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Chapter 35: Problem 55

A source \(S\) of monochromatic light and a detector \(D\) are both located in air a distance \(h\) above a horizontal plane sheet of glass and are separated by a horizontal distance \(x .\) Waves reaching \(D\) directly from \(S\) interfere with waves that reflect off the glass. The distance \(x\) is small compared to \(h\) so that the reflection is at close to normal incidence. (a) Show that the condition for constructive interference is \(\sqrt{x^{2}+4 h^{2}}-x=\left(m+\frac{1}{2}\right) \lambda,\) and the condition for destructive interference is \(\sqrt{x^{2}+4 h^{2}}-x=m \lambda\) (Hint: Take into account the phase change on reflection.) (b) Let \(h=24 \mathrm{cm}\) and \(x=14 \mathrm{cm} .\) What is the longest wavelength for which there will be constructive interference?

Short Answer

Expert verified
The longest wavelength for constructive interference is 72 cm.

Step by step solution

01

Understanding Path Difference

For constructive or destructive interference, the key is to understand the path difference between the two light rays: one traveling directly from the source to the detector, and the other reflecting off the glass. The path taken by the reflected ray can be expressed as the hypotenuse of a right triangle where the horizontal distance is \(x\) and the vertical distance is \(2h\). Hence, the path length is \(\sqrt{x^2 + (2h)^2}\).
02

Calculate Path Difference

The path difference between the reflected ray and direct ray is \(\Delta = \sqrt{x^2 + (2h)^2} - x\). This value accounts for the extra distance traveled by the reflected wave compared to the direct wave.
03

Phase Change Consideration

When light reflects off a surface like glass, it undergoes a phase shift of \(\pi\) rad (or equivalently half a wavelength). This phase change must be accounted for in the interference condition.
04

Constructive Interference Condition

Constructive interference occurs when the path difference is an odd multiple of half-wavelengths: \(\Delta = (m + 0.5)\lambda\), where \(m\) is an integer. Substituting the expression for \(\Delta\), we get \(\sqrt{x^2 + 4h^2} - x = \left(m + \frac{1}{2}\right)\lambda\).
05

Destructive Interference Condition

Destructive interference occurs when the path difference is an integer multiple of wavelengths: \(\Delta = m\lambda\). Hence, \(\sqrt{x^2 + 4h^2} - x = m\lambda\).
06

Solving for Longest Wavelength

To find the longest wavelength for constructive interference with given \(h = 24 \text{ cm}\) and \(x = 14 \text{ cm}\), we use the constructive interference condition: \(\sqrt{14^2 + 4 \times 24^2} - 14 = \left(m + \frac{1}{2}\right) \lambda\). Solving \(\sqrt{196 + 2304} = 50\), the equation becomes \(50 - 14 = \left(m + \frac{1}{2}\right) \lambda\). Therefore, \(36 = \left(m + \frac{1}{2}\right) \lambda\). Selecting \(m = 0\), yields the longest wavelength as \(\lambda = 72 \text{ cm}\).

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Key Concepts

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

constructive interference
In the world of optics, constructive interference is a fascinating phenomenon. It occurs when two or more light waves superimpose in such a way that they amplify each other, resulting in a stronger light intensity. This happens when the crest of one wave aligns perfectly with the crest of another wave. In practical terms, it means the path difference between these waves must be an odd multiple of half-wavelengths.

Specifically, for our context, constructive interference is reached when the path difference \( \Delta \) fulfills the condition \( \Delta = \left(m + \frac{1}{2}\right) \lambda \). Here, \( m \) is an integer, representing the order of the interference, and \( \lambda \) is the wavelength of the light.

By meeting this condition, the light waves combine to produce a brighter, more intense light at the point of interference. This principle is frequently utilized in various applications like antireflective coatings and optical instruments.
destructive interference
Destructive interference is quite the opposite of constructive interference. Instead of amplifying, it results in the cancellation or reduction of the light intensity where two or more light waves overlap. This happens when the crest of one wave aligns with the trough of another, effectively canceling each other out.

For destructive interference, the path difference \( \Delta \) should be an exact multiple of the wavelength: \( \Delta = m \lambda \). Here, \( m \) is again an integer denoting the order of interference.

The net effect is that the amplitude of the resultant wave at the interference point is minimized, often leading to darkness or reduced intensity. This concept is very crucial in fields like noise-canceling technologies and radio wave applications.
phase shift in light waves
When light waves reflect off certain surfaces, like glass, they can undergo a phase shift. This phase shift is essentially a change in the wave's position within its cycle. For light reflecting off a denser medium, this shift is equivalent to \( \pi \) radians, or half a wavelength.

