI unit for frequency is Hertz ( Hz), which is equivalent to per seconds ( written as 1 s or s 1). Find out on what factor that the intensity of a wave depends and relate it to the light. The intensity (or illuminance or irradiance) of light or other linear waves radiating from a point source (energy per unit of area perpendicular to the source) is inversely proportional to the square of the distance from the source, so an object (of the same size) twice as far away receives only one-quarter the energy (in the same time period). Hint.To solve the given question, we must know about the intensity of a wave because light is also a wave.
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