After an eye examination, you put some eyedrops on your...

After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of refraction of 1.38, while the eyedrops have a refractive index of 1.45. After you put in the drops, your friends notice that your eyes look red, because red light of wavelength 600 nm has been reinforced in the reflected light. (a) What is the minimum thickness of the film of eyedrops on your cornea? (b) Will any other wavelengths of visible light be reinforced in the reflected light? Will any be cancelled? (c) Suppose you had contact lenses, so that the eyedrops went on them instead of on your corneas. If the refractive index of the lens material is 1.50 and the layer of eyedrops has the same thickness as in part (a), what wavelengths of visible light will be reinforced? What wavelengths will be cancelled?

After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of refraction of 1.38, while the eyedrops have a refractive index of 1.45. After you put in the drops, your friends notice that your eyes look red, because red light of wavelength
600 nm has been reinforced in the reflected light. (a) What is the minimum thickness of the film of eyedrops on your cornea? (b) Will any other wavelengths of visible light be reinforced in the reflected
light? Will any be cancelled? (c) Suppose you had contact lenses, so that the eyedrops went on them instead of on your corneas. If the refractive index of the lens material is 1.50 and the layer of eyedrops
has the same thickness as in part (a), what wavelengths of visible light will be reinforced? What wavelengths will be cancelled?

Key Concepts

Thin Film Interference

Thin film interference is a phenomenon where light waves reflecting from different interfaces of a thin layer interfere with each other. The interference can be constructive or destructive based on the difference in the optical paths and any phase changes during reflection. This is a central concept in understanding how certain colors are reinforced or cancelled in thin films.

Optical Path Difference and Phase Shift

The optical path difference in a thin film is determined by the product of the film's thickness and its refractive index, and it plays a crucial role in interference. In addition, phase shifts may occur when light reflects from a medium with a higher refractive index, usually resulting in a half-wavelength phase change. The combined effect of the optical path difference and any phase shifts determines whether interference is constructive or destructive for different wavelengths.

Refractive Index

The refractive index of a material indicates how much the speed of light is reduced in that material compared to a vacuum. It influences the optical path length in thin films and thus affects the interference conditions by altering the phase difference between reflected waves. Understanding how differing refractive indices interact at boundaries is essential in predicting the behavior of light in multi-layer systems.

Interference Conditions (Constructive and Destructive)

Constructive interference occurs when the waves are in phase, leading to an increase in the intensity of reflected light at specific wavelengths, while destructive interference happens when the waves are out of phase, resulting in cancellation. The specific conditions for interference depend on the film's thickness, the wavelength of the light, and the presence of any phase shifts during reflection.

Wavelength Selection and Color Effects

The reinforcement or cancellation of specific wavelengths due to interference leads to the appearance of colors in reflected light. The selective enhancement of certain wavelengths can lead to the perception of particular colors, which is central to applications like anti-reflective coatings, color filters, and even the observed color changes in everyday phenomena like a thin layer of liquid on a surface.

Transcript

00:01 This is chapter 35 problem number 38.

00:03 We have air, eye drop, and cornea.

00:11 So indexed refraction for cornea is 1 .38.

00:14 Indexer reflection for eye drop is 1 .45.

00:17 So let's look at the reflected light.

00:20 What happens to the face of the light as it's reflected from this boundary and from the boundary of eye drop and cornea as it reflects from air.

00:32 Drop interface, it undergoes a pie phase change.

00:39 However, as we look at this interface of eye drop in cornea, there is no phase change.

00:45 So the overall face change is going to be pie.

00:48 So if you're looking at the constructive interference, because in part a, we're giving that information that the reinforced wavelength is 600 nanometers.

00:59 So then we're talking about the constructive interference.

01:01 Then the equations for thick equation for thickness that we're going to use is going to be two t equals m plus one half lambda over and film right so in order to calculate the minimum thickness um let me rewrite this minimum thickness then we need m equals zero so t minimum is going to be one half lambda over to an f so lambda four times um the index of fraction of the film.

01:35 Then we have 600 times centigrade of negative 9 meters divided by four times 1 .45.

01:43 Then we have 103 nanometer as the minimum thickness of the eye drop.

01:54 Now, let's go to part b.

01:59 In part b, we're asked if there is any other wavelength in the visible range that is reinforced.

02:06 So for this minimum thickness of 103 .4 .5 nanometers, let's see if there is any other wavelength that is reinforced.

02:18 So we got this minimum by applying m equals zero, remember? and now what happens if m equals 1? if m equals 1, remember, remember, lambda equals to 2n film divided by t, uh, times t, sorry, divided by m plus one half from the previous equation.

02:42 Now twice the film, the index of fraction of the film times this thickness 103 .4 .5 times centered or negative 9 meters divided by m is one.

02:56 So this is going to give us one plus one half, which is three over two.

03:02 Now, if we calculate this, you're going to find it to be 200 nanometers, which is not in visible range.

03:09 And i want you to pay attention to the fact that as we increase m, we're actually decreasing the wavelength.

03:16 So we're going deeper and deeper in the uv.

03:18 Therefore, this tells you that there is no other day length in the visible range, no other.

03:27 Invisible that is reinforced for the minimum thickness though for the minimum thickness but any problem it doesn't specify whether or not we can use the minimum thickness or the thickness of the next thickness right so let's check that out too so next thickness would be what would it be and so we're basically going to do what we did in part a but this time we're not going to set m equals 0.

04:05 We're going to set m equals 1, right? so repeating what we had in pretty much part, a only for m equals 1 this time, t equals m 1 half lambda over 2m film.

04:22 Now what we have then is 3 over 2 lambda 2m film.

04:36 So it's basically three times, right? three times the initial thickness.

04:43 So if you plug in the numbers, you're going to see that it's three times this value.

04:47 So it's going to be 310 .35 nanometers.

04:53 This is the next thickness.

04:55 Now, let's see if there is any reinforced wavelength that falls in the visible range using this thickness.

05:02 Right so again the thickness our thickness is now 310 .3 5 nanometers and now if m equals 0 for this particular wave length then t is going to be 1 half lambda over 2 and film lambda over 4m lambda then from here we're again calculating the lambda right lambda is going to be 4 and f times the thickness 4 times 1 .45 times the new thickness 310 .35.

05:38 Well, you can convert that to meters.

05:42 So then the answer is going to be 1 ,800 times 10 to 0 .9 meters, which means 1 ,800 nanometers.

05:51 Again, this is an infrared range, right? this is not invisible.

05:58 And let's look at what happens to the wavelength as we increase m.

06:02 So for m equals, one, for example, we're going to have the thickness of 3 over 2, lambda, 2, and f.

06:13 So this is basically 4nf, right? so 4 nf times thickness divided by 3 is going to give us lambda.

06:25 So we know this part, this was 1 ,800.

06:28 Then divided by 3 is going to give us 600 nanometers.

06:31 This is something that we already found.

06:33 This is the red light that we were talking about.

06:35 So we know that it's already reinforced...