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$\bullet$ Light of wavelength 633 nm from a distant source is inci-dent on a slit 0.750 mm wide, and the resulting diffraction pat-tern is observed on a screen 3.50 $\mathrm{m}$ away. What is the distancebetween the two dark fringes on either side of the centralbright fringe?

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5.91 mm

Physics 102 Electricity and Magnetism

Physics 103

Chapter 26

Interference and Diffraction

Electromagnetic Waves

Reflection and Refraction of Light

University of Michigan - Ann Arbor

University of Sheffield

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.

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.

01:09

Light of wavelength 633 nm…

02:58

02:28

03:01

02:47

5. Light of wavelength 589…

03:09

Light waves with two diffe…

02:12

Light that has a wavelengt…

04:26

Light that has a 650 -nm w…

01:34

Light with a wavelength of…

02:02

03:21

Light of a single waveleng…

in this problem, the small angle approximation is valid. It's valid because this ratio here is small. It's very small, in fact. And when this ratio is very small, then we can use why ohm is equal to our and when, UH over a which is only valid sometimes in cases where this is true and now they want the distance between the two dark fringes on either side of central maximum. And since the distance between the central maximum and one of them is, why one the distance we really want us to white one. So that's that's what we want here. Well, we could just sulfur. Why one and then double it from this formula. We know that why one is equal to our land over a Just playing into him was a good one, since we want that first ah, destructive interference and now we can plug in these values which were given in the problem. R is 3.5 meters lambda 6 33 times 10 to the minus nine meters and a is 0.750 times 10 to the minus 30 meters and so we get out 2.95 times 10 to the minus 30 meters. And then if we double that after converting this two millimeters, right, this is also 2.95 millimeters. So if we double this, we get to my one, which is our desire and, um is five point 9,000,000 years, and that's the answer.

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