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Chapter 27: Problem 15

Orange light \(\left(\lambda_{\text {vacuum }}=611 \mathrm{nm}\right.\) ) shines on a soap film \((n=1.33)\) that has air on either side of it. The light strikes the film perpendicularly. What is the minimum thickness of the film for which constructive interference causes it to look bright in reflected light?

Short Answer

Expert verified
The minimum thickness of the soap film for constructive interference is about 114.85 nm.

Step by step solution

01

Understand the Conditions for Constructive Interference

For constructive interference in a thin film with a higher index of refraction than the surrounding medium (like soap film surrounded by air), the condition is that the path difference must be an integral multiple of the wavelength. Since the air-film-air setup causes a \(\frac{1}{2}\) wavelength shift due to reflection at the air-film boundary, constructive interference occurs when\: \[ 2nt = (m + 0.5) \frac{\lambda}{n} \] where \(m\) is an integer representing the interference order, \(n\) is the refractive index of the film, \(t\) is the thickness, and \(\lambda\) is the wavelength of light in vacuum.
02

Rearrange the Equation for Minimum Thickness

To find the minimum thickness \(t\) for constructive interference (beginning with the first order where \(m = 0\)), rearrange the formula to solve for \(t\): \[ t = \frac{(0 + 0.5) \cdot \frac{\lambda}{n}}{2n} \] Substitute \(m = 0\), \(\lambda = 611\ nm\), and \(n = 1.33\) to find the solution for the smallest \(t\).
03

Calculate the Wavelength of Light Inside the Film

Calculate the wavelength of light inside the soap film using the refractive index \(n\):\[ \lambda_{film} = \frac{\lambda_{vacuum}}{n} = \frac{611 \text{ nm}}{1.33} \approx 459.4 \text{ nm} \]
04

Calculate the Minimum Thickness of the Film

Substitute \(\lambda_{film}\) and the known values into the rearranged formula to calculate the thickness \(t\):\[ t = \frac{(0.5) \cdot 459.4 \text{ nm}}{2} \approx 114.85 \text{ nm} \] Thus, the minimum thickness of the film for constructive interference is about 114.85 nm.

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Key Concepts

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

Thin Film Interference
Thin film interference is a fascinating phenomenon that occurs when light waves reflect off the surfaces of a thin film and interact with each other. This interaction leads to constructive or destructive interference, depending on the thickness of the film and the wavelength of the light. In the case of constructive interference, the result is a bright appearance, as the light waves combine to amplify certain wavelengths.

One classic example is a soap film. When light strikes the film, some of it reflects off the top surface, and some travels through the film, reflecting off the bottom surface. These two reflected waves can interfere with each other. If the thickness of the film is just right, the waves combine constructively, enhancing the light's brightness at specific wavelengths.

Factors influencing this effect include the wavelength of the light in a vacuum and the refractive index of the film. By adjusting these variables, you can achieve different colors and intensities in the reflected light.
Refractive Index
The refractive index, often denoted as \(n\), is a measure of how much a substance can bend light. It's a critical factor in thin film interference because it influences how the light propagates through a medium like a film.

When light enters a medium with a different refractive index, its speed and wavelength change, although its frequency remains constant. The refractive index can be expressed as the ratio of the speed of light in a vacuum to its speed in the medium:
\[ n = \frac{c}{v} \]

Here, \(c\) is the speed of light in a vacuum, and \(v\) is its speed in the medium. A higher refractive index means that light travels more slowly in that substance. For example, in the exercise involving a soap film, the refractive index of the film is given as 1.33, indicating that light travels slower through the soap film than through the air.
Wavelength in Medium
The wavelength of light changes when it transitions from a vacuum into a different medium, like a soap film. The refractive index of the medium reduces the wavelength inside the medium compared to its original wavelength in a vacuum.

To calculate the wavelength within the film, you divide the wavelength in a vacuum by the refractive index of the film:
\[ \lambda_{\text{medium}} = \frac{\lambda_{\text{vacuum}}}{n} \]

For instance, in the given exercise, the wavelength of 611 nm in vacuum becomes roughly 459.4 nm inside the soap film. This adjusted wavelength is crucial for determining the conditions for constructive interference, as the light waves need to meet specific conditions based on their new wavelength inside the film.

The understanding of wavelength in medium is essential for manipulating and predicting the behavior of light in various materials, allowing us to create desired visual effects or solve physics problems involving interference.

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Most popular questions from this chapter

In a Young's double-slit experiment, the seventh dark fringe is located \(0.025 \mathrm{~m}\) to the side of the central bright fringe on a flat screen, which is \(1.1 \mathrm{~m}\) away from the slits. The separation between the slits is \(1.4 \times 10^{-4} \mathrm{~m}\). What is the wavelength of the light being used?

Interactive LearningWare 27.2 at provides some pertinent background for this problem. A transparent film \((n=1.43)\) is deposited on a glass plate \((n=1.52)\) to form a nonreflecting coating. The film has a thickness that is \(1.07 \times 10^{-7} \mathrm{~m}\). What is the longest possible wavelength (in vacuum) of light for which this film has been designed?

A single slit has a width of \(2.1 \times 10^{-6} \mathrm{~m}\) and is used to form a diffraction pattern. Find the angle that locates the second dark fringe when the wavelength of the light is (a) 430 \(\mathrm{nm}\) and (b) \(660 \mathrm{~nm}\).

In a Young's double-slit experiment, the wavelength of the light used is \(520 \mathrm{nm}\) (in vacuum), and the separation between the slits is \(1.4 \times 10^{-6} \mathrm{~m}\). Determine the angle that locates (a) the dark fringe for which \(m=0,\) (b) the bright fringe for which \(m=1,\) (c) the dark fringe for which \(m=1,\) and \((\mathrm{d})\) the bright fringe for which \(m=2\)

Concept Questions Light shines on a diffraction grating, and a diffraction pattern is produced on a viewing screen that consists of a central bright fringe and higher-order bright fringes (see the drawing). (a) From trigonometry, how is the distance \(y\) from the central bright fringe to the second-order bright fringe related to the diffraction angle \(\theta\) and the distance \(L\) between the grating and the screen? (b) From physics, how is \(\theta\) related to the order \(m\) of the bright fringe, the wavelength \(\lambda\) of the light, and the separation \(d\) between the slits? (c) In this problem, the angle \(\theta\) is small (less than a few degrees). When the angle is small, \(\tan \theta\) is approximately equal to \(\sin \theta,\) or \(\tan \theta \approx \sin \theta .\) Using this approximation, obtain an expression for \(y\) in terms of \(L, m, \lambda,\) and \(d .\) (d) If the entire apparatus in the drawing is submerged in water, would you expect the distance \(y\) to increase, decrease, or remain unchanged? Why? Problem Light of wavelength \(480 \mathrm{nm}\) (in vacuum) is incident on a diffraction grating that has a slit separation of \(5.0 \times 10^{-7} \mathrm{~m}\). The distance between the grating and the viewing screen is \(0.15 \mathrm{~m}\). (a) Determine the distance \(y\) from the central bright fringe to the second-order bright fringe. (b) If the entire apparatus is submerged in water \(\left(n_{\text {water }}=1.33\right),\) what is the distance \(y\) ? Be sure your answer is consistent with part (d) of the Concept Questions.
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