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$\bullet$$\bullet$ Sensitive eyes. After an eye examination, you put someeyedrops on your sensitive eyes. The cornea (the front part ofthe eye) has an index of refraction of 1.38, while the eyedropshave a refractive index of $1.45 .$ After you put in the drops, yourfriends notice that your eyes look red, because red light ofwavelength 600 $\mathrm{nm}$ has been reinforced in the reflected light.(a) What is the minimum thickness of the film of eyedrops onyour cornea? (b) Will any other wavelengths of visible light bereinforced in the reflected light? Will any be cancelled?(c) Suppose you had contact lenses, so that the eyedrops wenton them instead of on your corneas. If the refractive index ofthe lens material is 1.50 and the layer of eyedrops has the samethickness as in part (a), what wavelengths of visible light willbe reinforced? What wavelengths will be cancelled?

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a) 103.5 nmb) No other wavelength in visible light is cancelled.c) No light is reinforced 600 $nm$ wavelength will be cancelled

Physics 102 Electricity and Magnetism

Physics 103

Chapter 26

Interference and Diffraction

Electromagnetic Waves

Reflection and Refraction of Light

Cornell University

University of Washington

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.

03:20

$\bullet$$\bullet$ Sensiti…

Sensitive Eyes. After an e…

11:53

After an eye examination, …

04:49

06:39

You have just put some med…

02:26

03:50

Coating Eyeglass Lenses. E…

0:00

Eyeglass lenses can be coa…

05:41

04:45

$\bullet$ Coating eyeglass…

02:07

BIO Treating Cataracts Whe…

07:16

The maximum resolution of …

04:44

Okey dokey. Our eyeballs hurt, so we're gonna put some eye drops in there. We're gonna make a nice little thin film where a cornea has on index of refraction of 1.38 The drops that we put on our I make a nice little film with an index of refraction of 1.45 and out here in the air, we have a little and next of a faction of one going to label those air drops, I Okay, so whenever we start one of these, we always wanted figure out if we're gonna have a phase shift or not. So at this surface blank point, we're going from low and to a higher end. So when we go from low to Hari, we pick up ah, half cycle phase shift. And on this surface, blink, blink. We're going from hi toe low, so higher in to a lower end when that has no phase shift. So by the time our light starts interfering with each other, they're half a cycle out of phase, which means our constructive interference is gonna happen with our M plus 1/2 Landis in particular, we have to times are thickness of our thin film is equal to M plus 1/2 Lambert. Now this is the lambda inside the films. That's where the drops are. But we only have the wavelength of light that are friends we're seeing, which is 600 nanometers. So we want to figure out how that wavelength gets shifted inside this index of perfection. So we can. Here's the relationship that the index of refraction is the wavelength in air slash vacuum divided by the wavelength in the medium. So here we go. Lambda it is lam not over. And take this, Stick it in there. All right, so now I have two times my thickness of my film. It's in plus 1/2 Lambda not over. And so thickness of my film. I just divide all that by two. All right, So if I plugged it in So I handed in the calculator where you We're going to say that M is equal to zero because that's the minimum thickness that we can get. So zero plus 1/2 divided by two is 1/4 lambda. Not is the wavelength outside in the air. So it's 600 a meters and end of our film is one point for size, so that gives us a thickness of 103.45 Nana meters. All right, so it's our minimum thickness. Okay. Okay. So next part wants to know Are there any other wavelengths that are gonna be enhanced by this or that? Have the constructive interference. So assuming that we have the same thickness, So our thickness waas ah, 103.45 millimeters. And the equation that we left off with waas um, plus 1/2 oh, to soup lambda, not over. And okay, so you just try different values of em for Lambda Knots. So let's solve for land enough. So Lamda not multiply my two over to t multiply my end over divide by impulse 1/2. So let's try em. You know, M equals zero gets us the 600. So let's try, um, equals one. So in that case, the wavelength that we would see is too sickness of drops times and waas. One point for five over one plus 1/2. So over 1.5 that gives me a wavelength of 200 native years, which is not in the visible spectrum. So if we look at our where M plus 1/2 is, if we increase this toe M equals two, there's just going to give us a smaller, even smaller wavelengths. There's no point plugging in numbers there. So 600 nanometers is tthe e on lee wavelength in the visible spectrum, that's gonna have constructive interference. So they're going to see creepy red eyes. All right, so now, for part C, Now we're wearing contacts. So you have our air equals one. You have our drops and equals 1.45 and we have our contact and equals 1.5. So we changed our index of refraction down here. We need to look at our raise again. So this surface we're going from low to high. It's gonna abbreviate of this time Low to high saw. It picks up 1/2 Wade's length shift on this surface. We're going from 1.45 to 1.5. So that's a little too high again. So that picks up 1/2 leaves link shift. So now are constructive. Things are back in face like a shifted by a whole wave lengths. Other back in phase So my constructive it's gonna happen at two. T equals and Landa and my d struck tive is gonna happen at the M plus 1/2 you ready? So let's go ahead and put that in terms of the wavelength that you see outside. So some land not over in it seemed from this one to t equals and plus 1/2 land and not over. And already so we want to know, Are there gonna be any visible wave links that are constructive? And are there any visible way of thanks that air destructive? So let's solve these for wave licks. So Liam did not well say, constructive to do multiply my end over here. That's two t and over and and for destructive my land and not ISS to tea and over and plus at Okay, let's switch to do so for M equals zero that doesn't exist for my constructive but for my destructive that gets me 600 nanometers. So now if we put on contact lenses that red light gets completely cancelled out for M equals one. I plug that into here with the same thickness. You get a wavelength of 300. Nana Mears which is non visible, and down here we get 202 meters, which is not visible and if you want to, you can do and vehicles to, but that just makes it even smaller.

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