Can someone please help me with the summary part and application part, please I have attached an example of what it should be as well as the notes.

I already did most of it

Question:

A 520 nm-thick soap film (n = 1.40) in air is illuminated with white light in a direction perpendicular to the film. For how many different wavelengths in the 400 nm to 700 nm range is there fully destructive interference in the reflected light?

Answer: 1

Solution:

Solution The destructive interference in a thin film of thickness L with higher refractive index n2 than the surrounding media (air) is modeled by 2L = m- n2 where m = 0, 1, 2... and ) is the wavelength of the light. Solving for 1 yields the wavelength that exhibit the constructive interference, i.e., intensification of the reflected light: 1 = 2n2 L m Using the lower wavelength to get an upper bound for m 2(1.4) (520 nm) m = ~ 3.6. 400 nm Using the higher wavelength to get lower bound for m 2(1.4) (520 nm) m = ~ 2.1. 700 nm There is only one integer between 2.1 and 3.6, m = 3, so there is only 1 possible value for m.A 520 nm-thick soap film (n = 1.40) in air is illuminated with white light in a direction perpendicular to the film. For how many different wavelengths in the 400 nm to 700 nm range is there fully destructive interference in the reflected light? 2 O 3 4Chapter 35 Summary and Resources X Robert Franz posted on Jul 31, 2023 1:20 PM Here are the main ideas in this chapter.1. Light is a wave. Huygens's principles for waves states that all points on the wavefront serve as point sources for secondary wavelets. After a time the new wavefront has advanced and its new position is the superposition of all of the wavefronts tangent from these sources. 2. As light moves through a medium, its wavelength changes depending on the index of refraction of the medium. This is given mathematically as In = where n = c/v the vacuum speed of light divided by the speed of the wave in the medium and, is the vacuum wavelength of the light. Its frequency remains unchanged. Because the wavelength changes in the medium, two identical beams of light passing through two different mediums may have different phases when they re-emerge. 3. Young's experiment showed that light is a wave. A monochromatic light source is first incident on a screen with a slit. The light diffracts from the first slit and spread out where it encounters a second screen with two slits. The light diffracts from each of these slits where it forms two identical sources which then interfere with each other forming an alternating light an dark fringe pattern on a third screen. The position of the light fringes is when the path different from a wave from each of the slits is equal to an integral number of wavelengths. In other words, these waves are in phase with each other. This can be written for constructive interference (bright fringes) as: d sin 0 = m )where d is the spacing between the slits and Gis the angle with respect to the normal to the slit and m is an integer m = 0, 1, 2, 3, 4, .. .. The position of the dark fringes (in between each of the light fringes is the position of destructive interference where the two light rays (1 from each slit) combine completely out of phase with each other. The position of the dark fringes can be written as: d sine = (m + ? ) where each symbol is defined as before. For fringes near the central maximum, we can use the small angle approximation where sin 0 ~ 0 ~ tan Oto make the calculations of position easier to do. 4. The combination of two waves of different phases at a single location will give an intensity pattern given by the following equation I = 410 cos ) where is the initial phase difference of the waves. The intensity pattern for Young's double-slit interference can be explained if = 2rd sin 0. 5. When light is incident on a thin film the reflected light from the front and back surfaces can combine and interfere creating an interference pattern. The reflected wave can experience a phase shift if the speed of the wave in the new medium is slower in the new medium. Thus if a wave moves into a medium where the index is higher (slower speed) there is a .5 wavelength phase shift for the reflected wave. Refracted waves don't have a phase shift. In order for waves from the reflection from the back side of the film to be in phase with those reflected from the front surface, the path length of the wave must be equal to the have the wavelength (or its odd multiples). The path length for the second reflected wave is twice the thickness of the film. For bright fringes, one can write: 2L = m +. . For dark fringes 2L = md and m = n2 n2 0, 1, 2Question: We wish to coat flat glass (n = 1.50) with a transparent material (n = 1.20) so that reflection of light at wavelength 480. nm is eliminated by interference. What minimum thickness can the coating have to do this? a) 100. nm b) 120. nm c) 80.0 nm d) 200. nm e) 160. nm Concept: It is crucial to understand the concept of interference when approaching this problem. Understanding how the index of refraction, thin- film interference, reflection/refraction, and film thickness relate to one another is also important. If the film has a lower index of refraction than the medium (glass in the case of this problem), the wavelength will be shifted by pi radians. In order to identify the relationship between these two aspects we must understand the path difference. This concept just highlights the difference in the distance that two separate light waves travel when they reflect from the two surfaces of the film. This is how film thickness is related to the refractive index of the film and the area around the problem. Solution: 2L = m+ Solving for L: L = (m+ ; ) x 2 The question is asking for the least film thickness, so therefore m = 0: L = ( X 2= 480 nm An 4 * 1.20 = 100 nm Application: Interference from thin films can be found a fair amount in everyday products. One application I found particularly useful is its use in anti-glare screen protectors. These screen protectors are designed to minimize the amount of glare and light reflection on the surface of the phone's screen. This makes it so that the desired light signals from the device can be received easily by the user. While this technology can be found in anti-glare coatings in various optical devices (such as glasses), they all follow the same premise of interference from thin films. A screen protector is made up of several layers of thin films that all have different refractive indices. As shown in the problem above, the thickness of these films is key in determining what wavelengths of light undergo destructive interference. Reference: The basics of anti-reflection optical thin film: Tech Times: Technical information media for engineers. TECH TIMES | Technical information media for engineers. (n.d.). https://techtimes.dexerials.jp/en/optics/basic-of-anti- reflection/