Wave Optics

85 Thin Film Interference

Learning Objectives

  • Explain how thin-film interference produces colorful patterns.
  • Determine when light undergoes a phase change upon reflection.
  • Describe how film thickness and refractive index affect constructive and destructive interference.
  • Recognize applications of thin-film interference in biology, medicine, and optical technology.

Thin-Film Interference

The brilliant colors seen in soap bubbles, oil floating on water, butterfly wings, and many optical coatings arise from thin-film interference. Unlike the colors produced by pigments, these colors are created because light waves reflected from different surfaces interfere with one another.

When white light strikes a very thin transparent layer, part of the light reflects from the upper surface while the remainder enters the film and reflects from the lower surface. These reflected waves travel slightly different distances before reaching your eye. Depending on their relative phase, they either reinforce each other (constructive interference) or cancel each other (destructive interference).

Because different wavelengths interfere differently, some colors are strengthened while others are suppressed. As the thickness of the film changes, the wavelengths that undergo constructive interference also change, producing the shifting rainbow colors often observed in soap bubbles and oil slicks.

A thin film is generally one whose thickness [latex]t[/latex] is comparable to only a few wavelengths of visible light. Under these conditions, even extremely small changes in thickness produce noticeable changes in color.

Soap bubbles displaying vivid blue, purple, orange, and yellow colors produced by thin-film interference.
Figure 85.1: Soap bubbles display brilliant colors because light reflected from the front and back surfaces of the thin soap film interferes. As the film thickness varies, different wavelengths undergo constructive interference, producing changing colors across the bubble. (Credit: Scott Robinson/Flickr.)

Healthcare Connection

Thin-film interference is widely used in medicine. Anti-reflective coatings on eyeglasses, surgical microscopes, endoscopes, ophthalmic instruments, and camera lenses improve image brightness by reducing unwanted reflections. Similar multilayer coatings are also used in fluorescence microscopes to maximize light transmission and improve image quality.

How Thin-Film Interference Occurs

Figure 85.2 illustrates the two reflected rays responsible for thin-film interference.

  • Ray 1 is reflected from the upper surface of the film.
  • Ray 2 enters the film, reflects from the lower surface, and exits through the top of the film.

Because Ray 2 travels farther than Ray 1, the two rays generally arrive at the observer with different phases. Whether they interfere constructively or destructively depends on two factors:

  1. the additional distance traveled inside the film, and
  2. whether either reflection introduces a phase change.
Light reflecting from the upper and lower surfaces of a thin transparent film. One reflected ray comes from the top surface, while the other travels through the film, reflects from the bottom surface, and exits to interfere with the first ray.
Figure 85.2: Thin-film interference results from the combination of two reflected rays. Ray 1 reflects from the upper surface, while Ray 2 reflects from the lower surface after traveling through the film. Their interference depends on the film thickness, wavelength, and refractive indices of the surrounding materials.

Phase Changes Upon Reflection

Reflection does more than redirect light—it can also change the phase of the wave.

Rule for Phase Changes

When light reflects from a material with a higher index of refraction than the medium in which it is traveling, the reflected wave undergoes a phase shift of [latex]\mathbf{180^\circ}[/latex] (equivalent to one-half wavelength, [latex]\mathbf{\lambda/2}[/latex]).

When light reflects from a material with a lower index of refraction, no phase shift occurs.

This rule is essential for analyzing thin-film interference because the additional half-wavelength phase shift may convert constructive interference into destructive interference, or vice versa.

Why Soap Bubbles Become Dark

A soap bubble consists of a thin layer of soapy water surrounded by air.

At the upper surface, light travels from air into water, so Ray 1 reflects from a material with a higher refractive index and undergoes a phase shift of [latex]\lambda/2[/latex].

Ray 2 reflects from the lower water–air boundary, where light reflects from a higher-index material toward a lower-index material. In this case, no phase shift occurs.

Near the top of a soap bubble, where the film becomes extremely thin, the extra distance traveled by Ray 2 is essentially zero. The only difference between the rays is therefore the half-wavelength phase shift produced during reflection.

As a result, the reflected waves cancel each other for nearly every visible wavelength, producing destructive interference. This is why the thinnest regions of a soap bubble appear black immediately before the bubble bursts.

