Monochromatic light of wavelength 580 nm passes through a single slit and the diffraction pattern is observed on a screen. Both the source and screen are far enough from the slit for Fraunhofer diffraction to apply. (a) If the first diffraction minima are at +-90.0°, so the central maximum completely fills the screen, what is the width of the slit? (b) For the width of the slit as calculated in part (a), what is the ratio of the intensity at u = 45.0° to the intensity at u = 0?

Video transcript

Hello, fellow physicists today, we're to 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. Two half razor blades are placed side by side with a narrow space between them to form a single slit of with a a monochromatic beam of wavelength 0.520 micrometers passes through the slit on a board placed very far from the blades. A from hofer diffraction pattern is observed. The first dark fringe is visible at theta equals plus or minus pi divided by two radiant. I determine the width of the formed slit and I I, the ratio of the intensity observed at theta equals pi divided by six to the intensity of the central bright fringe I subscript zero. OK. So we're given some multiple choice answers for I and I I, all the units for iron and micrometers and all the answers for I I are I divided by I subscript zero equals blank. So let's read off our multiple choice answers to see what our final answer set might be. A is 0. 0.260 and 0.20. B is 0.390 and 0.62. C is 0.520 and 0.41. And D is 0.780 and 0.62. OK. So first off, let us recall the equation for destructive interference which states that sign the equals N sum minimum integer value multiplied by the wavelength divided by the width of the slit, which is represented by a. Let's also note that the first dark fringe occurs at N equals plus one. And that the first visible dark fringe is at theta equals plus or minus pi divided by two radiant. And let's note that this is for the dark case dark and dark. OK. So since theta equals pi divided by two radians, we can solve for the for the width of the slit by plugging in pi divided by two into theta. So let's do that. So sign two divided sorry pi divided by two radiant. So sine of pi divided by two radiant equals N which equals one multiplied by the wavelength divided by the width of the slit. Let's note that A equals is equal to the wavelength which is equal to 0.5 20 micrometers. OK. So for I, the slit with has to be 0.520 micrometers. OK. So that is our first answer hooray. OK. So now let's start to solve for I I, OK. So to solve for I I, the intensity of I at any angle theta where theta is the angle of the line from the center of the slit to the position on the screen. In terms of I subscript zero, where I subscript zero is the intensity in the forward positive direction where theta equals zero can be written as follows. So the intensity is equal to I subscript zero multiplied by sign hi, multiplied by A which is the slit with multiplied by sign data divided by lambda, divided by pi multiplied by a multiplied by sine theta divided by lambda all squared. OK. So let's plug in our known variables using theta equals pi divided by six. And also note that A is equal to LAMBDA. So we can write that I divided by I subscript zero is equal to sign of hi, multiplied by Lambda multiplied by sine high divided by six, divided by the wavelength divided by pi multiplied by lambda multiplied by sign high divided by six divided by lambda which represents the wavelength. So when we plug that into a calculator, we should get 0.41 which is our answer for II I oh actually I I, OK. So the ratio of the intensity observed at theta equals pi divided by six to the intensity of the central bright fringes I divided by I subscript zero is 0.41. OK. So let's look at our multiple choice answers. So that means that our correct answer has to be I is a equals 0.520 micrometers and I I is I divided by I subscript zero equals 0.41. Awesome. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.