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Two identical horizontal sheets of glass have a thin film of air of thickness \(t\) between them. The glass has refractive index \(1.40 .\) The thickness \(t\) of the air layer can be varied. Light with wavelength \(\lambda\) in air is at normal incidence onto the top of the air film. There is constructive interference between the light reflected at the top and bottom surfaces of the air film when its thickness is \(650 \mathrm{nm} .\) For the same wavelength of light the next larger thickness for which there is constructive interference is \(910 \mathrm{nm}\). (a) What is the wavelength \(\lambda\) of the light when it is traveling in air? (b) What is the smallest thickness \(t\) of the air film for which there is constructive interference for this wavelength of light?

Short Answer

Expert verified
The wavelength \(\lambda\) of the light when it is traveling in air is 520 nm, and the smallest thickness \(t\) of the air film for which there is constructive interference for this wavelength of light is 260 nm.

Step by step solution

01

Finding the wavelength in air

We are given two thicknesses \(t_1 = 650 nm\) and \(t_2 = 910 nm\) that give constructive interference, hence we can write two conditions \(2t_1 = m_1\lambda_{air}\) and \(2t_2 = m_2\lambda_{air}\). However, since \(t_2\) corresponds to next order of constructive interference, \(m_2 = m_1 + 1\). Subtracting these two equations will remove \(m\), giving us the wavelength \(\lambda_{air}\).
02

Calculating the wavelength in air

We have \(2(t_2 - t_1) = \lambda_{air}\). Substituting the given values for \(t_2\) and \(t_1\), that is 910 nm and 650 nm respectively, calculates the wavelength \(\lambda_{air} = 2*(910-650) = 520 nm.\)
03

Finding the smallest thickness for constructive interference

For smallest thickness, we consider the first order of constructive interference, that is \(m = 1\). Substitute the values of \(m\) and \(\lambda_{air}\) into the equation \(2t = m\lambda_{air}\), then solve for \(t\).
04

Calculating the smallest thickness for constructive interference

Substituting \(m = 1\) and \(\lambda_{air} = 520 nm\) into the equation \(2t = m\lambda_{air}\), we solve for \(t = \frac{1 * 520 nm}{2} = 260 nm.\)

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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 occurs when light waves reflect off the upper and lower boundaries of a thin layer, like an air film between two sheets of glass. As the light reflects, waves can interfere with one another, either amplifying the light (constructive interference) or canceling it (destructive interference). The result is often a display of colorful patterns, familiar in soap bubbles or oil slicks on water.

For constructive interference specifically, the condition is that the path difference between the two reflecting waves should be an integral multiple of the wavelength, or half-wavelength in case of a phase shift due to reflection from a medium with higher refractive index. In the exercise example, an air gap creates this effect, resulting in visible bright spots at certain thickness measurements.
Wavelength Calculation
Calculating the wavelength of light involves understanding the properties of the waves involved, such as their frequency and speed. In the context of thin film interference, the wavelength calculation can be directly tied to interference patterns, often using the formula for constructive interference, which is \(2t = m\lambda\), where \(t\) is the thickness of the film, \(m\) is the order of the interference (a whole number), and \(\lambda\) is the wavelength of the light.

In the problem, we calculate \(\lambda_{air}\) by determining the difference in film thickness between two successive patterns of constructive interference. This method simplifies the calculation significantly because we don’t need to know the order \(m\); we only need the thickness difference. By considering the scenario that one order of interference differentiates the two provided thicknesses, we solve for the wavelength in air, which is essential for understanding wave behavior in this context.
Refractive Index
The refractive index of a medium, denoted typically by \(n\), is a critical value in optics. It measures how much the speed of light is reduced inside the medium compared to its speed in a vacuum. The index relates to the denser the medium, the slower the light travels, and the larger the refractive index. The refractive index influences various phenomena, including the bending of light at interfaces and, as seen in this exercise, the interference patterns in thin films.

In the given problem, we used the refractive index of the glass sheets (1.40) to realize that upon reflection at a boundary with a higher refractive index, a phase shift of half a wavelength occurs for the wave. This knowledge is vital when analyzing interference conditions and calculating the expected patterns of constructive interference.
Optics
Optics is the branch of physics that deals with the study of light and its interactions with matter. It encompasses the behavior of visible, ultraviolet, and infrared light. Optics theories explain phenomena such as reflection, refraction, interference, and diffraction. The topic of optics is fundamental in addressing problems like the one presented here, where the principles of wave interference are applied to determine the physical properties of materials and the nature of light passing through them.

By grasping the fundamental concepts of optics, one can predict the behavior of light in various environments, whether it's for practical applications such as designing lenses or understanding natural phenomena like rainbows. The interference of light waves is just one of the many fascinating aspects of optics that impacts the scientific and technological advances of today.

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Most popular questions from this chapter

Two slits spaced \(0.450 \mathrm{~mm}\) apart are placed \(75.0 \mathrm{~cm}\) from a screen. What is the distance between the second and third dark lines of the interference pattern on the screen when the slits are illuminated with coherent light with a wavelength of \(500 \mathrm{nm} ?\)

Consider two antennas separated by \(9.00 \mathrm{~m}\) that radiate in phase at \(120 \mathrm{MHz},\) as described in Exercise \(35.1 .\) A receiver placed \(150 \mathrm{~m}\) from both antennas measures an intensity \(I_{0} .\) The receiver is moved so that it is \(1.8 \mathrm{~m}\) closer to one antenna than to the other. (a) What is the phase difference \(\phi\) between the two radio waves produced by this path difference? (b) In terms of \(I_{0},\) what is the intensity measured by the receiver at its new position?

After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of refraction of 1.38 , while the eyedrops have a refractive index of \(1.45 .\) After you put in the drops, your friends notice that your eyes look red, because red light of wavelength \(600 \mathrm{nm}\) has been reinforced in the reflected light. (a) What is the minimum thickness of the film of eyedrops on your cornea? (b) Will any other wavelengths of visible light be reinforced in the reflected light? Will any be cancelled? (c) Suppose you had contact lenses, so that the eyedrops went on them instead of on your corneas. If the refractive index of the lens material is 1.50 and the layer of eyedrops has the same thickness as in part (a), what wavelengths of visible light will be reinforced? What wavelengths will be cancelled?

A radio transmitting station operating at a frequency of \(120 \mathrm{MHz}\) has two identical antennas that radiate in phase. Antenna \(B\) is \(9.00 \mathrm{~m}\) to the right of antenna \(A .\) Consider point \(P\) between the antennas and along the line connecting them, a horizontal distance \(x\) to the right of antenna \(A .\) For what values of \(x\) will constructive interference occur at point \(P ?\)

Two flat plates of glass with parallel faces are on a table, one plate on the other. Each plate is \(11.0 \mathrm{~cm}\) long and has a refractive index of \(1.55 .\) A very thin sheet of metal foil is inserted under the end of the upper plate to raise it slightly at that end, in a manner similar to that discussed in Example 35.4 . When you view the glass plates from above with reflected white light, you observe that, at \(1.15 \mathrm{~mm}\) from the line where the sheets are in contact, the violet light of wavelength \(400.0 \mathrm{nm}\) is enhanced in this reflected light, but no visible light is enhanced closer to the line of contact. (a) How far from the line of contact will green light (of wavelength \(550.0 \mathrm{nm}\) ) and orange light (of wavelength \(600.0 \mathrm{nm}\) ) first be enhanced? (b) How far from the line of contact will the violet, green, and orange light again be enhanced in the reflected light? (c) How thick is the metal foil holding the ends of the plates apart?

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