Key Concept

Thin-film interference depends on both:

  • the optical path difference created by the film thickness, and
  • any phase changes introduced when the light reflects from the film's surfaces.

Ignoring either effect leads to incorrect predictions of whether the reflected light will be bright or dark.

The Role of Film Thickness

Film thickness is the second key factor that determines the interference pattern.

For light striking the film perpendicular to its surface, Ray 2 travels approximately

[latex]2t[/latex]

farther than Ray 1, where [latex]t[/latex] is the film thickness.

Inside the film, the wavelength is shorter than it is in air or vacuum because light travels more slowly in the material. The wavelength in the film is

[latex]\lambda_n=\frac{\lambda}{n},[/latex]

where

  • [latex]\lambda[/latex] is the wavelength in vacuum (or approximately in air),
  • [latex]n[/latex] is the refractive index of the film, and
  • [latex]\lambda_n[/latex] is the wavelength inside the film.

Constructive or destructive interference occurs whenever the total phase difference between the two reflected rays corresponds to an integer or half-integer number of wavelengths. Which condition applies depends on both the film thickness and the phase changes that occur upon reflection.

Everyday Examples

Thin-film interference produces many familiar optical effects, including:

  • rainbow colors on soap bubbles,
  • iridescent oil films on wet pavement,
  • the changing colors of butterfly wings and some beetles,
  • anti-reflective coatings on eyeglasses and camera lenses, and
  • special optical coatings used in microscopes and medical imaging systems.

Example 85.1: Designing a Non-Reflective Lens Coating Using Thin-Film Interference

Modern cameras, microscopes, telescopes, and medical imaging systems contain many glass lenses. Each air-to-glass interface reflects a small fraction of the incoming light, reducing image brightness and producing unwanted glare or "ghost" images.

To minimize these reflections, manufacturers coat lenses with an extremely thin layer of magnesium fluoride (MgF2). The coating is chosen so that the reflected light waves interfere destructively.

Problem: What is the minimum thickness of a magnesium fluoride coating needed to minimize the reflection of 550-nm light? Assume the coating has an index of refraction of 1.38 and the glass lens has an index of refraction of 1.52.

Strategy

Identify the refractive indices of each material:

  • Air: [latex]n_1=1.00[/latex]
  • Magnesium fluoride coating: [latex]n_2=1.38[/latex]
  • Glass lens: [latex]n_3=1.52[/latex]

Because light reflects from a material with a higher refractive index at both interfaces (air → coating and coating → glass), both reflected rays undergo a phase shift of [latex]\lambda/2[/latex]. Since both reflections experience the same phase shift, these shifts cancel one another when comparing the two reflected rays.

To produce destructive interference, the second reflected ray must therefore travel an additional half wavelength inside the coating.

For light incident perpendicular to the surface, the additional distance traveled inside the film is

[latex]2t.[/latex]

Solution

The condition for destructive interference is

[latex]2t=\frac{\lambda_n}{2},[/latex]

where the wavelength inside the coating is

[latex]\lambda_n=\frac{\lambda}{n_2}.[/latex]

Substituting this expression gives

[latex]2t=\frac{\lambda/n_2}{2}.[/latex]

Solving for the film thickness,

[latex]t=\frac{\lambda}{4n_2}.[/latex]

Substituting the given values,

[latex]\begin{aligned} t&=\frac{550\ \text{nm}}{4(1.38)}\\ &=99.6\ \text{nm}. \end{aligned}[/latex]

Answer: The thinnest magnesium fluoride coating that minimizes the reflection of 550-nm light is

[latex]\boxed{t=99.6\ \text{nm}.}[/latex]

Discussion

Thin-film coatings are most effective when they are as thin as possible because the destructive interference remains effective over a wider range of viewing angles. Although these coatings are commonly called anti-reflective coatings, they cannot eliminate reflections for every wavelength of visible light simultaneously. Instead, they are optimized for one wavelength or a narrow range of wavelengths.

Modern optical systems often use several thin-film layers with different refractive indices. These multilayer coatings greatly reduce reflections across much of the visible spectrum, producing brighter images and improved contrast.

