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A soap bubble \((n=1.93)\) having a wall thickness of \(120 \mathrm{~nm}\) is floating in air. (a) What is the wavelength of the visible light that is most strongly reflected? (b) Explain how a bubble of different thickness could also strongly reflect light of this same wavelength. (c) Find the two smallest film thicknesses larger than the one given that can produce strongly reflected light of this same wavelength.

Short Answer

Expert verified
The wavelength of the most strongly reflected visible light is around \(928 \mathrm{~nm}\). A bubble of a different thickness can reflect the same wavelength by being an integer multiple of half the wavelength of that light. The two smallest film thicknesses larger than \(120 \mathrm{~nm}\) that can produce strongly reflected light of this same wavelength are roughly \(240 \mathrm{~nm}\) and \(360 \mathrm{~nm}\).

Step by step solution

01

Calculation of the Wavelength of Most Strongly Reflected Light

According to the thin-film refraction theory, a light wave reflects strongest when the thickness of the film is an integer number of half wavelengths of the incident light. So, using the formula \(2nt=mλ\), where \(n\) is the refractive index, \(t\) is the thickness of the film, \(m\) is an integer representing the order of interference, and \(λ\) is the wavelength of incident light, we can rearrange the formula to solve for \(λ\), hence: \(λ=\frac{2nt}{m}\). For condition of maximum reflection, constructive interference, we make \(m=1\) (first order), hence \(λ=4*120 nm*1.93\).
02

Explaining How A Bubble of Different Thickness Could Reflect The Same Wavelength

A bubble of a different thickness could reflect the same wavelength of light if the thickness of the bubble is such that it is an integer multiple of half the wavelength of the light. This happens because the constructive interference occurs when the path difference between two rays reflected by the thin film is an integer multiple of the wavelength. Hence, even if the thickness is different, as long as it allows the path difference to be an integer multiple of the wavelength, the same light will be strongly reflected.
03

Finding The Two Smallest Film Thicknesses Larger Than The One Given

We can find the two smallest film thicknesses larger than the one given using the same formula \(2nt=mλ\), where m is an integer, which we will increase in order to get larger film thicknesses that still cause constructive interference for the same light wavelength. By setting \(m=2\) and \(m=3\) in the equation, we will get the two next thickness which will reflect light of this same wavelength.

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Key Concepts

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

Constructive Interference
Constructive interference occurs when waves superimpose on each other in such a way that their crests (the highest points of the waves) and troughs (the lowest points) align to produce a wave of greater amplitude. This phenomenon is essential in understanding how thin films, like soap bubbles, create brilliant patterns of color.

To visualize this, imagine two light waves traveling through air and then hitting the surface of a soap bubble. One wave reflects off the top surface, while another enters the bubble, reflects off the back side, and exits the film. If the thickness of the soap film is just right, these two waves emerge in phase and their peaks and troughs align perfectly, resulting in a burst of color known as constructive interference. This effect is precisely what enables us to see vibrant colors in bubbles and oil films on water.

The condition for constructive interference in thin films depends on the film's thickness, the wavelength of the incident light, and the refractive index of the film material. Notably, the phase of light can change upon reflection, especially if the light reflects from a medium with a higher refractive index. This phase change must be accounted for when calculating the constructive interference conditions in thin films like soap bubbles.
Light Wavelength
The wavelength of light is the distance over which the light's wave shape repeats. It's a critical factor in the interference patterns of thin films, like those seen on soap bubbles. In a vacuum, light travels at a constant speed, so the wavelength and frequency are inversely related; as one increases, the other decreases.

