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Chapter 9: Problem 27

A thin film of ethyl alcohol \((n=1.36)\) spread on a flat glass plate and illuminated with white light shows a color pattern in reflection. If a region of the film reflects only green light \((500 \mathrm{nm})\) strongIy, how thick is it?

Short Answer

Expert verified
The film thickness is approximately 92 nm.

Step by step solution

01

Understand Thin Film Interference

The thin film interference occurs due to the phase change that happens when light reflects off surfaces. Constructive interference happens for a specific wavelength when the path length difference is a multiple of the wavelength.
02

Set Up Equation for Constructive Interference

For constructive interference in reflected light, the equation is \( 2nt = (m + \frac{1}{2}) \lambda \), where \( n \) is the refractive index, \( t \) is the thickness of the film, \( m \) is the order of interference, and \( \lambda \) is the wavelength.
03

Insert Known Values

Given \( n = 1.36 \), \( \lambda = 500 \) nm. For the smallest non-zero thickness, assume \( m = 0 \). Substituting, we have: \[ 2 \times 1.36 \times t = (0 + \frac{1}{2}) \times 500 \text{ nm} \].
04

Solve for Thickness \( t \)

Simplify the equation: \[ 2.72t = 250 \text{ nm} \]. Dividing both sides by 2.72 gives \[ t = \frac{250}{2.72} \approx 91.91 \text{ nm} \].
05

Round off the Answer

Round the calculated thickness to the nearest whole number. Thus, \( t \approx 92 \text{ nm} \).

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Key Concepts

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

Constructive Interference
Constructive interference is a fascinating phenomenon that occurs when two or more waves superpose to form a resultant wave of greater amplitude. In the context of thin film interference, this happens when light waves reflect off the upper and lower boundaries of a thin film, like a drop of oil on water or, as in our problem, a film of ethyl alcohol on glass.
\[ \] This occurs because the reflected light waves have traveled different path lengths. If the difference in path for two waves is a whole number plus a half wavelength, these waves will meet in-phase and amplify each other. This causes the particular wavelengths to appear prominently, which is observed as the colorful pattern on the film.
\[ \] In our exercise, only the green color is observed, indicating that the condition for constructive interference is satisfied for that specific wavelength (500 nm). By understanding this, we can use it to predict or calculate other properties of the thin film, such as its thickness.
Refractive Index
The refractive index, denoted as \( n \), is a measure of how much the speed of light is reduced inside a medium compared to vacuum. It is crucial in problems involving light and optics because it affects how light propagates through materials.
\[ \] In our thin film example, ethyl alcohol has a refractive index of 1.36. This tells us that light moves approximately 1.36 times slower in this medium compared to a vacuum. When light enters a medium of different refractive index, its speed and wavelength change, but its frequency remains constant.
\[ \] The refractive index influences the path difference of light traveling through the film and hence affects the patterns of interference seen. It is a key parameter in the equation for determining constructive interference, as it helps calculate the change in phase of light as it moves through different media.
Wavelength
Wavelength, represented by the Greek letter \( \lambda \), is the distance between successive peaks (or troughs) of a wave. It is a fundamental property of all wave phenomena, including light waves.
\[ \] In our problem, we're focusing on green light with a wavelength of 500 nm. This wavelength is where the condition for constructive interference is met, causing the green light to be strongly reflected from the thin film. Each wavelength of light corresponds to a different color in the visible spectrum, and changes in the wavelength due to interactions with materials (like thin films) lead to the colorful interference patterns we observe.
\[ \] The wavelength is crucial because it directly relates to the path difference that results in constructive interference. When calculating the required thickness of a film for constructive interference, this wavelength value is a key component of the interference equation. Thus, understanding its role is essential in solving such thin film interference problems.

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Most popular questions from this chapter

Red plane waves from a ruby laser \(\left(\lambda_{0}=694.3 \mathrm{nm}\right)\) in air impinge on two parallel slits in an opaque screen. A fringe pattern forms on a distant wall, and we see the fourth bright band \(1.0^{\circ}\) above the central axis. Kindly calculate the separation between the slits.

Imagine that you have an opaque screen with three horizontal very narrow parallel slits in it. The second slit is a center-to-center distance \(a\) beneath the first, and the third is a distance \(5 a / 2\) beneath the first. (a) Write a complex exponential expression in terms of \(\delta\) for the amplitude of the electric field at some point \(P\) at an elevation \(\theta\) on a distant screen where \(\delta=k a \sin \theta .\) Prove that $$I(\theta)=\frac{I(0)}{3}+\frac{2 I(0)}{9}(\cos \delta+\cos 3 \delta / 2+\cos 5 \delta / 2)$$ Verify that at \(\theta=0, I(\theta)=I(O)\)

A \(3 \times 5\) card containing two pinholes, \(0.08 \mathrm{mm}\) in diameter and separated center to center by \(0.10 \mathrm{mm}\), is illuminated by parallel rays of blue light from an argon ion laser \(\left(\lambda_{0}=487.99 \mathrm{nm}\right) .\) If the fringes on an observing screen are to be \(10 \mathrm{mm}\) apart, how far away should the screen be?

Using Lloyd's mirror, \(X\) -ray fringes were observed, the spacing of which was found to be \(0.0025 \mathrm{cm} .\) The wavelength used was 8.33 脌. If the source-screen distance was \(3 \mathrm{m}\), how high above the mirror plane was the point source of X-rays placed?

Suppose a wedge-shaped air film is made between two sheets of glass, with a piece of paper \(7.618 \times 10^{-5} \mathrm{m}\) thick used as the spacer at their very ends. If light of wavelength 500 nm comes down from directly above, determine the number of bright fringes that will be seen across the wedge.
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