Healthcare Connection

Anti-reflective coatings are essential in many medical devices. They improve image quality in surgical microscopes, endoscopes, ophthalmoscopes, retinal cameras, and fluorescence microscopes by increasing light transmission and reducing glare. Similar coatings are also applied to prescription eyeglasses to reduce distracting reflections and improve vision, especially under bright lighting or while driving at night.

Conditions for Thin-Film Interference

Thin-film interference is strongest when the additional distance traveled by one reflected ray corresponds to a whole-number or half-whole-number multiple of the wavelength inside the film.

For light incident perpendicular to the surface, the extra distance traveled through a film of thickness [latex]t[/latex] is approximately

[latex]2t.[/latex]

The wavelength inside the film is

[latex]\lambda_n=\frac{\lambda}{n},[/latex]

where [latex]\lambda[/latex] is the wavelength in air or vacuum and [latex]n[/latex] is the refractive index of the film.

Possible path differences include whole-number multiples of the wavelength,

[latex]2t=m\lambda_n,\qquad m=0,1,2,3,\ldots[/latex]

and half-whole-number multiples,

[latex]2t=\left(m+\frac{1}{2}\right)\lambda_n,\qquad m=0,1,2,3,\ldots[/latex]

However, these equations alone do not determine whether the interference is constructive or destructive. The phase changes that occur when the rays reflect must also be included.

How to Determine the Interference

  • First, determine whether each reflected ray undergoes a [latex]\lambda/2[/latex] phase shift.
  • Then compare the phase difference caused by reflection with the phase difference caused by the additional path length [latex]2t[/latex].
  • If the waves return in phase, they interfere constructively.
  • If the waves return one-half cycle out of phase, they interfere destructively.

Thin-film interference therefore depends on three main quantities:

  • the thickness of the film,
  • the wavelength of the light, and
  • the refractive indices of the film and surrounding materials.

When white light strikes a film whose thickness varies, different wavelengths undergo constructive interference at different locations. This produces the rainbow patterns seen in soap bubbles, oil films, and other thin layers.

Example 85.2: Multiple Soap-Bubble Thicknesses Produce the Same Color

A soap bubble is surrounded by air and has an index of refraction of [latex]n=1.333[/latex]. It is illuminated with red light of wavelength 650 nm.

  1. Find the three smallest film thicknesses that produce constructive interference in the reflected light.
  2. Find the three smallest film thicknesses that produce destructive interference.

Strategy

For a soap bubble, the refractive indices are approximately

[latex]n_1=n_3=1.00,\qquad n_2=1.333.[/latex]

At the upper surface, light reflects from air toward the higher-index soap film, so Ray 1 undergoes a [latex]\lambda/2[/latex] phase shift.

At the lower surface, light reflects from the soap film toward lower-index air, so Ray 2 undergoes no phase shift.

Because only one reflected ray experiences a phase reversal, the usual interference conditions are reversed:

  • Constructive reflection occurs when the path difference is a half-whole-number multiple of the wavelength in the film.
  • Destructive reflection occurs when the path difference is a whole-number multiple of the wavelength in the film.

The wavelength inside the soap film is

[latex]\lambda_n=\frac{\lambda}{n}.[/latex]

Solution for (a): Constructive Interference

For one phase reversal, constructive interference occurs when

[latex]2t_c=\left(m+\frac{1}{2}\right)\lambda_n,\qquad m=0,1,2,\ldots[/latex]

Equivalently, the first three constructive conditions are

[latex]2t_c=\frac{\lambda_n}{2},\qquad \frac{3\lambda_n}{2},\qquad \frac{5\lambda_n}{2}.[/latex]

First calculate the wavelength in the soap film:

[latex]\lambda_n=\frac{650\ \text{nm}}{1.333}=487.6\ \text{nm}.[/latex]

For the smallest constructive thickness,

[latex]\begin{aligned} 2t_1&=\frac{\lambda_n}{2}\\ t_1&=\frac{\lambda_n}{4}\\ &=\frac{487.6\ \text{nm}}{4}\\ &=121.9\ \text{nm}. \end{aligned}[/latex]
[latex]\boxed{t_1\approx122\ \text{nm}}[/latex]

