A series of parallel linear water wave fronts are traveling directly toward the shore at 15.0 cm/s on an otherwise placid lake. A long concrete barrier that runs parallel to the shore at a distance of 3.20 m away has a hole in it. You count the wave crests and observe that 75.0 of them pass by each minute, and you also observe that no waves reach the shore at +-61.3 cm from the point directly opposite the hole, but waves do reach the shore everywhere within this distance. (b) At what other angles do you find no waves hitting the shore?

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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, an oscillator in a water wave tank generates plain waves at a frequency of 2.5 Hertz. The waves propagate at a speed of 4.0 centimeters per second. The waves strike at normal incidents. The gap between two long rods fixed to a wall of the tank at 20 centimeters from the rods. A diffraction pattern is produced. The first diffraction minimum is 3. centimeters from the central maximum. Find the diffraction angle theta for the 2nd and 3rd diffraction minimums. OK. So our end goal is to find the diffraction angle theta for the 2nd and 3rd diffraction minimums. Awesome. So we're given some multiple choice answers. Let's read them off to see what our final answer might be. A, is the second diffraction minimum occurs at plus or minus 10.1 degrees. And the third diffraction minimum occurs at plus or minus 16.2 degrees. B is the second diffraction minimum occurs at plus or minus 10.1 degrees. And the third diffraction minimum occurs at plus or minus 18 point 0.4 degrees. C is, the second diffraction minimum occurs at plus or minus 20.2 degrees. And the third diffraction minimum occurs at plus or minus 31.1 degrees. And finally, for d the second diffraction minimum occurs at plus or minus 20.2 degrees. And the third diffraction minimum occurs at plus or minus 36.8 degrees. OK. So first off, let us recall the equation to determine the wavelength. Yes, we need to find the wavelength in order to use the equation for destructive interference in order to determine or final goal. So we need to use all those equations to, to help us get to our, our final goal which is to solve for the diffraction angle for the 2nd and 3rd diffraction animals. OK. So don't panic, we'll get there together. But let's start off by determining the wavelength. So let's recall that the equation for the to sol for wavelength states that lambda wavelength is equal to the speed of the propagation of the waves divided by the frequency of the water waves. OK. So let's plug in our known values the sol for the wavelength. So the speed of the propagation of the water waves was 4. centimeters per second divided by the frequency which was 2.5 Hertz. So when you plug that into a calculator, you should get 1.6 centimeters. OK. Now, like I mentioned before, we need to recall and use the destructive interference equation which states that sine theta, which represents the angle or theta represents the angle equals M which represents the minimum value which is some integer multiplied by the wavelength divided by the gap length, which is represented by a. OK. Let's also make a note that the first diffraction minimum occurs at plus or minus one. OK. So now we could start solving for the diffraction angle. So in order to solve for the diffraction angle, we need to recall and use the following equation that 10 gent of theta one is equal to Y one Y subscript one divided by X where Y one is the first diffraction minimum is 3.5 centimeters from the central maximum. So that's the distance between the diffraction minimum and the central maximum. So that's the value for Y one. And then X represents the 20 centimeters from the rods. A diffraction pattern is produced. So that's the distance from the rods to the diffraction pattern produced. OK. So in order to solve for beta one, we need to take the inverse of tangent. So inverse tangent multiplied by Y one divided by X equals tangent one. OK. So now we can plug in all of our knownn values to solve for tangent. Sorry, I mean theta one. OK. So in order to solve for theta one plugging in all of our known variables. So inverse tangent multiplied by and Y one is 3.5 centimeters divided by 20 centimeters as we mentioned above. So when you plug that into a calculator, you should get 9.92 degrees. So now at this stage, we need to determine the width of the opening. So we need to find a, so rearranging the destructive interference equation to solve for a using some algebra, we get that the width of the opening is equal to the wavelength divided by sine beta one. So when we plug in our known variables to solve for a, we know that the wavelength is 1.6 centimeters as we determined above divided by sign. And then this is the theta value we just found for theta one which was 9. degrees. So when you plug that into a calculator, you should get 9. centimeters. OK. Now, we must consider the ratio of the wavelength to the width. And that is the wavelength divided by the width opening which is equal to sign the one data subscript one is equal to sign 9. degrees, which is equal to 0.172. OK. So we can use this value that we got from the ratio to help us find the 2nd and 3rd diffraction minimum. So let's do that. OK. So let's start out by finding the second diffraction minimum which would. So the value for that would be M equals plus or minus two for the second diffraction minimum. So that states that sign data subscript to is equal to plus or minus two wave length divided by the width of the opening which is equal to plus or minus two, wait plus or minus two multiplied by the ratio value we just determined which was 0.172. OK. OK. So the soul for it, we need to take inverse sign multiplied by two, multiplied by a ratio value 0.1. OK. 0.172. And when we plug that into a calculator, we get plus or minus 20. degrees per two. OK. So using the similar methodology to solve for the third diffraction minimum, which would be M equals plus or minus three, let's note that sign theta subscript three, theta three is equal to plus or minus three wavelength divided by the width of the opening is equal two plus or minus three, multiplied by the ratio 0.172. OK. So like above inverse sign multiplied by plus or minus three, multiplied by the ratio value 0.172. When you plug that into a calculator, you should get that data three equals plus or minus 31.1 degrees. Awesome. So these are our final answers. Fantastic. So our final answer has to be see the second diffraction minimum occurs at plus or minus 20.2 degrees and the third diffraction minimum occurs at plus or minus 31.1 degrees. Thank you so much for watching. Hopefully that helped and see you next time. Bye.

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Key Concepts

Wave Interference

Diffraction

Wave Speed and Frequency