a. Consider a 1 kW AM radio transmitter at 700 kHz. Calculate the number of phot

Question

a. Consider a 1 kW AM radio transmitter at 700 kHz. Calculate the number of photons emitted from the antenna per second. b. The average intensity of sunlight on Earth's surface is about 1 kW m^-2. The maximum intensity is at a wavelength around 800 nm. Assuming that all the photons have an 800 nm wavelength, calculate the number of photons arriving on Earth's surface per unit time per unit area. What is the magnitude of the electric field in the sunlight? Yellow, cyan, magenta, and white Three primary colors, red, green, and blue (RGB), can be added together in various proportions to generate any color on various displays and light emitting devices in what is known as the additive theory of color. For example, yellow can be generated from adding red and green, cyan from blue and green, and magenta from red and blue. a. A device engineer wants to use three light emitting diodes (LEDs) to generate various colors in an LED-based color display that is still in the research stage. His three LEDs have wavelengths of 660 nm for red, 563 nm for green, and 450 nm for blue. He simply wishes to generate the yellow and cyan by mixing equal optical powers from these LEDs: optical power, or radiant power, is defined as the radiation energy emitted per unit time. What are the numbers of red and blue photons needed (to the nearest integer) to generate yellow and cyan, respectively, for every 100 green photons? b. An equi-energy white light is generated by mixing red, green, and blue light in equal optical powers. Suppose that the wavelengths are 700 nm for red, 546 nm for green, and 436 nm for blue (which is one set of possible standard primary colors). Suppose that the optical power in each primary color is 0.1 W. Calculate the total photon flux (photons per second) needed from each primary color. c. There are bright white LEDs on the market that generate the white light by mixing yellow (a combination of red and green) with blue emissions. The inexpensive types use a single blue LED to generate a strong blue radiation, some of which is absorbed by a phosphor in front of the LED which then emits yellow light. The yellow and the blue passing through the phosphor mix and make up the white light. In one type of white LED, the blue and yellow wavelengths arc 450 nm and 564 nm, respectively. White light can be generated by setting the optical (radiative) power ratio of yellow to blue light emerging from the LED to be about 1.74. What is the ratio of the number of blue to yellow photons needed? (Sometimes the mix is not perfect and the white LED light tends to have a noticeable slight blue tint.) If the total optical power output from the white LED is 100 mW, calculate the blue and yellow total photon fluxes (photons per second).

Explanation / Answer

3.2

a. Given three wavelengths, lambda1 = 600 nm ( red)

lambda2 = 563 nm ( green)

lambda3 = 450 nm ( blue)

yellow = red + green

cyan = blue + green

as equal optical powers have to be mixed

for yellow color, for 100 green photons, let number of red photons be n

then n*h*c/lambda1 = 100*hc/lambda2

n = 100*lambda1/lambda2 = 100*600/563 = 106.57 = 107 photons

similiarly

for cyan color

n = 100*lambda3/lambda2 = 100*450/563 = 79.92 = 80 photons

b. white = red + blue + green ( in equal powers )

Given three wavelengths, lambda1 = 700 nm ( red)

lambda2 = 546 nm ( green)

lambda3 = 436 nm ( blue)

optical power of each color = 0.1 W

photon flux of red color = n

n*hc/lambda1 = 0.1

h = 6.63*10^-34 ( planks constant)

c = 3*10^8 m/s ( speed of light)

n = 3.519*10^17 photons per second

similiarly

for green color, n = 2.745*10^17 photons per second

for blue color , n = 2.192*10^17 photons per second

c. blue, lambda1 = 450 nm

yellow, lambda2 = 564 nm

power of yellow : power of blue = 1.74

ratio of blue to yellow photons = nb/ny

now, nb*hc*lambda2/lambda1*ny*hc = 1/1.74

nb*lambda2/ny*lambda1 = 1/1.74

nb/ny = 0.4585

for power, P = 100 mW = 0.1 W

nb*hc/lambda1 + ny*hc//lambda2 = 0.1

ny*hc[0.4585/lambda1 + 1/lambda2] = 0.1

ny = 1.8007*10^17 photons per second

nb = 0.825655*10^17 photons per second

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