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$\bullet$ Radio interference. Two radio antennas $A$ and $B$ radiate inphase. Antenna $B$ is 120 $\mathrm{m}$ to the right of antenna $A .$ Considerpoint $Q$ along the extension of the line connecting the anten-nas, a horizontal distance of 40 $\mathrm{m}$ to the right of antenna $B$ .The frequency, and hence the wavelength, of the emittedwaves can be varied. (a) What is the longest wavelength forwhich there will be destructive interference at point $Q ?$ (b) Whatis the longest wavelength for which there will be constructiveinterference at point $Q ?$

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a) 240$m$b) 120$m$

Physics 102 Electricity and Magnetism

Physics 103

Chapter 26

Interference and Diffraction

Electromagnetic Waves

Reflection and Refraction of Light

Cornell University

Rutgers, The State University of New Jersey

Simon Fraser University

Lectures

02:30

In optics, ray optics is a geometric optics method that uses ray tracing to model the propagation of light through an optical system. As in all geometric optics methods, the ray optics model assumes that light travels in straight lines and that the index of refraction of the optical material remains constant throughout the system.

10:00

In optics, reflection is the change in direction of a wavefront at an interface between two different media so that the wavefront returns into the medium from which it originated. Common examples include the reflection of light, sound and water waves. The law of reflection says that for specular reflection the angle at which the wave is incident on the surface equals the angle at which it is reflected. Reflection may also be referred to as "mirror image" or "specular reflection". Refraction is the change in direction of a wave due to a change in its speed. The refractive index of a material is a measure of its ability to change the direction of a wave. A material with a higher refractive index will change the direction of a wave to a greater degree than a material with a lower refractive index. When a wave crosses the boundary between two materials with different refractive indices, part of the wave is refracted; that is, it changes direction. The ratio of the speeds of propagation of the two waves determines the angle of refraction, which is the angle between the direction of the incident and the refractive rays.

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Two radio antennas A and B…

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Two radio antennas $A$ and…

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Two radio antennas radiati…

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Antenna $B$ is $40.0 \math…

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Two radio antennas radiat…

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35.44. Two radio antennas …

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A radio transmitting stati…

All right. So in this problem we have two speakers a and be separated by 120 meters. And some observation point Q, which is 40 meters to the right of speaker be speaker and speaker be are admitting their sound in phase, which means our waves are leaving in the same pattern. So if it leaves from Speaker A like that, it leaves from speaker. Be like that. So all of our peaks and our Traves lineup is the best way to think about that. Eso for party. We want to know when is destructive interference going to happen? What is the largest wavelength that we're going to get? Destructive interferes. So since we're looking at destructive interference, we're gonna start here with our two minus are one is equal to end plus 1/2 times Linda, we want to solve for Lambda. So we're gonna divide both sides. I m plus 1/2. So Lamda is equal to our two minus r one. Divided by Mm plus. Uh huh. All right. So are too. I like to use the one that's furthest away. Ah, it actually doesn't matter which one. You put your just going to get either a positive value or negative value for Lambda. So I like to put the biggest one first. So since a is furthest away from our observation point, that's going to be, ah, 120 meters plus 40 meters for a total of 160 meters and our one which is between B and our observation point. It's 40 meters and n plus 1/2. So what should we use for N if we start with zero, we get 120 meters divided by half. We start with one. We get 100 and 20 meters divided by one and 1/2. We want the largest value. So the largest value that we get with our EMS is gonna be one Emmy this equal to zero. And if you don't trust me, you completely in different values of them and test it out. So if we put this into our handed in the calculator went or just do some asked, 1 60 minus 40 is 120 meters divided by 1/2. So it's times two that's gonna be a wavelength of 200 and 40 meters. All right, well, likewise. We want to find the largest wavelength where there's constructive interference. Who are you? So are two mice are one is equal to em. Lambda we saw for Lambda Make it are too minus are one divided by M. Now, in this case, we can't use M equals zero because that's gonna, you know, explode on infinite wavelength. The wavelength is does not exist, etcetera, etcetera So already is M equals one. It's only plug in. Our values are too remember was 100 and 60 meters for one was 40 meters. Do right by one That's going to give us 120 meters for our wavelengths. And again, that's the largest. We contest that out. Yes, we divided by two, which is our next value of M. You cut this in half and get 60 meters. So this one

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