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$\bullet$ Coating eyeglass lenses. Eyeglass lenses can be coated onthe inner surfaces to reduce the reflection of stray light to theeve. If the lenses are medium flint glass of refractive index1.62 and the coating is fluorite of refractive index 1.432 ,(a) what minimum thickness of film is needed on the lenses tocancel light of wavelength 550 nm reflected toward the eye atnormal incidence, and (b) will any other wavelengths of visi-ble light be cancelled or enhanced in the reflected light?

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a) 96 nmb) No other wavelength of visible light is enhanced.

Physics 102 Electricity and Magnetism

Physics 103

Chapter 26

Interference and Diffraction

Electromagnetic Waves

Reflection and Refraction of Light

University of Washington

Simon Fraser University

University of Winnipeg

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.

03:50

Coating Eyeglass Lenses. E…

0:00

Eyeglass lenses can be coa…

03:19

05:41

00:29

00:18

Eyeglass lenses of refract…

05:29

Antireflection coatings ca…

03:57

03:32

You're applying a 110…

02:25

Lenses are generally coate…

05:42

$\bullet$ Nonglare glass. …

02:17

An antireflection coating …

03:38

Calculate the minimum thic…

Before we do any math, let's think about the situation. We have a light ray entering a more dense medium, and so there's going to be an initial 180 degree a shift. But then the life that does enter that dancer medium is going to encounter glass, which is even more dense. And so there's gonna be another 180 degree phase shift so to 183 fishes creates no neff Asia. So we won't have to worry about the faces in this problem. And so the path difference is too t. And so the condition for destructive interference with no net phase shift is too t is equal to plus 1/2. That's a way of life solving this 40 gives t is equal to m plus one have I am over too. But this is the lambda in the media and we only know the Lambda and air. But we can relate them by this. Wherein is the index of refraction of the medium. And so playing this in gins own plus 1/2 Graham did not over two in and now we want the first minimum or the minimum thickness. So we want the smallest tea here. The way we get the small STI is by picking ILM to be zero. So we're gonna pick him to be zero. We could pick larger values of film, but we would get larger value of tea and we want the minimal thickness here. And so he is equal to landed, not over foreign and in it is equal to well, first off, Lam denies five fifteen 15 animators and in is also equal to 1.432 And so the minimal thickness is 96 nanometers and we move on to part B. Barbie wants to know if there are any other thickness is that we can have toe where there's the destructive or constructive interference. Well, for destructive interference, we still have This duty is equal to one plus 1/2 when did not over in and so solving this for landing on. We get to tea and over plus 1/2. But we figured out what t wass here. Bernie is a same tea and we know it isn't just the index of refraction of the medium. It's 1.432 and so we can simplify this 22 175. That's an animators. Thes air units, not variables. So it's Nana meters over l'm plus 1/2. Now, when we plug in and it was equal to zero, we get Lambda not is equal to 5 50 Nana Mears. So this one works. This is the one that we already saw. If it was equal to one, we get a Lambda Non of 183 years. Now this works, but this is not in the visible spectrum range. And so we don't count it, and subsequent in values will produce even smaller landed knots. So we're definitely not going to count those. And so this is the on ly lambda, not in the visible range that produces destructive interference. Now, for constructive interference, we're gonna have the same thing. Except it's changed ever so slightly. It's just m landed, not over in. All right, there's not in and there is not half imager. There's full editor. And so solving for lamb did not again gives us 2 75 except it's over. L not impulse 1/2 this time. And so if film is equal to one, we get Landon on is equal to 2 75 and then all other in values will produce even smaller. But even this is not in the visible range, so all the other ones definitely won't be in the visible range. So there are no Lambda nods in the visible range that produce construct of interference and that.

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