A slit 0.240 mm wide is illuminated by parallel light rays of wavelength 540 nm. The diffraction pattern is observed on a screen that is 3.00 m from the slit. The intensity at the center of the central maximum (u = 0°) is 6.00 x 10^-6 W/m2. (a) What is the distance on the screen from the center of the central maximum to the first minimum?

Video transcript

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem. A monochromatic light beam with a wavelength of 457 nanometers strikes a narrow single slit with a width of 150 micrometers. A diffraction pattern is formed on a paper graph 2.0 m away from the slip. The central maximum has a an intensity of I subscript zero. What would be the distance measured between the central maximum and the first dark fringe on the graph on the paper graph? OK. So we're given some multiple choice answers here. They're all in the same units of millimeters. Let's read them off to see what our final answer might be. A is 6.10 B is 12.2 C is 9.20 and D is 18.4. OK. So first off, let us recall the equation for destructive interference which states that sine theta where theta represents the angle is equal to M, which represents the minimum value which is some integer value multiplied by the wavelength represented by lambda divided by the slit width represented by a OK. So note that the first dark fringe occurs at M equals plus or minus one. So let's plug in our known variables into the destructive interference equation. So sign theta and this is to help solve for the first dark fringe. So we're gonna call it sine theta subscript one for the first dark fringe equals M equals one multiplied by the wavelength which we need to convert. Since wavelength is given to us as 457 nanometers, we need to convert that to meters, which is really easy. We just take 457 multiplied by 10 to the power of negative nine to to convert it to meters divided by the slit width which is given to us as 150 micrometers when we need to convert micrometers to meters. And so in order to do that, it's also a quick fix is 150 multiplied by 10 to the power of negative 6 m. So plugging this into a calculator noting that just to solve for theta one, we need to take the inverse of sign by one multiplied by multiplied by 10 to the power of negative 9 m divided by 150 multiplied by 10 to the power of negative 6 m. So solving for theta, when we plug that into a calculator? Or should I should, I should say theta one solving for theta one plugging that into a calculator, we get 3.05 multiplied by 10 to the power of negative three radiant. OK. So now we need to recall the equation for the distance from the central maximum to the first dark fringe which is given as tangent of theta one is equal to Y one Y subscript one divided by X where X is the distance of the slit on the paper graph. So we need to rearrange this equation to solve for Y one. So when we do that, we get that, we get that Y one equals X multiplied by tangent of theta one. So at this stage, we could plug in our known variables to solve for Y one. So the value for X which is the distance of the slit on the paper graph was 2.00 m. And this is given to us in the pro multiplied by tangent of the value we just found for theta one which was 3.05 multiplied by 10 to the negative three radiance. So we get, when we plug into a calculator that Y one equals 6.10 multiplied by 10 to the negative three power meters. But we need to convert meters to millimeters. So in order to do that, we can perform some dimensional analysis maneuvers to turn meters to millimeters. So let's recall that there's 1000 millimeters in 1 m, the meters cancel out and we're left with 6.10 millimeters, which is our final answer. Awesome. So that means that our final answer has to be a 6.10 millimeters. Thank you so much for watching. Hopefully, that helped and I can't wait to see you in the next video. Bye.