Diffraction occurs for all types of waves, including sound waves. High-frequency sound from a distant source with wavelength 9.00 cm passes through a slit 12.0 cm wide. A microphone is placed 8.00 m directly in front of the center of the slit, corresponding to point O in Fig. 36.5a . The microphone is then moved in a direction perpendicular to the line from the center of the slit to point O. At what distances from O will the intensity detected by the microphone be zero?

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 teacher is playing a 5.0 centimeter wavelength constant tone sound through a speaker. The sound wave passes through a 10 centimeter hole in the wall to the next room where it is intercepted by a sound level meter placed at a distance of 3.0 m from the wall. The sound level meter is moved along a perpendicular line from I the point that is aligned with the center of the hole towards the ceiling, determine the distances from I at which the wave intensity is zero. OK. So we're given some multiple choice answers. They're all in the same units of meters. Let's read them off to see what our final answer might be. A is 0.87 B is 1.10 C is 1.70 and D is 2.62. OK. So to begin to help us better visualize this problem. OK. Let's note really fast that the distance of 3.0 m from the wall is shown here and this is the ceiling where the sound level meters are located. Awesome. So now that we have a visualization for this problem, let us start to solve for our final answer, which our goal is to determine the distances from I at which the wave intensity is zero. So in order to do that, we need to first recall and use the destructive interference equation which states that sign Cedar is equal to the minimum for the order, which is some integer value, which usually it says like if it's first order, second order, third order for fringes, et cetera. But in this problem, it doesn't state one specifically. So let's just assume in this case that the order is one, but we'll talk more about that in a second. So N represents the minimum value multiplied by LAMBDA, which represents the wave length divided by the whole width or the whole size I should say in this case, which is represented by a. OK. So we need to rearrange the destructive interference equation to solve for theta. So when we do that, we get theta is equal to inverse sign multiplied by the integer or the minimum, the minimum value which is some integer for the order multiplied by the wavelength divided by the whole size. OK. So as I was saying before, let's make a quick little note that we're gonna assume that it's first order. So we're gonna state that M equals one. So taking that into consideration and then plugging in all of our known variables to solve for theta, we get, we, so to determine the, let's plug in our known variable. So theta equals the inverse sign multiplied by the wavelength which the wavelength in this case is five centimeters divided by the whole size which is centimeters. OK. So the equals inverse sine multiplied by five centimeters divided by centimeters. So when we plug that into a calculator, we should get 30 degrees. So like I said before, we need to note that the first minimum for M is plus one because it's in the direction towards the ceiling of a room which is in the positive Y direction. OK? To be clear. So it's plus one because it's moving towards the ceiling in the positive Y direction. Awesome. So now we need to recall the equation to determine the distance from the central maximum I which is the intensity in this case to the first destructive interference or minimum intensity. OK. So the equation to find that states that Y subscript one equals X multiplied by tangent data where X in this case is the interception interception distance, which is 3.0 m. So let's plug in our known values to solve for Y subscript one, Y one. So X as we said is 3.0 m multiplied by tangent which we learn to use the angle value we just found for theta which was 30 degrees. And when we plug that into a calculator, we should get 1. 1.73 m ray, we found our final answer. OK? So the answer is positive because plus 1. m, the because this is the case because the sound level meter is moved towards the ceiling in the upwards positive direction. So because the sound meter is moved towards the ceiling and it's moving in the upward positive direction. So that's why the answer is positive. Fantastic. So that means our final answer has to be C 1.73 m. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.