For the second constructive thickness,

[latex]\begin{aligned} t_2&=\frac{3\lambda_n}{4}\\ &=\frac{3(487.6\ \text{nm})}{4}\\ &=365.7\ \text{nm}. \end{aligned}[/latex]
[latex]\boxed{t_2\approx366\ \text{nm}}[/latex]

For the third constructive thickness,

[latex]\begin{aligned} t_3&=\frac{5\lambda_n}{4}\\ &=\frac{5(487.6\ \text{nm})}{4}\\ &=609.5\ \text{nm}. \end{aligned}[/latex]
[latex]\boxed{t_3\approx610\ \text{nm}}[/latex]

Solution for (b): Destructive Interference

For one phase reversal, destructive interference occurs when

[latex]2t_d=m\lambda_n,\qquad m=0,1,2,\ldots[/latex]

For [latex]m=0[/latex],

[latex]\boxed{t_0=0}[/latex]

This represents the limiting case of an extremely thin bubble. The path difference approaches zero, but the single reflection phase shift keeps the reflected waves out of phase.

For the first nonzero destructive thickness, [latex]m=1[/latex]:

[latex]\begin{aligned} 2t_1&=\lambda_n\\ t_1&=\frac{\lambda_n}{2}\\ &=\frac{487.6\ \text{nm}}{2}\\ &=243.8\ \text{nm}. \end{aligned}[/latex]
[latex]\boxed{t_1\approx244\ \text{nm}}[/latex]

For the next destructive thickness, [latex]m=2[/latex]:

[latex]\begin{aligned} 2t_2&=2\lambda_n\\ t_2&=\lambda_n\\ &=487.6\ \text{nm}. \end{aligned}[/latex]
[latex]\boxed{t_2\approx488\ \text{nm}}[/latex]

Discussion

If the bubble were illuminated only with red light, alternating bright and dark bands would appear as the film thickness increased:

  • dark near [latex]t=0[/latex],
  • bright at 122 nm,
  • dark at 244 nm,
  • bright at 366 nm,
  • dark at 488 nm, and
  • bright at 610 nm.

If the film thickness changed gradually, these bands would be evenly spaced. Under white light, different wavelengths satisfy the interference conditions at different thicknesses, so the bubble displays many colors instead of simple bright and dark red bands.

Pattern to Notice

For a soap bubble with one phase reversal, constructive and destructive thicknesses alternate at intervals of

[latex]\frac{\lambda_n}{4}.[/latex]

This regular spacing explains why a smoothly varying film can produce a sequence of repeated colored or dark bands.

Applications of Thin-Film Interference

Thin-film interference is more than an interesting optical phenomenon—it is used in scientific instruments, biological research, manufacturing, and many everyday technologies.

Interference Between Microscope Slides

A simple demonstration of thin-film interference can be created by placing two microscope slides together so that they touch at one end while remaining slightly separated at the other. The thin wedge of air between the slides gradually increases in thickness from one end to the other.

As light reflects from the two surfaces of the air wedge, the reflected waves interfere. Because the thickness changes smoothly, the interference condition changes continuously along the slides, producing alternating bright and dark regions or, under white light, a repeating sequence of rainbow colors.

At the point where the slides touch, the air layer has essentially zero thickness. Since only one reflected ray undergoes a phase reversal, destructive interference occurs and a dark band is observed.

As the air gap becomes thicker, the interference pattern repeats. Eventually the colored bands become more difficult to distinguish because small changes in viewing angle produce increasingly large changes in the optical path difference.

If monochromatic light is used instead of white light, the rainbow colors disappear and are replaced by regularly spaced bright and dark interference fringes.

Two microscope slides separated by a thin wedge of air. One photograph shows colorful interference bands, while the accompanying diagram illustrates the reflected light rays that create the pattern.
Figure 85.3: (a) Colorful interference fringes appear because the air gap between the microscope slides gradually increases in thickness. Different wavelengths satisfy the conditions for constructive interference at different locations. (b) The diagram shows the reflected rays responsible for the interference pattern in the thin wedge of air.

