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Chapter 28: Problem 86

A thin film of oil \((n=1.30)\) floats on water \((n=1.33) .\) When sunlight is incident at right angles to this film, the only colors that are enhanced by reflection are blue \((458 \mathrm{nm})\) and red \((687 \mathrm{nm})\). Estimate the thickness of the oil film.

Short Answer

Expert verified
The estimated thickness of the oil film is approximately 280 nm.

Step by step solution

01

Understanding Thin Film Interference

The oil film floating on water creates two surfaces where light can reflect: the air-oil surface and the oil-water surface. When light reflects off these surfaces, certain wavelengths (colors) might be enhanced if they undergo constructive interference. The key factors here include the refractive indices and the wavelengths of light involved.
02

Identify the Condition for Constructive Interference

For constructive interference in a thin film, the path difference needs to be a multiple of the wavelength. The general condition of constructive interference for reflected light is \(2nt = m\lambda\), where \(n\) is the refractive index, \(t\) is the thickness of the film, \(\lambda\) is the wavelength in air, and \(m\) is an integer representing the order of interference.
03

Calculate Effective Wavelengths in the Film

Since the light travels through the film, the effective wavelengths \(\lambda'\) inside the oil film are calculated by \(\lambda' = \frac{\lambda}{n}\). For blue light, \(\lambda_{blue} = \frac{458 \ nm}{1.30}\), and for red light, \(\lambda_{red} = \frac{687 \ nm}{1.30}\).
04

Calculate Thickness Using Enhanced Color Wavelengths

The respective conditions are \(2t = m\cdot \frac{458}{1.30}\) for blue and \(2t = m\cdot \frac{687}{1.30}\) for red. We solve these equations to find the thickness. To find an estimate, we can equate these two conditions under the assumption they satisfy nearby orders \((m)\).
05

Find Common Thickness for Both Colors

By starting with a common \(m\) or finding a least thick possibility satisfying both conditions for small integer values, we can approximate:\(\frac{2t}{458/1.30} 鈮� m_1\) and \(\frac{2t}{687/1.30} 鈮� m_2\). Calculating values for small integers leads to a thickness estimation.

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Key Concepts

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

Constructive Interference
When we discuss thin film interference, constructive interference is a crucial concept. It occurs when two or more waves overlap in such a way that they reinforce one another, resulting in a higher amplitude. Within the context of a thin film, like oil floating on water, waves reflecting off both the top and bottom surfaces can interfere.
  • For this interference to be constructive, the path difference between these reflected waves must be an integer multiple of the wavelength of the light.
  • This is mathematically represented as \(2nt = m\lambda\), where \(n\) is the refractive index, \(t\) the thickness of the film, and \(m\) is the interference order (an integer).
Understanding constructive interference helps explain why certain light wavelengths are intensified when passing through a medium, like how we see bright blue and red reflections on the oil film.
Refractive Index
The refractive index, often denoted by \(n\), is a measure of how much the speed of light is reduced inside a medium, compared to its speed in a vacuum. This property is essential when determining the behavior of light as it enters or exits different materials, such as in our thin film problem.
  • When light moves from one medium to another (like air to oil), it changes speed and direction based on the refractive index.
  • In our oil on water scenario, the refractive index of oil is 1.30, while the refractive index of water is 1.33.
By knowing these indices, we can calculate how light behaves, specifically how the phases of light waves change, affecting whether they will produce constructive or destructive interference. It's especially useful for calculating the effective wavelength of light within the film.
Wavelength in Film
In thin film interference, understanding the concept of wavelength within the film itself is pivotal. The wavelength of light changes as it moves through mediums with different refractive indices.
  • This effective wavelength in the film is given by \(\lambda' = \frac{\lambda}{n}\), where \(\lambda'\) is the wavelength within the film, \(\lambda\) is the wavelength of light in the air, and \(n\) is the refractive index of the film.
  • With this calculation, we determine how light of different colors reflects at surfaces beneath the film.For instance, blue light with \(\lambda = 458 \, \text{nm}\) has an effective wavelength within the oil film of \(\frac{458 \, \text{nm}}{1.30}\), and red with \(\lambda = 687 \, \text{nm}\) becomes \(\frac{687 \, \text{nm}}{1.30}\).
Knowing these wavelengths allows us to figure out which colors of light are enhanced due to constructive interference, leading us to estimate the thickness of the film accurately.

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Most popular questions from this chapter

\- The headlights of a pickup truck are \(1.32 \mathrm{m}\) apart. What is the greatest distance at which these headlights can be resolved as separate points of light on a photograph taken with a camera whose aperture has a diameter of \(12.5 \mathrm{mm} ?\) (Take \(\lambda=555 \mathrm{nm} .)\)

Green light \((\lambda=546 \mathrm{nm})\) strikes a single slit at normal incidence. What width slit will produce a central maximum that is \(2.50 \mathrm{cm}\) wide on a screen \(1.60 \mathrm{m}\) from the slit?

A set of parallel slits for optical interference can be made by holding two razor blades together (carefully!) and scratching a pair of lines on a glass microscope slide that has been painted black. When monochromatic light strikes these slits at normal incidence, an interference pattern is formed on a distant screen. The thickness of each razor blade used to make the slits is \(0.230 \mathrm{mm}\), and the screen is \(2.50 \mathrm{m}\) from the slits. If the centerto-center separation of the fringes is \(7.15 \mathrm{mm},\) what is the wavelength of the light?

White light reflected at normal incidence from a soap bubble \((n=1.33)\) in air produces an interference maximum at \(\lambda=575 \mathrm{nm}\) but no interference minima in the visible spectrum. (a) Explain the absence of interference minima in the visible. (b) What are the possible thicknesses of the soap film? (Refer to Example \(25-3\) for the range of visible wavelengths.)

A thin soap film \((n=1.33)\) suspended in air has a uniform thickness. When white light strikes the film at normal incidence, violet light \(\left(\lambda_{\mathrm{V}}=420 \mathrm{nm}\right)\) is constructively reflected. (a) If we would like green light \(\left(\lambda_{G}=560 \mathrm{nm}\right)\) to be constructively reflected, instead, should the film's thickness be increased or decreased? (b) Find the new thickness of the film. (Assume the film has the minimum thickness that can produce these reflections.)
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