Coherent light of frequency 6.32 * 1014 Hz passes through two thin slits and falls on a screen 85.0 cm away. You observe that the third bright fringe occurs at ±3.11 cm on either side of the central bright fringe. (a) How far apart are the two slits? (b) At what distance from the central bright fringe will the third dark fringe occur?

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 beam of light frequency 5.41 multiplied by 10 to the power of 14 Hertz from a coherent source illuminates a distant panel. When it passes through two parallel placed narrow slits, the observation panel is positioned 75 centimeters from the source. Determine the slit separation if the second bright fringe occurs at plus or minus 2. centimeters on either side of the central bright fringe. Also find how far the second dark fringe from the central bright fringe is. OK. So we have two goals here. We need to determine the slit separation the distance if the second bright fringe occurs at plus or minus 2.42 centimeters. And then our second goal is to find how far, which is another distance value, how far the second dark fringe is from the central bright fringe. OK. So we're given mo some multiple choice answers here and we're given answers for representing D for the slit separation. And then Y subscript two for how far the second dark fringe is from the, from the central bright fringes. And they're all in the same units of meters. So let's read them off to see what our final answer pair might be. A is 3.44 multiplied by 10 to the power of negative five and 1.81 multiplied by 10 to the power of negative two B is 1.4 multiplied by 10 to the power of negative five and 3. multiplied by 10 to the power of negative two C is 2.1 multiplied by 10 to the power of negative six and 2.7 multiplied by 10 to the power of negative four. And D is 1.6 multiplied by 10 to the power of negative six and 3.4 multiplied by 10 to the power of negative four. OK. So to solve for D, so let's solve for the slit separation. First, let us recall and use the equation to find the wavelength which states that the velocity is equal to the frequency multiplied by the wavelength. So we can rewrite this equation to isolate and solve for wavelength. So wavelength is equal to the velocity divided by the frequency. OK. So now we need to recall the equation for the nth bright fringe from the central maxima which states that Y subscript N is equal to N where, where N is some nth is the nth value. So the N is some integer value for the order number. So first order, second order, third order et cetera multiplied by the wavelength multiplied by capital D were capital D is the distance of the, that the panel is positioned from the source. OK. Divided by little D or little D is the value that we're trying to solve for the slit separation. Awesome. OK. So when N equals two, since this is for the second order bright fringe, it will be, it will state that Y two. So Y subscript two is equal to two multiplied by the wavelength multiplied by capital D divided by lower case D. So now we can plug in the value for lambda. So let's rewrite why subscript to plugging in the value for lambda? So when we do that, we'll get, we'll get Y two is equal to two multiplied by the velocity divided by the frequency multiplied by capital D all divided by D. Now we need to rearrange this equation to solve for D. So let's do that. So when we rearrange this equation to sulfur D, we get that D is equal to V divided by F. So the velocity divided by the frequency multiplied by two multiplied by capital D divided by Y subscript two Y two. OK. Now we can plug in our known variables. OK. So let's do this. So let's plug it over on all of our known variables to solve for D. So we're gonna use for the velocity, we're gonna use the speed of light since the speed of light is equal to the velocity. So we know that the speed of light is 3.0 multiplied by 10 to the power of 8 m per second divided by the frequency which is given to us in the problem as 5.41 multiplied by 10 to the power of 14 art multiplied by two, multiplied by capital D which is 75 centimeters. But we need to convert centimeters to meters. So all we do is take 75 multiplied by 10 to the power of negative two meters divided by, we're gonna use the positive value of 2.42. And then now you need to convert 2.42 centimeters, 2 m. So always you take 2.42 multiplied by 10 to the power of negative two meters. Since our Y two values given to us in the problem is it says that if the second bright fringe occurs at 2.42 so that is our Y two value, they give it to us. OK? Just a minute just in case. That was confusing. OK? So now when we plug in all of this into a calculator, we get that D should be equal to 3. multiplied by 10 to the power of negative five meters. And that is our first answer. So D equals 3.44 multiplied by 10 to the power of negative 5 m. Now, we need to solve for Y two. So to solve for Y two, let's make the following notes really quick. So we need to note that the distance of the nth bright fringe from the central Maxima is YN equals N multiplied by LAMBDA multiplied by capital D divided by D and the distance of the NTH dark fringe. OK. So I'm gonna make a note this is bright and then the one for the dark, since we're talking about the dark fringe, so the nth dark fringe from the central maxima can be written as YNY subscript N is equal to and minus one plus one divided by two multiplied by LAMBDA multiplied by capital D all divided by lower case D. OK. So now we need to note that N equals two for the bright fringe since it's a second order bright fringe, so we can rewrite the dark fringe equation as YN equals two N minus one divided by two multiplied by LAMBDA multiplied by capital D all divided by lowercase D. OK. But we know with the value for lambda, the wavelength, so we need to rewrite this equation to plug in the value for the wavelength. So let's do that. So why subscript N is equal to two N minus one divided by two multiplied by velocity divided by frequency multiplied by capital D all divided by lower case D OK. So now we can plug in all of our known variables to solve or Y two. So N, so let's remember that N equals two. So Y two equals. So now we need to plug in all of our known values that two multiplied by two minus one, divide it by two multiplied by the velocity which was the speed of light 3.0 multiplied by 10 to the power of 8 m per second divided by the frequency which is 5.41 multiplied by 10. The power of 14 Hertz. OK. Multiplied by capital D which was centimeters. But we need to convert that to meters. So it's 75 multiplied by 10. The power of negative 2 m divided by the value or little D which we determine that to be 3.44 multiplied by 10 to the power of negative five meters. OK? So we should get when we plug that into a calculator that Y two is equal to 1.81 multiplied by 10. OK. So 1.81 multiplied by 10 is the power of negative 2 m which we wrote this in scientific notation. You probably saw something like this when you type it into your calculator, 0.0181 m. But we can write it in scientific notation because our multiple choice answers are written in scientific notation. OK? So when we go back to look at our multiple choice answers. That means that the correct answer has to be the letter ad equals 3.44 multiplied by 10 to the power of negative 5 m. And Y subscript two Y two is equal to 1.81 multiplied by 10 to the power of negative 2 m. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.