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Chapter 27: Problem 19

Orange light \(\left(\lambda_{\text { vacuum }}=611 \text { nm) shines on a soap film }\right.\) \((n=1.33)\) that has air on either side of it. The light strikes the film perpendicularly. What is the minimum thickness of the film for which constructive interference causes it to look bright in reflected light?

Short Answer

Expert verified
The minimum thickness for constructive interference is approximately 114.9 nm.

Step by step solution

01

Understanding the Problem

We need to find the minimum thickness of a soap film that will cause constructive interference with orange light when it reflects off the film. The soap film has a refractive index of 1.33 and is surrounded by air.
02

Identify Conditions for Constructive Interference

For constructive interference in thin films, the condition is a path difference of multiple wavelengths. This can be represented as \( 2t = (m+\frac{1}{2}) \frac{\lambda}{n} \) for minimum thickness, where \( t \) is the thickness of the film, \( m \) is an integer representing the order of interference, and \( n \) is the refractive index.
03

Solve for Minimum Thickness

We start with the smallest non-zero value for \( m \, (m = 0)\), which gives us the minimum thickness for constructive interference. Rewriting the formula for \( t \), we have:\[ 2t = (0+\frac{1}{2}) \frac{\lambda}{n} \]\[ 2t = \frac{1}{2} \times \frac{611 \text{ nm}}{1.33} \]\[ 2t = \frac{305.5 \text{ nm}}{1.33} \]\[ t = \frac{152.75 \text{ nm}}{1.33} \]\[ t \approx 114.9 \text{ nm} \]
04

Conclusion

The minimum thickness required for constructive interference, causing the soap film to appear bright, is approximately 114.9 nm.

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Key Concepts

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

Thin Film Interference
Thin film interference is a fascinating optical phenomenon that occurs when light waves reflect off surfaces of a thin film. In this situation, the film is thinner than the coherence length of the light source, meaning the light can interfere with itself. An everyday example of this is seen in soap bubbles or oil films on water, which often display bright, iridescent colors.
Here鈥檚 how it works: light hits the film, and part of it is reflected back into the air, while another part refracts inside, reflects off the back surface, and eventually exits the film.
This creates two sets of waves鈥攐ne from the top surface and another from the bottom surface of the film. Depending on the thickness of the film and the wavelength of the incoming light, these waves interfere constructively or destructively.
  • Constructive interference occurs when the path difference between the waves results in them combining to amplify the overall intensity of the reflected light.
  • Destructive interference occurs when the path difference results in them canceling each other out, reducing the intensity.
The exact nature of this interference depends on several factors: the thickness of the film, the wavelength of the incident light, and the refractive indices of the media involved.
Refractive Index
The refractive index is a measure of how much a material can bend light. Represented as 'n', it compares the speed of light in a vacuum to the speed of light in that material. For instance, if a material has a refractive index of 1.33, it means that light travels 1.33 times slower in that material compared to in a vacuum.
This property is crucial in thin film interference as it affects how light waves change speed and direction when entering the film. When light enters a material of higher refractive index from a material of lower refractive index (like air to soap film), it slows down and bends, which can cause a change in the path length of the light wave.
In our exercise, the soap film has a refractive index of 1.33, meaning that light waves are significantly slowed down and bent within the film. This change influences the wavelength of light in the film and, subsequently, the type of interference perceived.
  • Higher refractive index means greater bending of light.
  • Changes the effective wavelength of light within the medium, essential for calculating interference conditions.
Constructive Interference
Constructive interference happens when multiple waves combine to form a wave of greater amplitude. In optics, it leads to processes where brighter regions are formed in the pattern of reflected or transmitted light.
For thin films, constructive interference specifically occurs when the path difference between the two light beams reflects off the top and bottom surfaces of the film, arriving in phase with each other. This phase alignment leads to the amplification of the wave.
Mathematically, it is represented by the equation: \[ 2t = (m+\frac{1}{2}) \frac{\lambda}{n} \]where:
  • \( t \) is the thickness of the film
  • \( m \) is the order of interference (an integer)
  • \( \lambda \) is the wavelength in vacuum
  • \( n \) is the refractive index of the film
For the minimum constructive interference in a soap film reflecting orange light, the smallest non-zero \( m \) is selected, such that the calculation leads to the minimum thickness required for the film to appear bright. This is the foundation of understanding how various colors can be enhanced or diminished due to the interference of light in thin films.

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Most popular questions from this chapter

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)?

The central bright fringe in a single-slit diffraction pattern has a width that equals the distance between the screen and the slit. Find the ratio \(\lambda / W\) of the wavelength \(\lambda\) of the light to the width \(W\) of the slit.

Astronomers have discovered a planetary system orbiting the star Upsilon Andromedae, which is at a distance of \(4.2 \times 10^{17} \mathrm{m}\) from the earth. One planet is believed to be located at a distance of \(1.2 \times 10^{11} \mathrm{m}\) from the star. Using visible light with a vacuum wavelength of \(550 \mathrm{nm},\) what is the minimum necessary aperture diameter that a telescope must have so that it can resolve the planet and the star?

A hunter who is a bit of a braggart claims that from a distance of 1.6 km he can selectively shoot either of two squirrels who are sitting ten centimeters apart on the same branch of a tree. What鈥檚 more, he claims that he can do this without the aid of a telescopic sight on his rifle. (a) Determine the diameter of the pupils of his eyes that would be required for him to be able to resolve the squirrels as separate objects. In this calculation use a wavelength of 498 nm (in vacuum) for the light. (b) State whether his claim is reasonable, and provide a reason for your answer. In evaluating his claim, consider that the human eye automatically adjusts the diameter of its pupil over a typical range of 2 to 8 mm, the larger values coming into play as the lighting becomes darker. Note also that under dark conditions, the eye is most sensitive to a wavelength of 498 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.
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