Healthcare Connection

Glass microscope slides and coverslips are manufactured to extremely high optical quality. Thin-film interference is often used during quality control to verify that these surfaces are flat and uniform. High-quality slides reduce image distortion and improve the performance of bright-field, fluorescence, and confocal microscopes used in biological and medical laboratories.

Newton's Rings

Thin-film interference is also used to test the quality of precision optical components.

When the curved surface of a lens is placed against an extremely flat reference surface, a very thin air film forms between them. Because the thickness of this air layer changes gradually with distance from the point of contact, concentric circular interference fringes appear. These rings are known as Newton's rings.

Each successive ring corresponds to only a one-wavelength change in the thickness of the air gap. Consequently, even extremely small deviations from the desired lens shape become visible. This allows manufacturers to polish lenses and mirrors with extraordinary precision.

If the optical surface were perfectly shaped, the interference rings would disappear because the air gap would match the desired geometry everywhere.

Concentric colored interference rings produced between two optical surfaces separated by a very thin air gap.
Figure 85.4: Newton's rings are produced by interference between light reflected from two closely spaced optical surfaces. The circular fringe pattern reveals tiny variations in the separation between the surfaces and is widely used to evaluate the quality of precision lenses and mirrors. (Credit: Ulf Seifert/Wikimedia Commons.)

Other Examples of Thin-Film Interference

Thin-film interference appears throughout nature and technology.

  • Butterflies and moths: Many species produce brilliant iridescent colors because microscopic layers within their wing scales selectively reinforce certain wavelengths of light.
  • Automotive paints: Multilayer coatings create colors that change depending on the viewing angle, producing the shimmering appearance seen on some vehicles.
  • Security features: Thin-film coatings, diffraction gratings, and holograms are incorporated into banknotes, passports, credit cards, and identification documents to make counterfeiting much more difficult.
  • Optical coatings: Modern cameras, telescopes, microscopes, binoculars, laser systems, and medical imaging instruments rely on multilayer thin films to maximize light transmission and minimize reflections.

Did You Know?

The brilliant colors of many insects are not produced by pigments. Instead, they are structural colors, created by microscopic layers that manipulate light through interference. Because the colors depend on viewing angle, they often appear to shimmer or change as the observer moves.

Healthcare Connection: Biomedical Optics

Thin-film coatings are essential components of modern biomedical optics. Anti-reflective coatings improve image quality in microscopes, endoscopes, retinal cameras, surgical imaging systems, and optical coherence tomography (OCT). Specialized multilayer coatings are also used in fluorescence microscopes to selectively reflect or transmit specific wavelengths, allowing different fluorescent dyes to be imaged simultaneously.

Take-Home Investigation: Exploring Thin-Film Interference

Many everyday objects display thin-film interference or related optical effects. Observe several examples and note how their appearance changes as you change your viewing angle.

Possible examples include:

  • soap bubbles,
  • oil on wet pavement,
  • compact discs or DVDs (diffraction grating),
  • holograms on credit cards or passports, and
  • anti-reflective coatings on eyeglasses.

If you have access to two microscope slides, press them together and place a thin hair or a small strip of paper under one end to create a narrow wedge of air. Observe the interference pattern under white light and compare the colors with those shown in Figure 85.3. Notice how the pattern changes as you tilt the slides or change your viewing angle.

Problem-Solving Strategy for Wave Optics

Wave-optics problems often involve interference, diffraction, or both. The following steps provide a systematic way to analyze these situations.

Step 1: Identify the Optical Phenomenon

Determine whether the problem involves:

  • double-slit interference,
  • a diffraction grating,
  • single-slit diffraction,
  • the diffraction limit of an optical instrument, or
  • thin-film interference.

The physical arrangement and the appearance of the pattern usually indicate which phenomenon is involved.

Step 2: Recognize the Pattern

Double slits and diffraction gratings produce a series of interference maxima. Diffraction gratings generally produce narrower and more sharply defined maxima than double slits.

A single slit produces a broad central maximum with weaker maxima on either side. Resolution problems often involve the Rayleigh criterion.

Step 3: Analyze Thin Films Carefully

For thin-film interference, identify the two reflected rays that interfere and determine their path-length difference.

