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$\bullet$ Conserving energy. You want to coat the inner surfaces ofyour windows (which have refractive index of 1.51$)$ with afilm in order to enhance the reflection of light back into theroom so that you can use bulbs of lower wattage than usual.You find that $\mathrm{MgF}_{2},$ with $n=1.38,$ is not too expensive, soyou decide to use it. Since incandescent home lightbulbs emitreddish light with a peak wavelength of approximately 650 $\mathrm{nm}$ ,you decide that this wavelength is the one to enhance in thelight reflected back into the room. (a) What is the minimumthickness of film that you will need? (b) If this layer seems toothin to be able to put on accurately, what other thicknesseswould also work? Give only the three thinnest ones.

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a) 235.5 nmb) $471 nm, 706 nm, 942 nm$

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 Washington

Hope College

University of Sheffield

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.

06:38

A uniform film of TiO$_2$ …

06:10

35.31. A uniform film of $…

03:48

A uniform film of $\mathrm…

06:27

When viewing a piece of ar…

03:16

04:05

06:44

You want to coat a pane of…

05:09

35.28. Nonglare Glass. Whe…

05:42

$\bullet$ Nonglare glass. …

07:06

Nonglare Glass. When viewi…

So let's think about the situation. We have a liar, a coming in, and it's hitting this material here and reflecting off some of it transmits through and eventually hits the glass behind it and then reflects off that and then goes out. We want both these rays after they come out to be in face for constructive interference. Now, this first layer here's air, so and is a good one. The second layer is a layer that we're applying and we're told it's in is equal to 1.38 and then this layer here is glass Oh, and is equal to 1.51 And so if the distance here is T, then what we can say is if the light Ray comes in and equals one and hits, it is equal to 1.38 since 1.3 is higher than one, there's going to be a 1 80 degree phase shift here. But then this wave coming in and 1.3 a is gonna hit 1.51 which is larger, and so there's also gonna be a 1 80 degree phase shift. So in total, there's a net zero phase shift between these and so the condition for constructive interference is too T too, because it's t this way and t that way. This is Eagle two uninjured times the wavelength of the light. We have to be careful about playing from the wavelength here. We're told the wavelength out here is six fifteen 15 Anna Meares. But that's not the wavelength and hear the wavelength. And here we have to divide by this and next of refraction. Just 1.3. And so this is what we're actually gonna plug in for Landa. Not 6 50 In reality, we want t value. So solving that we get him land over to and for part A we want we want the thinnest layer. So you want the smallest tea, and we're going to get the smallest e by playing in the small stone. Serena, don't be little one. So for part, a T is equal to land over to remember, This is our landa. So plugging that in gives, uh, to 36 Nana meters for part B B One thing next, Then its layers the next three finish Thinnest, actually. And so we're gonna plug in and was equal to 23 and four since those air the next smallest and values playing them into this formula here and using the same Lambda we get T values of for 72 seven away and 9 44 and those are all in animators and that completes the

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