In the context of our soap bubble, the wavelength of the light most strongly reflected is determined by the film's thickness and its refractive index. According to the exercise, the soap bubble with a refractive index of 1.93 and a thickness of 120 nm will reflect light of a certain wavelength more strongly due to constructive interference. Mathematically, the relation is given by the modified formula \(\lambda = \frac{2nt}{m}\), where \(m\) stands for the order of interference. For maximum constructive interference, low integer values of \(m\) such as 1 (first order) are used to calculate the wavelength of light that is most prominently reflected. In practical terms, this means the soap bubble will shimmer with a color corresponding to that particular wavelength of light.
Refractive Index
The refractive index, symbolized by \(n\), indicates how much a material can bend (or refract) light. It's a dimensionless number representing the ratio of the speed of light in a vacuum to the speed of light in the material. For instance, the refractive index of air is close to 1, since light travels through air almost as fast as it does in a vacuum.

In our example, the soap bubble has a refractive index of 1.93. This value tells us that light travels significantly slower inside the bubble than in the air, causing the light to bend when transitioning between air and the bubble's surface. The refractive index is essential in thin-film interference calculations as it directly influences the wavelength of light inside the film. Using it in the formula \(\lambda = \frac{2nt}{m}\) helps to determine which wavelengths will constructively interfere for a given thickness—resulting in the reflection of specific light colors from the soap bubble. Understanding how refractive index affects light's behavior enables scientists and engineers to design optical devices like lenses, glasses, and even bubble-like structures with specific reflective properties.

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Most popular questions from this chapter

\(\mathrm{A}\) pair of narrow, parallel slits separated by \(0.250 \mathrm{~mm}\) is illuminated by the green component from a mercury vapor lamp \((\lambda=546.1 \mathrm{~nm})\). The interference pattern is observed on a screen \(1.20 \mathrm{~m}\) from the plane of the parallel slits. Calculate the distance (a) from the central maximum to the first bright region on either side of the central maximum and (b) between the first and second dark bands in the interference pattern.

Take red light at \(700 \mathrm{~nm}\) and violet at \(400 \mathrm{~nm}\) as the ends of the visible spectrum and consider the continuous spectrum of white light formed by a diffraction grating with a spacing of \(d\) meters between adjacent lines. Show that the interval \(\theta_{i 2} \leq \theta \leq \theta_{r 2}\) of the continuous spectrum in second order must overlap the interval \(\theta_{\mathrm{m} 3} \leq \theta \leq \theta_{\mathrm{r} 3} \mathrm{of}\) the third-order spectrum. Note: \(\theta_{n / 1}\) is the angle of the violet light in second order, and \(\theta_{r 2}\) is the angle made by red light in second order.

Waves from a radio station have a wavelength of \(900 \mathrm{~m}\). They travel by two paths to a home receiver \(20.0 \mathrm{~km}\) from the transmitter. One path is a direct path, and the second is by reflection from a mountain directly behind the home receiver. What is the minimum distance from the mountain to the receiver that produces destructive interference at the receiver? (Assume that no phase change occurs on reflection from the mountain.)

A coating is applied to a lens to minimize reflections. The index of refraction of the coating is \(1.55\) and that of the lens is \(1.48\). If the coating is \(177.4 \mathrm{~nm}\) thick, what wavelength is minimally reflected for normal incidence in the lowest order?

A diffraction grating has \(4.200 \times 10^{3}\) rulings per centimeter. The screen is \(2.000 \mathrm{~m}\) from the grating. In parts (a) through (e), round each result to four digits, using the rounded values for subsequent calculations. (a) Compute the value of \(d\), the distance between adjacent rulings. Express the answer in meters. (b) Calculate the angle of the second-order maximum made by the \(589.0-\mathrm{nm}\) wavelength of sodium. (c) Find the position of the secondary maximum on the screen. (d) Repeat parts (b) and (c) for the second-order maximum of the \(589.6-\mathrm{nm}\) wavelength of sodium, finding the angle and position on the screen. (e) What is the distance between the two secondary maxima on the screen? (f) Now find the distance between the two secondary maxima without rounding, carrying all digits in calculator memory. How many significant digits are in agreement? What can you conclude about rounding and the use of intermediate answers in this case?

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