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Light of wavelength \(624 \mathrm{nm}\) is incident perpendicularly on a soap film \((n=1.33)\) suspended in air. What are the (a) least and (b) second least thicknesses of the film for which the reflections from the film undergo fully constructive interference?

Short Answer

Expert verified
The least thickness is approximately 176.27 nm, and the second least thickness is approximately 352.53 nm.

Step by step solution

01

Identify the Conditions for Constructive Interference

Constructive interference occurs when the path difference between the reflected waves from the top and bottom surfaces of the film is an integer multiple of the wavelength in the medium. The formula for the path difference is given by \(2nt = m\lambda/n_\text{air}\), where \(t\) is the thickness of the film, \(n\) is the refractive index of the film, and \(m\) is an integer (0, 1, 2,...). Here, since the light is incident from air, \(n_\text{air} = 1\).
02

Convert Wavelength to Wavelength in Medium

The wavelength of light in a medium is given by \(\lambda_n = \lambda/n\), where \(n\) is the refractive index of the medium. Substitute the values to find \(\lambda_n\): \(\lambda_n = 624 \text{ nm} / 1.33 = 469.17 \text{ nm}\).
03

Calculate Least Thickness (m=0)

For the least thickness, use \(m = 0\): \(2nt = 0\cdot\lambda_n\). However, since \(t\) cannot be zero for constructive interference caused by reflection, the effective smallest \(m\) we use is \(m = 1\) to obtain a non-zero result: \[t = \frac{m \lambda_n}{2n} = \frac{1 \times 469.17 \text{ nm}}{2 \times 1.33} \approx 176.27 \text{ nm}\].
04

Calculate Second Least Thickness (m=1)

For the second least thickness, use \(m = 2\) since \(m = 1\) has already been addressed: \[t = \frac{m \lambda_n}{2n} = \frac{2 \times 469.17 \text{ nm}}{2 \times 1.33} \approx 352.53 \text{ nm}\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wavelength in Medium
When light travels through different mediums, its speed changes, affecting its wavelength. The familiar wavelength value we often hear, such as 624 nm for a common scenario in air, is not the same when the light enters a new material like a soap film. This change in wavelength is due to the refractive index of the new medium. Understanding this concept is crucial when tackling problems that involve light interaction, like interference patterns. To find the new wavelength, we use the formula \[ \lambda_n = \frac{\lambda}{n} \]where \(\lambda\) is the original wavelength in a vacuum, and \(n\) is the refractive index of the medium. Here, for the soap film with a refractive index \(n = 1.33\), the wavelength in the film is:\[ \lambda_n = \frac{624 \, \text{nm}}{1.33} \approx 469.17 \, \text{nm} \]This adjusted wavelength is what interacts with the film to create interference patterns. Understanding the impact of a medium's refractive index on wavelength helps us predict and analyze phenomena like constructive interference.
Soap Film
A soap film is a thin layer of soapy water that forms when a bubble is created. This thin film acts as a fascinating stage for the drama of interference due to its ability to reflect light from both its front and back surface. When light hits a soap film, some light is reflected from the top surface, and some pass through and reflect from the bottom surface. These reflected light waves can interfere with each other, leading to a phenomenon called constructive interference, where the waves align perfectly to amplify the light intensity. Soap films are great studying tools for understanding interference and thin film physics because they visibly demonstrate how light interacts with surfaces and materials. Such films help in practical applications, like understanding how coatings work on lenses or even the layers of a butterfly's wing!
Refractive Index
The refractive index is a number that tells us how much a material slows down and bends light as it passes through it. It's an essential concept in optics as it affects how light travels in various mediums. The refractive index \(n\) is defined as the ratio of the speed of light in a vacuum to the speed of light in the medium: \[ n = \frac{c}{v} \]where \(c\) is the speed of light in a vacuum, and \(v\) is the speed of light in the medium. For a soap film, we used \(n = 1.33\). This means light travels slower in the soap film than in the air.
  • A higher refractive index means light bends more upon entering the medium.
  • It's crucial in designing optical devices, helping us compute how light behaves.
Understanding refractive index helps us solve problems related to light interference, like in soap films, where it varies depending on the medium.
Film Thickness
Film thickness plays a vital role in interference effects, especially in thin films like soap bubbles. Thickness determines the path difference between the light rays reflecting from the top and bottom surfaces of the film. This path difference is essential in calculating interference effects. The formula for finding film thickness for constructive interference is:\[ 2nt = m\lambda_n \]where \(t\) is the film thickness, \(n\) is the refractive index, and \(m\) is the order of interference (integer values). This condition must be met for constructive interference where reflected waves reinforce each other.In our example, we found:
  • Least thickness \( (m = 1) \):\[ t = \frac{1 \times 469.17 \text{ nm}}{2 \times 1.33} \approx 176.27 \text{ nm} \]
  • Second least thickness \( (m = 2) \):\[ t = \frac{2 \times 469.17 \text{ nm}}{2 \times 1.33} \approx 352.53 \text{ nm} \]
Calculating film thickness is crucial for predicting interference colors in soap films and other thin film scenarios.

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Most popular questions from this chapter

A thin film of acetone \((n=1.25)\) coats a thick glass plate \((n=1.50) .\) White light is incident normal to the film. In the reflections, fully destructive interference occurs at \(600 \mathrm{nm}\) and fully constructive interference at \(700 \mathrm{nm}\). Calculate the thickness of the acetone film.

A plane wave of monochromatic light is incident normally on a uniform thin film of oil that covers a glass plate. The wavelength of the source can be varied continuously. Fully destructive interference of the reflected light is observed for wavelengths of 500 and \(700 \mathrm{nm}\) and for no wavelengths in between. If the index of refraction of the oil is 1.30 and that of the glass is \(1.50,\) find the thickness of the oil film.

In a double-slit arrangement the slits are separated by a distance equal to 100 times the wavelength of the light passing through the slits. (a) What is the angular separation in radians between the central maximum and an adjacent maximum? (b) What is the distance between these maxima on a screen 50.0 cm from the slits?

57 through 68 Transmission through thin layers. In Fig. \(35-43,\) light is incident perpendicularly on a thin layer of material 2 that lies between (thicker) materials 1 and 3 . (The rays are tilted only for clarity.) Part of the light ends up in material 3 as ray \(r_{3}\) (the light does not reflect inside material 2 ) and \(r_{4}\) (the light reflects twice inside material 2). The waves of \(r_{3}\) and \(r_{4}\) interfere, and here we consider the type of interference to be either maximum (max) or minimum (min). For this situation, each problem in Table \(35-3\) refers to the indexes of refraction \(n_{1}, n_{2},\) and \(n_{3},\) the type of interference, the thin-layer thickness \(L\) in nanometers, and the wavelength \(\lambda\) in nanometers of the light as measured in air. Where \(\lambda\) is missing, give the wavelength that is in the visible range. Where \(L\) is missing, give the second least thickness or the third least thickness as indicated. Table 35-3 Problems 57 through 68: Transmission Through Thin Layers. See the setup for these problems. $$\begin{array}{llllllll} & n_{1} & n_{2} & n_{3} & \text { Type } & L & \lambda \\ \hline 62 & 1.68 & 1.59 & 1.50 & \max & 2 n \mathrm{~d} & 342 \end{array}$$

Monochromatic green light, of wavelength \(550 \mathrm{nm},\) illuminates two parallel narrow slits \(7.70 \mu \mathrm{m}\) apart. Calculate the angular deviation \((\theta\) in Fig. \(35-10)\) of the third-order \((m=3)\) bright fringe (a) in radians and (b) in degrees.

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