For perpendicular incidence, the additional physical distance traveled inside a film of thickness [latex]t[/latex] is approximately

[latex]2t.[/latex]

Use the wavelength inside the material:

[latex]\lambda_n=\frac{\lambda}{n}.[/latex]

Also determine whether either reflected ray undergoes a phase shift of [latex]\lambda/2[/latex]. A phase shift occurs when light reflects from a medium with a higher refractive index.

Step 4: Identify the Unknown

Write down exactly what the problem asks you to determine. Common unknowns include wavelength, film thickness, slit spacing, angle, fringe position, or resolving power.

A labeled diagram is often useful, especially when several materials or reflected rays are involved.

Step 5: List the Known Quantities

Record the values given in the problem, including units. For thin films, be sure to identify:

  • the wavelength in air or vacuum,
  • the refractive index of each material,
  • the film thickness, if known, and
  • the angle of incidence.

Step 6: Choose and Rearrange the Appropriate Equation

Select the equation that matches the physical situation and solve it algebraically for the unknown before substituting numerical values.

Step 7: Determine Whether the Interference Is Constructive or Destructive

Compare the waves after including both the path-length difference and any phase changes caused by reflection.

  • Waves that return in phase interfere constructively.
  • Waves that return one-half cycle out of phase interfere destructively.

Do not assume that a whole-wavelength path difference is always constructive. A single phase reversal exchanges the constructive and destructive conditions.

Step 8: Check the Result

Confirm that the answer is physically reasonable. For example:

  • the film thickness should usually be comparable to the wavelength of light,
  • the sine of an angle cannot exceed 1,
  • an interference angle cannot exceed [latex]90^\circ[/latex], and
  • the units must match the requested quantity.

Section Summary

  • Thin-film interference occurs when light reflected from the upper and lower surfaces of a thin layer combines.
  • The interference depends on the film thickness, the wavelength of light, the refractive indices of the materials, and the angle of incidence.
  • For perpendicular incidence, the second reflected ray travels approximately [latex]2t[/latex] farther than the first ray.
  • The wavelength of light inside a material is
[latex]\lambda_n=\frac{\lambda}{n}.[/latex]
  • When light reflects from a material with a higher refractive index, the reflected wave undergoes a [latex]180^\circ[/latex] phase change, equivalent to a shift of [latex]\lambda/2[/latex].
  • When light reflects from a material with a lower refractive index, no phase change occurs.
  • If neither reflected ray or both reflected rays undergo a phase reversal, the reflection phase shifts do not create a relative phase difference.
  • If only one reflected ray undergoes a phase reversal, the usual constructive and destructive interference conditions are exchanged.
  • White light produces colored thin-film patterns because different wavelengths interfere constructively at different film thicknesses.
  • Thin-film interference is used in anti-reflective coatings, microscopes, optical testing, security features, and biomedical imaging systems.

Conceptual Questions

  1. Two glass slides form a wedge-shaped air gap. What happens to the spacing between neighboring interference fringes if the wedge angle is increased? Why might the fringes become impossible to distinguish if the angle becomes too large?
  2. Two light waves begin in phase but travel different distances before reaching the same point. Explain how their path-length difference determines whether they interfere constructively or destructively. How can reflection or refraction change your conclusion?
  3. A contact lens with an index of refraction of approximately 1.50 floats on a tear layer. The upper surface of the lens is exposed to air. At which surface or surfaces does reflected light undergo a [latex]180^\circ[/latex] phase shift? Explain your reasoning by comparing refractive indices.
  4. A biological sample is placed in a drop of water on a glass microscope slide and covered with a glass coverslip. Light shines downward from air. Reflections can occur at the upper and lower surfaces of the coverslip and at the water–slide boundary. At which reflections does a phase reversal occur?
  5. Repeat Question 24 if the liquid between the two pieces of glass is carbon disulfide, whose index of refraction is greater than that of crown glass. How does changing the liquid alter the phase shifts?
  6. A thin moist layer on the surface of a piece of meat displays rainbow colors under white light. Explain how the colors can be produced even if the liquid itself is nearly colorless.
  7. The thinnest region of a soap bubble appears dark because nearly all visible wavelengths undergo destructive interference. Could the same idea be used to create a single-layer lens coating that eliminates reflection at every visible wavelength? Discuss the required thickness and refractive index and explain why such a coating would be difficult or impossible to produce.
  8. The anti-reflective coating in Example 85.1 is optimized for one wavelength at perpendicular incidence. Describe what happens when light of a different wavelength strikes the coating or when light arrives at an angle.
  9. Why are interference fringes generally easier to observe from a thin film than from a thick sheet of glass? Would using monochromatic light make fringes from thick glass easier to observe? Explain.

