Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License. Use the information below to generate a citation. derived for light that propagates mainly in one. Then you must include on every digital page view the following attribution: monochromatic light by means of the three-dimensional intensity. If you are redistributing all or part of this book in a digital format, Then you must include on every physical page the following attribution: If you are redistributing all or part of this book in a print format, Want to cite, share, or modify this book? This book uses the This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. Since the arc subtends an angle ϕ ϕ at the center of the circle, To calculate the intensity at an arbitrary point P on the screen, we return to the phasor diagram of Figure 4.7. In solving that problem, you will find that they are less than, but very close to, ϕ = 3 π, 5 π, 7 π, … rad. The exact values of ϕ ϕ for the maxima are investigated in Exercise 4.120. As a result, E 1 E 1 and E 2 E 2 turn out to be slightly larger for arcs that have not quite curled through 3 π 3 π rad and 5 π 5 π rad, respectively. Since the total length of the arc of the phasor diagram is always N Δ E 0, N Δ E 0, the radius of the arc decreases as ϕ ϕ increases. These two maxima actually correspond to values of ϕ ϕ slightly less than 3 π 3 π rad and 5 π 5 π rad. The proof is left as an exercise for the student ( Exercise 4.119). In part (e), the phasors have rotated through ϕ = 5 π ϕ = 5 π rad, corresponding to 2.5 rotations around a circle of diameter E 2 E 2 and arc length N Δ E 0. The amplitude of the phasor for each Huygens wavelet is Δ E 0, Δ E 0, the amplitude of the resultant phasor is E, and the phase difference between the wavelets from the first and the last sources is The phasor diagram for the waves arriving at the point whose angular position is θ θ is shown in Figure 4.7. This distance is equivalent to a phase difference of ( 2 π a / λ N ) sin θ. If we consider that there are N Huygens sources across the slit shown in Figure 4.4, with each source separated by a distance a/N from its adjacent neighbors, the path difference between waves from adjacent sources reaching the arbitrary point P on the screen is ( a / N ) sin θ. To calculate the intensity of the diffraction pattern, we follow the phasor method used for calculations with ac circuits in Alternating-Current Circuits. Calculate the intensity relative to the central maximum of an arbitrary point on the screen.Calculate the intensity relative to the central maximum of the single-slit diffraction peaks.Therefore, the light from the sun is not monochromatic as it does not have purely one colour.By the end of this section, you will be able to: Sunlight is white but it consists of 7 different colours. The name mono-chrome means having only one colour. The colour of the light depends on the wavelength of the light and the energy level determines the frequency. The monochromatic light is the electromagnetic radiation obtained from a single photon emission from atoms. Therefore, the correct answer is option A. Therefore, energy of beam A can be given by, Where, n is the number of photoelectrons. Therefore, using these two conditions, we shall calculate our answer. We know that the formula for energy or Planck’s equation depends inversely on the wavelength. Therefore, the intensity of the light is equal for both. It is given that all the light is used to eject photoelectrons. Hint: In this question we have been asked to calculate the ratio of number of photoelectrons ejected from beam A to beam B.
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