Two 0.010-mm-wide slits are 0.030 mm apart (center to center)....

Question

Two 0.010-mm-wide slits are 0.030 mm apart (center to center). Determine (a) the spacing between interference fringes for 520-nm light on a screen 1.0 m away and (b) the distance between the two diffraction minima on either side of the central maximum of the envelope.

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem in a physics laboratory. A group of students is conducting a double slit experiment to observe the interference and diffraction patterns produced by light. They set up two narrow slits each measuring 0.015 millimeters in width separated by a distance of 0.045 millimeters. The students use a light source emitting light with a wavelength of 600 nanometers and project the light onto a screen located 2.0 m away from the slits. A calculate the spacing between the interference fringes observed on the screen due to the light passing through the two slits B determine the distance between the first diffraction minima on either side of the central maximum in the resulting diffraction pattern. Awesome. So it appears for this particular problem we're asked to solve for two separate answers. For part A our first answer we're asked to figure out the spacing between the interference fringes that are observed on the screen due to the light passing through these two slits. That's our first answer. So we're trying to figure out the spacing distance between these interference fringes. And then for b our second answer, we're asked to figure out the distance between the first diffraction minima on either side of the central maximum that results in a diffraction pattern. Awesome. So now that we have an idea of what we're solving for, let's read off our multiple choice answers to see what our final answer pair might be. Noting that our first answers for the spacing between the interference fringes for A and then for B, it's the distance between the first diffraction minima on either side of the central maximum. And I'm not going to read that every single time. So we're good. That's for B. So A is the space in between the interference fringes and B is the distance between the first diffraction minima on either side of the central maximum. OK. So that will mean reading them off in order to see what our final answer pair might be. A is 27 millimeters and 120 millimeters B is 33 millimeters and 160 millimeters C is 33 millimeters and 120 millimeters. And finally D is 27 millimeters and 160 millimeters. OK. Let's start solving this problem, shall we? So first off, let us write down all of our known variables all the variables that are given to us by the problem itself. So we're told that the wavelength of light, which we're gonna call LAMBDA and the wavelength of light lambda is equal to 600 nanometers, which is also equal to 600 multiplied by 10 to the power of negative 9 m. So for this particular prompt, I am gonna perform all the conversions for you in an effort to save some time solving this problem. But in order to do any of the conversions that I'm gonna do for you, you can use dimensional analysis to solve for them on your own or you just quickly look them up. Awesome. We'll be right along here. We also know the distance between the slits which we're gonna call D and D is equal to 0.045 millimeters which is also equal to 0.045 multiplied by 10 to the power of negative 3 m. Awesome. And once again, that's a value for the distance between the slits and then the distance to the screen which we're gonna call L or the length to the screen, if that helps you remember is equal to 2.0 m. And then finally, we know the width of each slit and we're gonna call the width of the slit A and A is equal to 0.015 millimeters which is also equal to 0.015 multiplied by 10 to the power of negative 3 m and once again, A denotes the width of the slits. Awesome. So now we could start solving for part A all right. OK. So we need to calculate the spacing between the interference fringes. So in order to do that, we need to recall and use the formula for fringe spacing in a double slit experiment. And the fringe spacing between or this fringe spacing between slits and a double slit experiment. The equation states that delta Y which is the change in the distance Y delta denotes change. So delta Y is equal to lambda multiplied by L all divided by D Fantastic. So now all we need to do is plug in all of our known variables. And so for the numerical value of delta Y, and that's gonna be our first answer for part A Awesome. So when we plug in our known variables, we know that our wavelength is 600 multiplied by 10 to the power of negative 9 m. Then we need to multiply it by L which is 2.0 m. Then we need to divide everything by our distance D which is 0.045 multiplied by 10 to the power of negative 3 m. Fantastic. So when we plug that into our calculator, we're gonna get our answer that is approximately equal to 0.0267 meters when we round to four decimal places, which we have a bit of a problem. All of our multiple choice sensors are in units of millimeters. So we need to convert our final answer for delta Y into units of millimeters. And when we do that, and we round to the nearest whole number, we will find that delta Y is equal to 27 millimeters. And that is our final answer, Henry, we did it. So now we could start solving for part B array, we are moving right along here. So to solve for part B, we need to determine the distance between the first diffraction minima. So in order to do that, we need to recall and use the formula to help us solve for the position of the first diffraction minima, which is why min is equal to M multiplied by lambda multiplied by L all divided by A. So now all we need to do is plug in all of our known variables into our equation and solver the numerical value of Y min, which is our final answer are almost our final answer. We're gonna have to do one step which we need to solve for ym first in order to help us solve for our final answer for part B. OK. So let's not get too ahead of ourselves. So let's plug in all of our known variables. So we know that M and M in this case denotes the order which is some integer value. And since we're dealing with a first order diffraction or sorry, the first order minima that means M equals one. So we need to plug in one for M and then LAMBDA, of course, our lambda, our wavelength is gonna be equal to 600 multiplied by 10 to the power of negative 9 m. And then we need to multiply it by L which is 2.0 m. And then we need to divide everything by our value for A which is the width of the slits which is 0.015 multiplied by 10 to the power of negative 3 m. Fantastic. So when we plug that into our calculator, we will find that YM is approximately equal to 0.08 m, which is approximately equal to when we round to the nearest whole number. And we convert meters to millimeter units, we will get 80 millimeters and that's it. That is our final answer for Wyman Hooray. We did it. So we need to do one last step here as we should recall and note the distance between the two diffraction minima on either side of the central maximum, which we'll call this capital D for the distance is equal to two multiplied by Y min, which is equal to two multiplied by 80 millimeters. So we need to plug in our value for YM which is 80 millimeters. And when we plug that into our calculator, we're gonna get 160 millimeters. And that is our final answer for part B, hooray, we did it. So part A is equal to 27 millimeters and part B is equal to 160 millimeters. Hooray, we did it. So let's go back to the top and look at our multiple choice answers to see which answer corresponds with the answers that we just found together. And it appears the final answer has to be the multiple choice answer letter D which states once again that a the spacing between the interference fringes is 27 millimeters. And for part B, the distance between the first diffraction minima on either side of the central maximum is 160 millimeters. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

Two 0.010-mm-wide slits are 0.030 mm apart (center to center). Determine (a) the spacing between interference fringes for 520-nm light on a screen 1.0 m away and (b) the distance between the two diffraction minima on either side of the central maximum of the envelope.

Key Concepts

Double-Slit Interference

Diffraction

Interference and Diffraction Minima

Watch next