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Chapter 35: Problem 30

A researcher measures the thickness of a layer of benzene (\(n\) = 1.50) floating on water by shining monochromatic light onto the film and varying the wavelength of the light. She finds that light of wavelength 575 nm is reflected most strongly from the film. What does she calculate for the minimum thickness of the film?

Short Answer

Expert verified
The minimum thickness of the benzene film is approximately 191.67 nm.

Step by step solution

01

Understand the Phenomenon

The problem involves thin film interference. When light reflects off the top and bottom surfaces of a thin film, constructive or destructive interference occurs based on the thickness of the film and the wavelength of light. In this problem, we are dealing with constructive interference for maximum reflection.
02

Apply the Condition for Constructive Interference

For constructive interference in a thin film where the film is on a denser medium, the formula is given by \(2nt = m \lambda\), where \(n\) is the refractive index of the film, \(t\) is the thickness, \(m\) is the order of interference (an integer), and \(\lambda\) is the wavelength of light in a vacuum.
03

Consider Minimum Thickness and Simplify

For the minimum thickness that gives maximum reflection, consider the first order of interference \(m = 1\). This gives \(2nt = \lambda\). Solving for \(t\), the thickness, we have \[t = \frac{\lambda}{2n}\].
04

Plug in Given Values

Using the given wavelength \(\lambda = 575\, \text{nm} = 575 \times 10^{-9} \text{m}\) and the refractive index \(n = 1.50\), substitute into the formula: \[t = \frac{575 \times 10^{-9}}{2 \times 1.50}\].
05

Calculate the Thickness

Calculate \[t = \frac{575 \times 10^{-9}}{3} \approx 191.67 \times 10^{-9} \text{m}\]. Therefore, the thickness \(t\) is approximately \(191.67\, \text{nm}\).

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Key Concepts

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

Constructive Interference
When light hits a thin film, like the benzene floating on water in the problem, it can undergo constructive interference. Imagine light waves as ripples on a pond; when two identical waves meet, they can pile up to make a bigger ripple. This is constructive interference.

In constructive interference, the reflected light waves from the top and bottom surfaces of the film align perfectly. This alignment strengthens the light waves. The key is getting the waves "in phase," which means their peaks and troughs match up. For this to happen, certain conditions like the thickness of the film and the wavelength of light must be met.

In our exercise, because the light of wavelength 575 nm is reflected most strongly, we know constructive interference occurred. The formula, \[2nt = m \lambda\] helps us determine the right conditions. Here, "\( n \)" is the refractive index, "\( t \)" is the thickness of the film, and "\( m\)" is the interference order, usually starting with 1 for minimum thickness.
Refractive Index
The refractive index shows us how much light slows as it enters a material. Think of it as comparing running on a smooth path versus through mud; one is much slower. This property of materials affects how light behaves at the boundary, where the light transitions between materials, like air to benzene.

In our example, benzene has a refractive index of 1.50. This value tells us how much the speed of light is reduced in benzene compared to air or vacuum. The higher this number, the more it bends or slows the light.

When light strikes the benzene film, part of it reflects off the surface, while the rest travels through, reflects off the backside, and then exits the film. The refractive index is crucial because it affects the path length within the film, which in turn influences interference patterns. The thickness calculation depends on knowing how different the medium's refractive index is from the surrounding areas.
Wavelength of Light
The wavelength of light is a key player in interference phenomena. Wavelength refers to the distance between consecutive peaks of a light wave. It determines the color of the light; for instance, 575 nm (nanometers) that we see in the problem reflects a specific hue.

This characteristic is crucial for calculating interference in thin films because it defines how the light waves add or cancel out. In our exercise, light of 575 nm wavelength undergoes perfect reinforcement at a certain film thickness, leading to maximum brightness.

For thin film interference calculations, the wavelength in the medium is not the same as in air. It shortens based on the refractive index. Knowing the wavelength helps us use the equation \[t = \frac{\lambda}{2n}\] to figure out how the thickness contributes to constructive interference. In this case, calculating the minimum film thickness requires transforming the wavelength when the refractive index is considered. This results in optimal conditions for the waves to reinforce each other.

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Most popular questions from this chapter

Eyeglass lenses can be coated on the \(inner\) surfaces to reduce the reflection of stray light to the eye. If the lenses are medium flint glass of refractive index 1.62 and the coating is fluorite of refractive index 1.432, (a) what minimum thickness of film is needed on the lenses to cancel light of wavelength 550 nm reflected toward the eye at normal incidence? (b) Will any other wavelengths of visible light be cancelled or enhanced in the reflected light?

In your research lab, a very thin, flat piece of glass with refractive index 1.40 and uniform thickness covers the opening of a chamber that holds a gas sample. The refractive indexes of the gases on either side of the glass are very close to unity. To determine the thickness of the glass, you shine coherent light of wavelength \(\lambda_0\) in vacuum at normal incidence onto the surface of the glass. When \(\lambda_0\) = 496 nm, constructive interference occurs for light that is reflected at the two surfaces of the glass. You find that the next shorter wavelength in vacuum for which there is constructive interference is 386 nm. (a) Use these measurements to calculate the thickness of the glass. (b) What is the longest wavelength in vacuum for which there is constructive interference for the reflected light?

What is the thinnest film of a coating with \(n\) = 1.42 on glass (\(n\) = 1.52) for which destructive interference of the red component (650 nm) of an incident white light beam in air can take place by reflection?

Two slits spaced 0.260 mm apart are 0.900 m from a screen and illuminated by coherent light of wavelength 660 nm. The intensity at the center of the central maximum (\(\theta\) = 0\(^\circ\)) is \(I_0\) . What is the distance on the screen from the center of the central maximum (a) to the first minimum; (b) to the point where the intensity has fallen to \(I_0\) /2?

A compact disc (CD) is read from the bottom by a semiconductor laser with wavelength 790 nm passing through a plastic substrate of refractive index 1.8. When the beam encounters a pit, part of the beam is reflected from the pit and part from the flat region between the pits, so these two beams interfere with each other (Fig. E35.31). What must the minimum pit depth be so that the part of the beam reflected from a pit cancels the part of the beam reflected from the flat region? (It is this cancellation that allows the player to recognize the beginning and end of a pit.)
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