Problems & Exercises

  1. A soap bubble is 100 nm thick and illuminated by white light incident perpendicular to its surface. What wavelength and color of visible light is most constructively reflected, assuming the same index of refraction as water?
  2. An oil slick on water is 120 nm thick and illuminated by white light incident perpendicular to its surface. What color does the oil appear (what is the most constructively reflected wavelength), given its index of refraction is 1.40?
  3. Calculate the minimum thickness of an oil slick on water that appears blue when illuminated by white light perpendicular to its surface. Take the blue wavelength to be 470 nm and the index of refraction of oil to be 1.40.
  4. Find the minimum thickness of a soap bubble that appears red when illuminated by white light perpendicular to its surface. Take the wavelength to be 680 nm, and assume the same index of refraction as water.
  5. A film of soapy water ([latex]n=1\text{.}\text{33}[/latex]) on top of a plastic cutting board has a thickness of 233 nm. What color is most strongly reflected if it is illuminated perpendicular to its surface?
  6. What are the three smallest non-zero thicknesses of soapy water ([latex]n=1\text{.}\text{33}[/latex]) on Plexiglas if it appears green (constructively reflecting 520-nm light) when illuminated perpendicularly by white light? Explicitly show how you follow the steps in Problem Solving Strategies for Wave Optics.
  7. Suppose you have a lens system that is to be used primarily for 700-nm red light. What is the second thinnest coating of fluorite (magnesium fluoride) that would be non-reflective for this wavelength?
  8. (a) As a soap bubble thins it becomes dark, because the path length difference becomes small compared with the wavelength of light and there is a phase shift at the top surface. If it becomes dark when the path length difference is less than one-fourth the wavelength, what is the thickest the bubble can be and appear dark at all visible wavelengths? Assume the same index of refraction as water. (b) Discuss the fragility of the film considering the thickness found.
  9. A film of oil on water will appear dark when it is very thin, because the path length difference becomes small compared with the wavelength of light and there is a phase shift at the top surface. If it becomes dark when the path length difference is less than one-fourth the wavelength, what is the thickest the oil can be and appear dark at all visible wavelengths? Oil has an index of refraction of 1.40.
  10. Figure 85.3 shows two glass slides illuminated by pure-wavelength light incident perpendicularly. The top slide touches the bottom slide at one end and rests on a 0.100-mm-diameter hair at the other end, forming a wedge of air. (a) How far apart are the dark bands, if the slides are 7.50 cm long and 589-nm light is used? (b) Is there any difference if the slides are made from crown or flint glass? Explain.
  11. Figure 85.3 shows two 7.50-cm-long glass slides illuminated by pure 589-nm wavelength light incident perpendicularly. The top slide touches the bottom slide at one end and rests on some debris at the other end, forming a wedge of air. How thick is the debris, if the dark bands are 1.00 mm apart?
  12. Repeat problem 1, but take the light to be incident at a [latex]\text{45º}[/latex] angle.
  13. Repeat problem 2, but take the light to be incident at a [latex]\text{45º}[/latex] angle.
  14. Unreasonable Results To save money on making military aircraft invisible to radar, an inventor decides to coat them with a non-reflective material having an index of refraction of 1.20, which is between that of air and the surface of the plane. This, he reasons, should be much cheaper than designing Stealth bombers. (a) What thickness should the coating be to inhibit the reflection of 4.00-cm wavelength radar? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

Glossary

thin film interference
interference between light reflected from different surfaces of a thin film
definition

License

Icon for the Creative Commons Attribution 4.0 International License

Introductory Physics for the Health and Life Sciences II Copyright © 2012 by OSCRiceUniversity is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

Share This Book