Laser light of wavelength 500.0 nm illuminates two identical slits, producing an interference pattern on a screen 90.0 cm from the slits. The bright bands are 1.00 cm apart, and the third bright bands on either side of the central maximum are missing in the pattern. Find the width and the separation of the two slits.

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Hello, fellow physicists say we're gonna solve the following practice problem together. So first off, let's read the problem and highlight all the key piece of information that we need to use in order to solve this problem, consider a pair of identical slits through which parallel rays of monochromatic light pass. The normally incident light has a wavelength of 540 nanometers. As light waves can interfere constructively and destructively an observing screen 85 centimeters away displays bright and dark fringes. A closer inspection of the patterns reveal the second bright bands of the central maximum to be missing. How wide should the slit be and what should be their separation distance if the adjacent bright bands are 1.3 centimeters away? OK. So that is our end goal. Our end goal is to find two different answers. We're trying to find an answer for how wide the slit should be and what the separation should be or the separation distance I should say should be if the adjacent bright bands are 1.3 centimeters away. Awesome. So our multiple choice answers reflect that it gives us an answer for the slit width or the width of the slit or how wide the slit should be. And it also has an answer for the separation distance D. So the slit, how wide the slit is, is represented by A and all the units for both A and D are in micrometers. So let's read them off to see what our final answer might be. A is 28.3 and 55.9. B is 10 and 25.1 C is 17.7 and 35.4 and D is 19.4 and 45.9. OK. So we're given a diagram to help us better visualize this problem. It shows us where the second bright bands of the central maximum are missing and it shows where the missing ones are with this little bracket. Here. It also shows the separation distance D between the two slits with A that are wide to amount A. So this is the slit width, how wide it is at A for here and here. And it also shows the distance it is of how far it is away from the screen, which is 85 centimeters. Awesome. So first off, just so we can start solving this problem. Let us recall and use the equation for single slit diffraction and that will help us to determine the diffraction min. OK. So let's call it equation one. So the single slit diffraction equation states that the slit with a multiplied by sine of theta. So a multiplied by sin of theta is equal to N minimum and min, which is where N is some integer value. And it represents like the order like if it's first order, second order et cetera multiplied by the wavelength. OK. So we can rearrange equation one to solve for A which is one of our N goal answers. So when we do that using a little bit of algebra, we get that A is equal to nm multiplied by wavelength divided by sine theta. Awesome. So now we need to recall and use the equation for the double slit interference where there is two slit interference maxima. So let's call that equation three shall we? And let's call the equation that we just found equation two. So equation three states that the separation distance D multiplied by sine of theta is equal to N maximum multiplied by the wavelength. So like before we need to rearrange the sulfur D. So let's do that. So D using a little bit of algebra D is equal to N max multiplied by the wavelength divided by sine of the fantastic. So now we need to divide equations three and one. So equation three states that D sine theta divided by equation one which states a sine theta is equal to in max multiplied by the wavelength divided by N minimum divide multiplied by wavelength. So N max N subscript max multiplied by the wavelength divided by N min multiplied by the wavelength is the same as saying the separation distance divided by the slit width is equal to N max divided by nm, which is also the same as saying D divided by A is equal to two divided by one because the maximum value is since we're doing with the second order, second bright bands of second order, so that means that N max will be two and, and men, since we're dealing with a single slit, we're dealing with bright and dark. So it's a first order fringe on that part, it's a first order band I should say. So that's where the one comes from. So we can rearrange, let's box it in that we can rearrange this equation to solve for D. So let's do that. So D is equal to two multiplied by a Awesome. Now, we need to use the equation to determine the distance from the central maximum to the second minimum fringe. So the equation that we need to use for that is why subscript two Y two is equal to two multiplied by delta Y. Let's also note, make a note here that tangent of theta two is equal to two multiplied by delta Y divided by R. And we can get this by looking at our diagram that's given to us. And let's remember that tangent is opposite over adjacent. OK. So when we rearrange this to isolate and solve for tangent for the tangent of theta So we reran rearrange to solve for just theta. We'll get that theta two is equal to inverse tangent of two multiplied by delta Y divided by capital R. OK. Let's also note that delta Y is given to us in the prom itself and it's delta Y represents the adjacent bright bands that are 1.3 centimeters away. So 1.3 centimeters, but we need to convert centimeters to meters. So all we have to do is just take 1.3 and multiply it by 10 to the power of negative two. Let's also note that we're given capital R which capital R is 85 centimeters. And this is the how far away it is the distance from, you know what we are observing on the screen. So it's the distance away from where the source of light is being projected. And we also need to convert that to meters. So like before what we do to take 85 multiplied by 10 to the power of negative 2 m. And we're also given the wavelength which the wavelength is 540 nanometers. So we need to convert nanometers to meters. So all we do is take 540 multiply it by 10 to the power of negative nine to get it to meters. OK. So now we just need to plug in all of our known variables and solve for theta two. So let's do that. So theta two is equal to oh you can't forget that we have inverse tangent here. So theta two is equal to tangent or inverse tangent. I should say of two multiplied by 1.3 multiplied by 10 to the power of negative 2 m divided by capital R which was 85 multiplied by to the power of negative 2 m. So when we plug that into a calculator, we should get 1.75 degrees. OK. So now we can use the equation two to plug in the value we just found for theta two. And let's also note that N minimum is one. So let's make a note of that. So N minimum is one. So let's remember that equation two stated that A is equal to wavelength divided by sine theta. So let's plug in all of our known variables to solve for A. So A is equal to the wavelength which was 540 multiplied by 10 to the power of negative 9 m divided by sign of 1.75 which is the theta two value, we just found 1.75 degrees. So when we plug it into a calculator, the value for A you will get is 1.768 multiplied by 10 to the power of negative 5 m. But we need to convert it to micrometers because all of our answers in the multiple choice are in micrometers. So all we have to do is just multiply it by 10 to the power of six. When we do that, we'll get 17.7 micrometers. And that is our final answer or a. OK. So now let's solve for D shall we? OK. So to solve for D, we just need to plug in all of our known variables into equation four. So let's do that. So let's also make a quick note that N max is equal to two in that equation four states that D is equal to N max divided by sign of beta or sign data, I should say. So let's plug in our known variables and software D so D is equal to N max. Oh We forgot N max multiplied by wavelength that's important. So N max is two multiplied by the wavelength which was 540 multiplied by 10 to the power of negative nine meters divided by sine of theta two, which was 1.75 degrees. So when we plug that into a calculator, we should get D is equal to 3.536 multiplied by 10 to the power of negative 5 m. But we need to convert to micrometers. So multiplied by 10 to the power of six. So when we do that, we should get 35.4 micrometers. And that is our final answer for D ray. We did it. So let's go back up to the top and look at our multiple choice answers. To see what our final answer pair is. So that means the correct answer has to be C A equals 17.7 micrometers and D equals 35.4 micrometers. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

Verified Solution

Key Concepts

Interference of Light

Young's Double-Slit Experiment

Wavelength and Slit Separation