This concept is pivotal in understanding interference because it alters how the waves align when they meet. In the context of our exercise, this phase shift must be factored into interference conditions.
  • For the reflected wave, the effective path difference must account for this shift.
  • This is why constructive interference occurs at path differences of half-integral wavelengths, rather than integral multiples.
Correctly accounting for phase shifts is crucial for accurate predictions of the interference pattern in practical optics setups.
monochromatic light
Monochromatic light refers to light that consists of a single wavelength or color. Think of it as light with a singular "note" in the spectrum, like the clear tone of a tuning fork.

Using monochromatic light simplifies the study of interference patterns because variations in color (which are due to variations in wavelength) are eliminated. This set wavelength allows for precise calculations and observations.

In laboratory settings or theoretical exercises, monochromatic sources, such as lasers, are often employed to achieve clear and consistent interference patterns without the complexity introduced by multiple wavelengths. Understanding monochromatic light is essential when analyzing and predicting interference, making it a fundamental concept in the study of optics.

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Most popular questions from this chapter

Two slits spaced 0.260 \(\mathrm{mm}\) apart are placed 0.700 \(\mathrm{m}\) from a screen and illuminated by coherent light with a wavelength of 660 nm. The intensity at the center of the central maximum \(\left(\theta=0^{\circ}\right)\) is \(I_{0}\) . (a) What is the distance on the screen from the center of the central maximum to the first minimum? (b) What is the distance on the screen from the center of the central maximum to the point where the intensity has fallen to \(I_{0} / 2 ?\)

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

Radio Interference. Two radio antennas \(A\) and \(B\) radiate in phase. Antenna \(B\) is 120 \(\mathrm{m}\) to the right of antenna \(A .\) Consider point \(Q\) along the extension of the line connecting the antennas, a horizontal distance of 40 \(\mathrm{m}\) to the right of antenna \(B\) . The frequency, and hence the wavelength, of the emitted waves can be varied. (a) What is the longest wavelength for which there will be destructive interference at point \(Q ?\) (b) What is the longest wave-length for which there will be constructive interference at point \(Q ?\)

Points \(A\) and \(B\) are 56.0 \(\mathrm{m}\) apart along an east-west line. At each of these points, a radio transmitter is emitting a 12.5 -MHz signal horizontally. These transmitters are in phase with each other and emit their beams uniformly in a horizontal plane. A receiver is taken 0.500 \(\mathrm{km}\) north of the \(A B\) line and initially placed at point \(C,\) directly opposite the midpoint of \(A B .\) The receiver can be moved only along an east-west direction but, due to its limited sensitivity, it must always remain a range so that the intensity of the signal it receives from the transmitter is no less than \(\frac{1}{4}\) of its maximum value. How far from point \(C\) (along an east-west line) can the receiver be moved and always be able to pick up the signal?

Reflective Coatings and Herring. Herring and related fish have a brilliant silvery appearance that camouflages them while they are swimming in a sunlit ocean. The silveriness is due to platelets attached to the surfaces of these fish. Each platelet is made up of several alternating layers of crystalline guanine \((n=1.80)\) and of cytoplasm \((n=1.333,\) the same as water), with a guanine layer on the outside in contact with the surrounding water (Fig. \(\mathrm{P} 35.56\) ). In one typical platelet, the guanine layers are 74 nm thick and the cytoplasm layers are 100 \(\mathrm{nm}\) thick. (a) For light striking the platelet surface at normal incidence, for which vacuum wavelengths of visible light will all of the reflections \(R_{1}\) , \(R_{2}, R_{3}, R_{4},\) and \(R_{5},\) shown in Fig. P35.56, be approximately in phase? If white light is shone on this platelet, what color will be most strongly reflected (see Fig. 32.4\() ?\) The surface of a herring has very many platelets side by side with layers of different thickness, so that all visible wavelengths are reflected. (b) Explain why such a "stack" of layers is more reflective than a single layer of guanine with cytoplasm underneath. (A stack of five guanine layers separated by cytoplasm layers reflects more than 80\(\%\) of incident light at the wavelength for which it is "tuned.") (c) The color that is most strongly reflected from a platelet depends on the angle at which it is viewed. Explain why this should be so. (You can see these changes in color by examining a herring from different angles. Most of the platelets on these fish are oriented in the same way, so that they are vertical when the fish is swimming.)
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