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. (b) What is the intensity at a point on the screen midway between the center of the central maximum and 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 helium neon laser beam of wavelength 632.8 nanometers is directed to a vertical slit of width 315 micrometers. The diffraction pattern composed of bright and dark fringes is viewed on a board at 2.00 m from the slit. The bright central fringe occurs at theta equals zero degrees. A photo sensor moves across the viewing board measuring the intensity of the diffracted light as a function of the horizontal distance, an intensity of 4.25 multiplied by 10 to the power of negative six watts per meter. Meter squared is measured at theta equals zero degrees. The detector is moved to point P point P is located three quarters of the distance between the bright central fringe and the first dark fringe beginning from the bright central fringe determine the measured intensity at point P. OK. So our end goal is to determine the measured intensity at point P. OK. So we're given some multiple choice answers. They're all in the same units of watts per meter square. So let's read them off to see what our final answer might be. A is 1.62 multiplied by 10 to the power of negative seven. B is 3.83 multiplied by 10 to the power of negative seven C is 0.90 multiplied by 10 to the power of negative six. And D is 2.13 multiplied by 10 to the power of negative six. OK. So first off, let us recall and use the equation for destructive interference which states that sine theta equals and some minimum value which is an integer value multiplied by the wavelength divided by the slit width. So note that the first dark fringe occurs at N equals plus or minus one. So this is for the dark case. So for the dark fringe, OK. So considering N equals one, we can rewrite our destructive interference equation as sin theta subscript one equals. So when we plug in numerical values N equals one multiplied by the wavelength which is given to us in nanometers, but we need to convert it to meters. So 300 0 sorry, 632 0.8. And all we have to do to, to convert nanometers to meters is we multiply it by 10 to the power of negative nine divided by the slit width which is given to us is 350 micrometers. So we need to convert that to meters. So to do that, we take 350 we multiply it by 10 to the power of negative six. OK. So when we plug that into a calculator, we should get that theta one equals 1.81 multiplied by to the power of negative three radiant. OK. So we need to note and recall that we can determine the distance between the center of the bright central fringe. And the first dark fringe. With the following equation, Y subscript one equals X multiplied by tangent beta subscript one where X is the distance from the slit. OK. So let's plug in our known variables to solve for A Y one. So X is given to us in the prom as 2.00 m multiplied by tangent. And we just found the, you just found theta one which theta one is 1. multiplied by 10 to the power of negative three radiance. So when we plug into a calculator, we should find that Y one is equal to 3.62 multiplied by 10 to the power of negative m. So to find the measured intensity at point P, you can think that if point P is located three quarters, so 3/4 of the distance between the bright central fringe and the first dark fringe beginning from the bright central fringe. It implies that we can write the following. You can say that Y equals 3/4 multiplied by 3.6 multiplied by 10 to the power of negative three M equals 0.0027. OK. And we use our uh the Y one value that we just determined is this value OK. So the diffraction angle at point P must be. So let's solve for tangent theta equals Y divided by X. So when we plug in our known values, so Y is equal to 0.0027 m can't bring the meters divided by 2. m is equal to theta which is equal to 1. five multiplied by 10 to the power of negative three radiant. OK. So now we must use the phase difference and the intensity equations. OK. So beer equals so this is the phase difference equation. So beta equals so beta, the phase difference is equal to two multiplied by pi divided by the wavelength multiplied by the slit width multiplied by sine data. So when we plug in our known values, the two pi divided by 600 sorry 632. multiplied by 10 to the power of negative 9 m is our wavelength multiplied by the slit width which was 350 multiplied by 10 to the power of negative m multiplied by sign of the, the value we just found which is 1. 575 multiplied by 10 to the negative three radiance. OK. So when we plug it into a calculator, we should determine that the phase difference beta is equal to 4.717600351. OK. So now we can solve for the intensity. So intensity is equal to I excuse me, I zero multiplied by sign phase, the sign of the phase difference divided by two divided by the phase difference divided by two oh squared, all squared. So now we need to plug in all of our known values to solve for the intensity. So let's do that. Let's give us a little space first. So I equals the. So I zero is given to us in the problem as 4.25 multiplied by 10 to the power of negative six watts per meter square multiplied by sign of which is the value we just found for the phase difference. So it's 4. 51 divided by two, divided by the phase difference. divided by two squared. So that means when you plug this into a calculator, the intensity I should equal 3.83 multiplied by 10 to the power of negative seven watts per meter squared, which is our final answer. Ray we did it. So looking at our multiple choice answers. Our final answer has to be a letter B 3.83 multiplied by 10 to the power of negative seven watts per meter squared. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.