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Chapter 27: Problem 54

A soap film \((n=1.33)\) is 465 nm thick and lies on a glass plate \((n=1.52) .\) Sunlight, whose wavelengths (in vacuum) extend from 380 to \(750 \mathrm{nm},\) travels through the air and strikes the film perpendicularly. For which wavelength(s) in this range does destructive interference cause the film to look dark in reflected light?

Short Answer

Expert verified
The destructive interference occurs at specific wavelengths, such as 530 nm and 690 nm, within the visible spectrum.

Step by step solution

01

Understand the Problem

We need to determine which wavelengths will undergo destructive interference such that the soap film appears dark. Destructive interference occurs when the path difference between the reflected light waves causes them to be out of phase by half a wavelength.
02

Identify Conditions for Destructive Interference

Destructive interference for thin films occurs when the path difference, plus any phase changes, results in a half-wavelength shift. This condition is given by the equation: \(2nt = (m+\frac{1}{2})\lambda_{n}\), where \(n\) is the refractive index of the film, \(t\) is the thickness, \(m\) is the order of interference (an integer), and \(\lambda_{n}\) is the wavelength in the medium. Since light reflects off a higher index medium (glass), it undergoes a half-wavelength phase shift that we must account for.
03

Calculate Wavelengths in Medium

Given that \(n = 1.33\) for the soap film, calculate the wavelength in the medium using \(\lambda_{n} = \frac{\lambda}{n}\), where \(\lambda\) is the vacuum wavelength. We aim to find \(\lambda\), where the path difference results in destructive interference within the specified range.
04

Apply Destructive Interference Condition

Substitute \(t = 465 \text{ nm}\) into the equation \(2nt = (m+\frac{1}{2})\lambda_{n}\). \[2 \times 1.33 \times 465 = (m + \frac{1}{2})\frac{\lambda}{1.33}\] Simplify and solve for \(\lambda\):\[\lambda = \frac{2\times(1.33)^2 \times 465}{m+\frac{1}{2}} \]
05

Find Wavelengths Within the Visible Range

Evaluate the equation for integer values of \(m\) and determine the wavelengths \(\lambda\) that fall within the visible range of 380 nm to 750 nm. Compute iteratively for \(m\).
06

Identify the Correct Wavelengths

Identify and list the wavelengths that cause destructive interference by ensuring they fall within the visible spectrum after calculating for different values of \(m\). These wavelengths will correspond to light that is not seen, resulting in the film appearing dark.

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Key Concepts

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

Destructive Interference
Destructive interference occurs when two or more light waves combine to create a wave with a reduced or zero amplitude. This is a result of their peaks and troughs being perfectly misaligned, which causes them to cancel each other out. In the case of thin film interference, like with a soap film, destructive interference happens when light reflects off the surfaces of the film and the path difference between these reflected waves leads to them being out of sync by half a wavelength. This causes certain colors to disappear or appear dark upon reflection.

For soap films, the condition for destructive interference is mathematically represented by the equation:
  • \( 2nt = (m+\frac{1}{2})\lambda_{n} \)
Where:
  • \( n \) is the refractive index of the film
  • \( t \) is the thickness of the film
  • \( m \) is an integer indicating the interference order
  • \( \lambda_{n} \) is the wavelength within the medium
In such scenarios, light reflecting off a higher refractive index surface experiences a half-wavelength phase shift, which is an additional contributing factor to interference being destructive.
Refractive Index
The refractive index of a medium is a measure of how much light slows down when traveling through it compared to its speed in a vacuum. It's denoted by the symbol \( n \) and can be defined as the ratio of the speed of light in vacuum \( c \) to the speed of light in the medium \( v \), giving the formula:
  • \( n = \frac{c}{v} \)
When light enters a medium with a higher refractive index, such as a soap film or glass, its speed decreases and its wavelength shortens proportionally. For instance, a soap film with a refractive index of 1.33 causes incoming light to slow down and compress in wavelength within the medium. This leads to the adjustment of the wavelength as \( \lambda_n = \frac{\lambda}{n} \), where \( \lambda \) is the original vacuum wavelength.

The importance of refractive index in thin film interference cannot be overstated. It directly influences the calculation of the conditions under which constructive or destructive interference will occur, and thus determines which wavelengths are reinforced or suppressed to cause the observed color patterns.
Visible Spectrum
The visible spectrum is the portion of the electromagnetic spectrum that is visible to the human eye, ranging from approximately 380 nm to 750 nm. Within this spectrum, different wavelengths correspond to different colors. For example, shorter wavelengths around 380 nm appear violet, while longer wavelengths near 750 nm appear red.

In the context of thin film interference, like with a soap film, the visible spectrum plays a critical role. Light waves within this spectrum are affected by the film's thickness and refractive index, leading to constructive or destructive interference at certain wavelengths. When destructive interference occurs, certain wavelengths (and thus colors) are canceled out, making the film appear dark at those specific wavelengths.

The calculation of destructive interference relies on determining which wavelengths fall within the visible spectrum once they have been adjusted for the film's refractive properties. These calculations determine the wavelengths that are not reflected back to the observer's eyes, causing the film to appear dark where these wavelengths are missing.

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Most popular questions from this chapter

The wavelength of the laser beam used in a compact disc player is 780 \(\mathrm{nm}\) . Suppose that a diffraction grating produces first-order tracking beams that are 1.2 \(\mathrm{mm}\) apart at a distance of 3.0 \(\mathrm{mm}\) from the grating. Estimate the spacing between the slits of the grating.

In a single-slit diffraction pattern, the central fringe is 450 times as wide as the slit. The screen is 18 000 times farther from the slit than the slit is wide. What is the ratio \(\lambda / W,\) where \(\lambda\) is the wavelength of the light shining through the slit and \(W\) is the width of the slit? Assume that the angle that locates a dark fringe on the screen is small, so that \(\sin \theta \approx \tan \theta\)

A diffraction grating is 1.50 cm wide and contains 2400 lines. When used with light of a certain wavelength, a third-order maximum is formed at an angle of \(18.0^{\circ} .\) What is the wavelength (in nm)?

You are standing in air and are looking at a flat piece of glass \((n=1.52)\) on which there is a layer of transparent plastic \((n=1.61)\) . Light whose wavelength is 589 nm is vacuum is incident nearly perpendicularly on the coated glass and reflects into your eyes. The layer of plastic looks dark. Find the two smallest possible nonzero values for the thickness of the layer.

As Section 17.3 discusses, high-frequency sound waves exhibit less diffraction than low-frequency sound waves do. However, even highfrequency sound waves exhibit much more diffraction under normal circumstances than do light waves that pass through the same opening. The highest frequency that a healthy ear can typically hear is \(2.0 \times 10^{4} \mathrm{Hz}\) Assume that a sound wave with this frequency travels at 343 m/s and passes through a doorway that has a width of 0.91 m. Determine the angle that locates the first minimum to either side of the central maximum in the diffraction pattern for the sound. This minimum is equivalent to the first dark fringe in a single-slit diffraction pattern for light. (b) Suppose that yellow light (wavelength \(=580 \mathrm{nm}\) in vacuum) passes through a doorway and that the first dark fringe in its diffraction pattern is lougted at the angle determined in part (a). How wide would this hypothetical doorway have to be?
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