A horizontal beam of laser light of wavelength 585 nm passes through a narrow slit that has width 0.0620 mm. The intensity of the light is measured on a vertical screen that is 2.00 m from the slit. (a) What is the minimum uncertainty in the vertical component of the momentum of each photon in the beam after the photon has passed through the slit? (b) Use the result of part (a) to estimate the width of the central diffraction maximum that is observed on the screen.

Video transcript

Hey everyone. So this problem is dealing with wave particle duality. It's a two part question, let's see what they give us and see what they're asking us for. So consider a vertical screen is placed at 1.5 m from a narrow slit. So we're gonna call distance is 1.5 m. We know that there is a horizontal beam of light. The wavelength of the light given λ is 560 nm. And so we're gonna rewrite that as 5. times 10 to the negative seven m. And the slit width, we'll just call that W for now is 0.625 millimeters. We're gonna rewrite that 6.25 times 10 to the negative fifth meters. First part is asking us to determine the least uncertainty in the vertical component of momentum. And the second problem is second part of the problem is asking us to calculate the width of the diffraction pattern for which it backs. The number of photons will pass through the slip. So this is a pretty the first part is pretty straightforward as long as we remember. Heisenberg's uncertainty principle. Um It's not as fun. We're not working with cats today, but it is pretty straightforward if we recall Heisenberg's uncertainty principle is delta Y, times delta P equals H over four pi. So H is Planck's constant. So let's recall it that 6.63 times 10 to the negative 34 jewels per second. And here, the uncertainty in the vertical component of momentum. This delta Y is the slit width. So we're actually going to this slip with here is equal to delta Y. So from there it's just a plug and chug for this problem vertical is going to be Y. Um horizontal will be denoted by X. So what we have is the um delta P. Y. The vertical component of momentum is H over four pi delta Y. And we can plug in all of those um known numbers and we will get our answer. So four pi times 6.25 times 10 to the negative fifth meters for our bottom turn, plug that in and we get 8.44 times 10 to the negative 31 killer gram meters per second. So that's part one, you take a look at our answers are potential answers and only one of them actually matches part one the answer for part one. But let's still work through part Two to make sure that we are not missing anything. And that our answer aligns with the answer for part two. So part two is asking us to calculate the width of the diffraction pattern for which a maximum number of photons will pass through the slit. So here we need to first recall broccoli's equation. We which is um the wavelength is equal to Planck's constant over the momentum. So in this case that's the momentum in the X direction. So we can solve for um that momentum. So let's do that quickly quickly, wavelength So we have this concept in our wavelength is 5.6 Times 10 to the negative seven m. And from there we get 1.18 times 10 to the negative 27 kilogram meters per second. Okay, so now we know the change in momentum in the Y direction. We know the momentum in the X. Direction. So in the vertical and the horizontal. And we need the width of the diffraction pattern. So we need to recall our angular deviation formulas. So that's data equals delta P. Y over P. X. And then why sir, slit width is given by data de So from here we actually do know recall from the problem that D. is 1.5 m. So from here we can solve for our slip with just the the goal of the second part of the problem. So we're gonna plug and chug Using the answer we found in part one and then this kind of interim term, Her interim answer we found here for part two. And we get 7. times 10 to the negative 4th radiance. And so for why? That gives us times the distance And why is 1.1 times 10 to the - m. So if we go back up to our possible answers, yes, that does align with the answer from C for part two of 1.1 mm. And so that's it. The answer for this problem is see that's all for this video, we'll see you in the next one.