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Chapter 35: Problem 6

Two light sources can be adjusted to emit monochromatic light of any visible wavelength. The two sources are coherent, 2.04\(\mu \mathrm{m}\) apart, and in line with an observer, so that one source is 2.04\(\mu \mathrm{m}\) farther from the observer than the other. (a) For what visible wavelengths \((380\) to 750 \(\mathrm{nm})\) will the observer see the brightest light, owing to constructive interference? (b) How would your answers to part (a) be affected if the two sources were not in line with the observer, but were still arranged so that one source is 2.04\(\mu \mathrm{m}\) farther away from the observer than the other? (c) For what visible wavelengths will there be destructive interference at the location of the observer?

Short Answer

Expert verified
Constructive interference at 408 nm, 510 nm, 680 nm; no change if not in line; destructive at 582.86 nm.

Step by step solution

01

Understand the Concept of Constructive and Destructive Interference

Constructive interference occurs when the path difference between two light sources is an integer multiple of the wavelength, leading to bright light visibility. Mathematically, it is given by \( \Delta x = m \cdot \lambda \) where \( \Delta x \) is the path difference, \( m \) is an integer (\( 0, 1, 2, \ldots \)), and \( \lambda \) is the wavelength. Destructive interference occurs when the path difference is an odd multiple of half wavelengths, \( \Delta x = (m + 0.5) \cdot \lambda \).
02

Calculate Constructive Interference Wavelengths

The path difference \( \Delta x \) is given as \( 2.04 \mu m \) or \( 2040 \ nm \). For constructive interference, solve for \( \lambda \) using \( \Delta x = m \cdot \lambda \), rearranging gives \( \lambda = \frac{\Delta x}{m} \). Substitute \( \Delta x = 2040 \ nm \) and determine valid \( \lambda \) within the visible range \( 380 \ nm \leq \lambda \leq 750 \ nm \).- For \( m = 5 \), \( \lambda = \frac{2040}{5} = 408 \ nm \).- For \( m = 4 \), \( \lambda = \frac{2040}{4} = 510 \ nm \).- For \( m = 3 \), \( \lambda = \frac{2040}{3} \approx 680 \ nm \).Therefore, visible wavelengths for constructive interference are \( 408 \ nm, 510 \ nm, \) and \( 680 \ nm \).
03

Consider Alignment Impact on Constructive Interference

If the sources are not in line with the observer but \( \Delta x \) remains the same (2.04 碌m), the condition for constructive interference is not altered. The wavelengths calculated previously would still apply.
04

Calculate Destructive Interference Wavelengths

For destructive interference, use \( \Delta x = (m + 0.5) \cdot \lambda \). Rearranging gives \( \lambda = \frac{2 \cdot 2040 }{2m + 1} \). Find valid solutions within the visible range for appropriate values of \( m \).- For \( m = 1 \), \( \lambda = \frac{2040}{1.5} \approx 1360 \ nm \) (not visible).- For \( m = 2 \), \( \lambda = \frac{2040}{2.5} = 816 \ nm \) (not visible).- For \( m = 3 \), \( \lambda = \frac{2040}{3.5} \approx 582.86 \ nm \).The visible wavelength for destructive interference is \( 582.86 \ nm \).

