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Chapter 26: Problem 16

\(\bullet\) What is the thinnest soap film (excluding the case of zero thickness) that appears black when viewed by reflected light with a wavelength of 480 \(\mathrm{nm}\) ? The index of refraction of the film is \(1.33,\) and there is air on both sides of the film.

Short Answer

Expert verified
The thinnest film thickness is approximately 90 nm.

Step by step solution

01

Understand the Problem

The problem asks for the thinnest possible soap film thickness that appears black when illuminated with light of wavelength 480 nm. This happens due to destructive interference of the reflected light rays.
02

Recall the Conditions for Destructive Interference

Destructive interference occurs for soap films when the path difference between the two reflected light rays is equal to an odd multiple of half the wavelength of the light in the medium (the soap film).
03

Calculate the Wavelength in the Film

The wavelength of light in the film is less than in a vacuum, calculated as \( \lambda_{film} = \frac{\lambda_0}{n} \) where \( \lambda_0 \) is the wavelength in air (480 nm) and \( n \) is the refractive index (1.33). So, \( \lambda_{film} = \frac{480}{1.33} \approx 360.9 \ nm \).
04

Apply the Condition for Minimum Thickness

The thinnest film for destructive interference (black appearance) is when the film thickness \( t \) is \( \frac{\lambda_{film}}{4} \) due to an extra half-wavelength phase shift on reflection. Hence, \( t = \frac{360.9}{4} \approx 90.22 \ nm \).
05

Conclusion

Thus, the thinnest soap film that appears black has a thickness of approximately 90 nm.

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Key Concepts

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

Destructive Interference
Destructive interference occurs when two waves combine to produce a smaller amplitude. In the case of thin films, such as a soap film, this effect creates colors or makes the film appear black. This happens when the path difference between the reflected rays is an odd multiple of half the wavelength. So, you ensure that the waves from the top and bottom surfaces of the film cancel each other out.
This effect is essential because it explains why certain colors vanish when observed at particular angles.
When viewing the soap film, if the thickness is just right, the path difference causes the light of specific wavelengths to destructively interfere, leading to the disappearance of that color.
Soap Film Thickness
Soap films are fascinating layers of liquid, often showing vibrant colors due to their thinness. The thickness of a soap film determines which colors are visible or vanish. It results from the interplay of light within the film.
In the exercise, we calculate the minimum thickness at which the film appears black by finding when destructive interference occurs. For the given wavelength of 480 nm in air and the refractive index, we find that the film should be about 90 nm thick.
Typically, the soap film sits between two layers of air, and as light reflects off its surfaces, the conditions for interference become evident. Even a small change in thickness can dramatically change the visual appearance of the film.
Wavelength in a Medium
The wavelength of light changes as it travels from one medium into another. This alteration is due to the change in the speed of light as it enters a different medium. The equation \( \lambda_{\text{film}} = \frac{\lambda_0}{n} \) helps determine this new wavelength, where \( \lambda_0 \) is the wavelength in air, and \( n \) is the refractive index of the medium.
For the given exercise, the wavelength in the soap film, calculated from a wavelength of 480 nm in air and a refractive index of 1.33, becomes 360.9 nm. Recognizing how to calculate this change is fundamental when studying interference patterns within thin films.
Refractive Index
The refractive index, represented by \( n \), quantifies how much light slows as it enters a medium. Materials with higher refractive indices bend light more. The refractive index influences the wavelength within the medium, and by extension, the interference patterns observed.
In the exercise, the soap film has a refractive index of 1.33. This means that, compared to the speed of light in air, light travels slower within the film. Calculating how this slowing affects the wavelength is critical for determining film appearances, as seen with the calculated wavelength of about 360.9 nm for light initially measured at 480 nm in air.
Understanding the refractive index allows one to predict how materials will behave under light and why thin films may show varying interference colors.

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Most popular questions from this chapter

\( (f鈥� # \)\bullet$$\bullet\( Two small loudspeakers that are 5.50 \)\mathrm{m}\( apart are emitting sound in phase. From both of them, you hear a singer singing \)\mathrm{C} \\#\( (frequency 277 \)\mathrm{Hz} )\( , while the speed of sound in the room is 340 \)\mathrm{m} / \mathrm{s}\( . Assuming that you are rather far from these speak- ers, if you start out at point \)P\( equidistant from both of them and walk around the room in front of them, at what angles (measured relative to the fine from \)P$ to the midpoint between the speakers) will you hear the sound (a) maximally enhanced, (b) cancelled? Neglect any reflections from the walls.

\(\bullet$$\bullet\) A wildlife photographer uses a moderate telephoto lens of focal length 135 \(\mathrm{mm}\) and maximum aperture \(f / 4.00\) to photo- graph a bear that is 11.5 \(\mathrm{m}\) away. Assume the wavelength is 550 \(\mathrm{nm}\) . (a) What is the width of the smallest feature on the bear that this lens can resolve if it is opened to its maximum aperture? (b) If, to gain depth of field, the photographer stops the lens down to \(f / 22.0,\) what would be the width of the small- est resolvable feature on the bear"?

\(\bullet$$\bullet\) Resolution of the eye, I. Even if the lenses of our eyes functioned perfectly, our vision would still be limited due to diffraction of light at the pupil. Using Rayleigh's criterion, what is the smallest object a person can see clearly at his near point of 25.0 \(\mathrm{cm}\) with a pupil 2.00 \(\mathrm{mm}\) in diameter and light of wavelength 550 \(\mathrm{nm}\) ? (To get a reasonable estimate without having to go through complicated calculations, we'll ignore the effect of the fluid in the eye.) Based upon your answer, does it seem that diffraction plays a significant role in limiting our visual acuity?

\(\bullet\) A diffraction grating has 5580 lines/cm. When a beam of monochromatic light goes through it, the second pair of bright spots occurs at 26.3 \(\mathrm{cm}\) from the central spot on a screen 42.5 \(\mathrm{cm}\) past the grating. (a) What is the wavelength of this light? (b) How far from the central spot does the next pair of bright spots occur on the screen?

\(\bullet\) A person with a radio-wave receiver starts out equidistant from two FM radio transmitters \(A\) and \(B\) that are 11.0 \(\mathrm{m}\) apart, each one emitting in-phase radio waves at 92.0 \(\mathrm{MH} z\) . She then walks so that she always remains 50.0 \(\mathrm{m}\) from transmitter \(B\) . See Fig- ure \(26.48 .\) ) For what values of \(x\) will she find the radio signal to be (a) maximally enhanced, (b) can- celled? Limit your solution to the cases where \(x \geq 50.0 \mathrm{m.}\)
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