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$\bullet$ Coherent light of frequency $6.32 \times 10^{14}$ Hz passes throughtwo thin slits and falls on a screen 85.0 $\mathrm{cm}$ away. You observethat the third bright fringe occurs at $\pm 3.11 \mathrm{cm}$ on either side ofthe central bright fringe. (a) How far apart are the two slits?(b) At what distance from the central bright fringe will the thirddark fringe occur?

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a) 0.0389 mmb) 26 cm

Physics 102 Electricity and Magnetism

Physics 103

Chapter 26

Interference and Diffraction

Electromagnetic Waves

Reflection and Refraction of Light

Rutgers, The State University of New Jersey

University of Winnipeg

McMaster 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.

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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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already. So in this problem we have another handy dandy double slit experiment. So this one's gonna be a little bit different than problem with number seven. So here we do not know what the distance between the slit. Sir, we know that the screen that were viewing is 85 centimeters away, which is point 85 meters basis. I is your friend. You're telling us that we're looking at the third bright fringe. So em is going to be three because this is, um, equal zero. Our center them equals one second bright fringes to vehicles. Three So so M equals three because it's the third bright fringe away from the center already. And we know that the distance between the center and the third bright fringe. So this distance here So why three is equal to 3.11 centimeters or 0.311 meters based us I for the win and we want to know what is deep. So we're gonna use our handy dandy equation 26.6 again 26.6, which tells us that the MP's constructive fringe so the bright spot is equal to our I'm Lambda over deep, but here they don't give us Lambda. They give us the frequency off the light instead, which is 6.32 I'm sent to the 14 hurts. So we get to do the extra step relating Theo Wavelength to the frequency via this formula here. Now, in this case, because we're dealing with light, we know what V is without them. Telling us V is the speed of light, which is a known quantity three times 10 to the eighth meters per second. So any time you're dealing with light, this is the value for the speed that you're gonna use so we can rewrite our equation. Wise of him. R m Landa news v mover f over deep. We want to solve for deep. So let's go ahead and do that Algebraic Lee. So I'm going to multiply besides body cancels here and then divide both sides by y seven cancels here. So d is equal to are and me older frequency times the wife or the pimps fringe. Yeah, What us? Plug in our numbers. Okay, So d is equal to our which was 0.85 meters times m. We're looking at the third bright fringe. So that's three times the which is the speed of light three times 10 to the eighth meters per second, divided by why some three, which was 0.311 meters. So 3.11 centimeters times the frequency of our light, which was 6.32 times 10 in the 14 hurts. And if we plug that into our handy dandy calculator and we get 38 0.9 times 10 to the negative 60 meters so you can report it like that, you can read it as 38.9 micro meters because that's what 10 to the negative six means or you can scooch it. Three more spots to the left to get 0.389 Millie Mears. Either one of these. There's an acceptable answer. So it's your choice. Okay, so for a part be, they want to know what is the distance between the center fringe and the third dark fringe. So if we come back to our beautiful fringy diagram over here, let's used blue. Our first dark fringe happens here 2nd 3rd so notice the our third dark fringe occurs between them equals two nm equals three. So remember our dark things occur at half wavelength differences. So our third dark fringe is going to be M equals two plus 1/2. Cool. All right, so we come back to our handy dandy equation will do this and rid so similarly to her, wise of them equals r M lambda over D For constructive fringes, you can do D stroke different juice or dark spots with our in plus 1/2. So bright spots, dark spots. All right. And just like before, if we plug in Lambda equals B over F, Yeah, we get wise of em equals R and plus 1/2 V over D Yes, but so let's carry that over plug in our values. So we want the third dark spot which was m equals two from our beautiful picture Little trick you s O r waas 0.85 meters. We want to plus 1/2 for em. V was three times 10 to the eight meters per second, diverted by D, which we just solved for, So it's going to be our 38.9 times 10 to the minus six meters and times are frequency which is 6.32 times 10 to the 14. It's well, all right if we plug that into her handy dandy calculator at the end of the day, we get 2.6 centimeters, which checks, So if we go back to our picture or 1/3 bright spot was at 3.11 centimeters. So since our dark spot is closer to the centre than that one, we expect to get smaller distance, so 2.6 isn't acceptable.

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