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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 two waves combine to make a larger wave, leading to an increase in intensity. This happens when the path difference between the waves is an integer multiple of their wavelength. In simple terms, waves are "in sync" at this point.
For example:
  • The crests of two waves line up perfectly.
  • The combined effect is a wave with much greater amplitude, resulting in brighter light.
In the context of optics, when light from two coherent sources meets, constructive interference occurs under the condition \( \Delta x = m \cdot \lambda \). Here, \( \Delta x \) is the path difference, \( \lambda \) is the wavelength, and \( m \) is any integer (0, 1, 2, ...).
This formula shows us that by correctly adjusting \( \lambda \) or \( m \), we can see specific wavelengths of bright light. For instance, if the path difference \( \Delta x \) is 2040 nm, the visible wavelengths for constructive interference might be 408 nm, 510 nm, and 680 nm depending on the integer values of \( m \). These calculations help determine the instances where light appears its strongest in experiments or practical applications.
Destructive Interference
In contrast to constructive interference, destructive interference occurs when two waves come together to reduce or cancel each other's effects, resulting in decreased intensity or darkness.
Destructive interference happens when the path difference is an odd half multiple of the wavelength.
  • This means the crest of one wave aligns with the trough of another.
  • The waves effectively cancel each other out, which causes the light to dim or disappear at certain spots.
The condition for destructive interference is given by the formula \( \Delta x = \left(m + 0.5\right) \cdot \lambda \), where \( m \) is an integer. Solving this for one light source being 2040 nm further than the other, effective wavelengths for destructive interference are usually not visible, except for calculated cases like approximately 582.86 nm.
These principles are crucial in many scientific and technical fields, helping to create precise optical instruments, improve communication technologies, or study physical conditions where wave interference is relevant.
Path Difference in Waves
Path difference is a key factor in wave interference. It refers to the difference in distance traveled by two waves arriving at a point. This distance difference causes shifts in the wave phases, impacting whether they reinforce or negate one another.
The path difference is represented mathematically as \( \Delta x \). This value determines the occurrence of constructive or destructive interference.
**Key insights include:**
  • If \( \Delta x \) is a whole number multiple of the wavelength \( \lambda \), constructive interference results.
  • If \( \Delta x \) is a half-integer multiple, destructive interference happens.
Adjusting the path difference can influence the resulting wave pattern. For example, by altering the position of the sources or the slits in a double-slit experiment, the observed interference pattern can be manipulated.
Understanding path difference helps us apply these concepts in practical settings, such as creating specific patterns in holography and laser technology or testing materials' properties through non-invasive techniques.

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Most popular questions from this chapter

GPS Transmission. The GPS (Global Positioning System) satellites are approximately 5.18 \(\mathrm{m}\) across and transmit two low-power signals, one of which is at 1575.42 \(\mathrm{MHz}\) (in the UHF band). In a series of laboratory tests on the satellite, you put two 1575.42 - MHz UHF transmitters at opposite ends of the satellite. These broadcast in phase uniformly in all directions. You measure the intensity at points on a circle that is several hundred meters in radius and centered on the satellite. You measure angles on this circle relative to a point that lies along the centerline of the satellite (that is, the perpendicular bisector of a line that extends from one transmitter to the other). At this point on the circle, the measured intensity is 2.00 \(\mathrm{W} / \mathrm{m}^{2}\) . (a) At how many other angles in the range \(0^{\circ}<\theta<90^{\circ}\) is the intensity also 2.00 \(\mathrm{W} / \mathrm{m}^{2} ?\) (b) Find the four smallest angles in the range \(0^{\circ}<\theta<90^{\circ}\) for which the intensity is 2.00 \(\mathrm{W} / \mathrm{m}^{2} .\) (c) What is the intensity at a point on the circle at an angle of \(4.65^{\circ}\) from the centerline?

Light with wavelength 648 \(\mathrm{nm}\) in air is incident perpendicularly from air on a film 8.76\(\mu \mathrm{m}\) thick and with refractive index \(1.35 .\) Part of the light is reflected from the first surface of the film, and part enters the film and is reflected back at the second surface, where the film is again in contact with air. (a) How many waves are contained along the path of this second part of the light in its round trip through the film? (b) What is the phase difference between these two parts of the light as they leave the film?

Sensitive Eyes. 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 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?

One round face of a 3.25 -m, solid, cylindrical plastic pipe is covered with a thin black coating that completely blocks light. The opposite face is covered with a fluorescent coating that glows when it is struck by light. Two straight, thin, parallel scratches, 0.225 \(\mathrm{mm}\) apart, are made in the center of the black face. When laser light of wavelength 632.8 \(\mathrm{nm}\) shines through the slits perpendicular to the black face, you find that the central bright fringe on the opposite face is 5.82 \(\mathrm{mm}\) wide, measured between the dark fringes that border it on either side. What is the index of refraction of the plastic?

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 \(\mathrm{nm} ?\)
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