An argon ion laser puts out 5.0 W of continuous power at a wavelength of 532 nm. The diameter of the laser beam is 5.5 mm. If the laser is pointed toward a pinhole with a diameter of 1.2 mm, how many photons travel through the pinhole per second? Assume that the light intensity is equally distributed throughout the entire cross-sectional area of the beam. (1 W = 1 J/s)

Video transcript

Hey everyone in this example we have a beam of light giving off 3.2 jewels per second of power. We're told the beam of light has a 4.3 centimeter diameter, A 452 pick a meter wavelength. And we need to figure out how many photons of this light are going through a 0. centimeter hole per second. So what we should recognize is that because we're given diameter values of two types of objects that are circular in shape. We're going to want to recall our formula for the area of a circle. And so we should recall that the area of the circle is equal to pi R squared. So what we also should recognize is that this is going to be a four part solution process where in part one we want to find the fraction of light through the hole. So what we should have is we're going to take the area of our whole and we're going to divide that by the area of our beam of light. And what we should have is in our numerator we should have pi times r radius and so based on the radius of our whole, we're going to go ahead and find that By taking the given diameter .89cm. Because we should recall that the radius is half of the diameter. So we're taking this and we're dividing it by two And this is going to give us our radius equal to a value of .445 cm. So we can plug that into the numerator and let me just make that part little neater. And this is squared in our denominator. We follow the same process pi times our radius of our beam of light. So we'll find that here below Again, based on our given diameter of the beam of light as 4.3 cm. We're going to recall the radius is half of this. So we divide this by two and we get our radius of our beam of light equal to 2. cm. And this is squared in the denominator. So now that we have our centimeter units aligned in the numerator and the denominator, we can cancel both out, leaving us with no units which is what we want for the fraction of light through the beam. The fraction of the beam of light through the hole and this is going to give us a value equal to 0.042-84. The next step in this problem is Part two. Where we're going to find the fraction of power of the light. And so in order to do so we're going to take the fraction of the beam of light through the hole which we found above. 0.04284. And we're multiplying this by the power that they give us above. So they say that this light is giving off 3.2 jewels per second of power. So we're going to multiply by this value and this is going to give us a value of 0.1371 jules per second. And so this is our fraction of power for the light. Our third step in this example is going to be finding the energy of our photon of light or of a photon of light. And so in order to do so we want to recall our formula for energy which is equal to Planck's constant times the speed of light and divided by our wavelength. And we should recall that this symbol in the denominator is lambda for wavelength. So in our numerator we're going to recall that plank's constant is 6. times 10 to the negative 34th power jewels times seconds. And then plank's constant. Again we're going to recall is multiplied by the speed of light, which we recall is three times 10 to the eighth Power meters per second. In our denominator. We're going to plug in the given wavelength from the problem which has given us 452. Pick a meters. I'm sorry we don't really need those parentheses. So we plug in 452 PICO meters. But we should recognize that because we're given our wavelength in PICO meters. We actually want to convert from PICO meters. two m because meters are the proper unit for wavelength. So we should recall that our prefix PICO tells us that we have 10 to the negative 12 power of our base unit meter. And this allows us to cancel out PICO meters leaving us with meters for our wavelength. Our next step is to focus on canceling out the rest of the units. So as you can see we have meters in the numerator so we can cancel that out and we have seconds in the numerator here and then in the denominator over here. So we're only left with jewels which is what we want for our unit of energy. And we can go ahead and calculate this quotient in our calculators. So this is going to give us a value equal to 4. times 10 to the negative 16th power jewels per photon of light. And so lastly we can move on to part four of this question which is finding the number of photons per second as the question asks us. So for this setup we're going to take the fraction of power of our light Which we found to be 0.1371 jewels per second. And we're going to divide this by our energy of our photon of a photon of light which above we found to be 4.397, 8 Times 10 to the -16 power jewels per photon. And so we're going to be able to cancel out our units of jewels because we have one in the numerator and one in the denominator. And we are left with seconds per photon, which is exactly what we want. So this is going to give us a value in our calculators equal to 3.1175 times 10 to the 14th power photons per second. And so this will complete this example as our final answer for the amount of photons going through the 0.89 centimeter hole per second. And this will correspond to choice see in our multiple choice. So I hope that everything be reviewed was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.