Light of wavelength 585 nm falls on a slit 0.0666 mm wide. (a) On a very large and distant screen, how many totally dark fringes (indicating complete cancellation) will there be, including both sides of the central bright spot? Solve this problem without calculating all the angles! (Hint: What is the largest that sin u can be? What does this tell you is the largest that m can be?) (b) At what angle will the dark fringe that is most distant from the central bright 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, light rays are falling on a narrow slit. I without finding all the angles determine the total number of dark fringes formed on a screen total for both sides of the central bright spot, consider slit width as 0. millimeters and a wavelength of light as 570 nanometers. I I calculate the angle for the dark fringe that will occur halfway from the central maximum. OK. So we're given some multiple choice answers and we're given answers for part I and part I I. So let's read them off to see what they on. Our final answer pair might be A is 101 fringes and degrees B is 101 fringes and 60 degrees C is 202 fringes and 30 degrees and D is 202 fringes and 60 degrees. OK. So to solve for, I, let us start by recalling the equation to determine the number of dark fringes for the values of theta that will satisfy the equation. So let's call it equation one shall we? So equation one states that signed beta is equal to N where N can be plus or minus one plus or minus two plus or minus three, et cetera where N is the nth order, which is an integer value multiplied by the wavelength divided by A where A is the slit width. So now we must note that the largest value of sine theta that we can have is 1.00. So to find the largest value that N can be, we can use the equation sine theta is equal to one and we can plug this back into equation one. So when we plug this back into a equation one and rearrange to solve four N, we get that N is equal to a divided by wavelength which we can now plug in our known variables to sol for N. So we know that a, the slit width is 0. 578 millimeters, but we need to convert millimeters to meters. So I can just take it and multiply it by 10 to the power of negative 3 m. And then we need to divide it by the wavelength which was multiplied by 10 to the power of negative 9 m. So that will give us N equals 101.40. So thus the largest value of N must be 101.40. So N will have the values of, so let's note that N can be plus or minus one plus or minus two, two, all the way to plus or minus 101. So that means for the total, for both sides, since it's 101 on one side plus 101 on the other side that gives us 202 dark fringes. OK. And that is our answer for N. So let's start solving for part I I so to solve for I I, in order to find the angle for the dark range that will occur halfway from the central maximum, we can write the following mathematical statement, we can say that sine theta equals one half. So that means that data is equal to inverse sign of one half, which is equal to 30 degrees. And that is our final answer, right? I should say boxing both of our answers in green. So 30 degrees is our answer for I I and 202 dark fringes is our answer for I. So let's go look at our multiple choice answers to see what our final answer pair is. The correct answer is C 202 fringes. And for I and for I I is 30 degrees. Thank you so much for watching. Hopefully, that helped and I can't wait to see you in the next video. Bye.