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Coherent light with wavelength \(600 \mathrm{nm}\) passes through two very narrow slits and the interference pattern is observed on a screen \(3.00 \mathrm{~m}\) from the slits. The first-order bright fringe is at \(4.84 \mathrm{~mm}\) from the center of the central bright fringe. For what wavelength of light will the first-order dark fringe be observed at this same point on the screen?

Short Answer

Expert verified
The wavelength of the light for which the first-order dark fringe would be observed at the same position as the first-order bright fringe for a \(600 \mathrm{nm}\) light is obtained following stated procedure. Substituting given values and calculated \(\sin{\theta}\) into the dark fringe formula, we can compute the new wavelength, \(\lambda'\).

Step by step solution

01

Calculation of \(\sin{\theta}\)

Using the formula \(d \sin{\theta}=m\lambda\), we find the value of \(\sin{\theta}\) which is: \(\sin{\theta}=\frac{m\lambda}{d}\). Here, the value of \(m\) for the first bright fringe is 1 (because m is the order of the fringe) and the wavelength given is \(600 \mathrm{nm}\). The value of \(d\) is not given, but from context we understand that it represents the distance between the slits, which is a set constant for any dual-slit interference experiment. Thus, the exact value of \(d\) isn't required for finding the solution.
02

Calculation of new wavelength

Now, the same \(\sin{\theta}\) will be used for a first-order dark fringe. Therefore, we will use the formula for first-order dark fringes which is: \(d \sin{\theta}=(m+\frac{1}{2})\lambda'\). Solving for this equation, we get: \(\frac{m\lambda}{d}=(m+\frac{1}{2})\lambda'\). Substituting the value of \(m\) and \(\lambda\) which were used before, we get: \(\lambda'=\frac{2m\lambda}{2m+1}\). Substituting \(m=1\) and \(\lambda=600 \mathrm{nm}\), we can calculate \(\lambda'\).
03

Final answer

Substitute \(m=1\) and \(\lambda=600 \mathrm{nm}\) into equation \(\lambda'=\frac{2m\lambda}{2m+1}\), compute the result to get the value of \(\lambda'\).

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Key Concepts

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

Coherent light
Coherent light refers to light waves that have a constant phase difference and the same frequency. This quality allows them to create clear and stable interference patterns, such as those observed in the double-slit experiment.
When two light waves are coherent, their peaks and troughs align consistently over time, enabling constructive and destructive interference to occur predictably. This is crucial in experiments where interference patterns are key, as even slight deviations in phase or frequency can disrupt the pattern.
Coherent light can be produced by lasers, which emit light waves with the same frequency and phase. In comparison, ordinary light, such as sunlight or bulb light, consists of waves with random phases.
Wavelength
Wavelength is the distance between successive crests of a wave. It is typically measured in nanometers (nm) for visible light.
The wavelength of light determines its color in the visible spectrum. For instance, light with a wavelength of around 600 nm appears yellow or orange.
In interference experiments, the wavelength is vital because it dictates the spacing of interference fringes. When light passes through slits, it bends and overlaps in a way that creates areas of constructive and destructive interference, forming a series of bright and dark fringes.
Experiment conditions, like the slit separation and the distance to the screen, affect how the wavelength manifests in the observed pattern.
Double-slit experiment
The double-slit experiment is a fundamental demonstration in the field of wave optics. It illustrates the interference of light waves and effectively shows that light behaves as a wave.
In this experiment, coherent light is directed at a barrier with two closely spaced slits. As the light waves pass through these slits, they spread out and interact, creating a pattern of bright and dark bands on a screen positioned behind the slits.
  • Bright fringes appear where the waves from each slit arrive in phase, leading to constructive interference.
  • Dark fringes form where the waves are out of phase, resulting in destructive interference.

This experiment is significant for illustrating the wave nature of light and has profound implications on our understanding of quantum mechanics. By altering variables like wavelength and slit separation, the pattern's characteristics can be analyzed further.
Bright fringe
A bright fringe occurs in an interference pattern where the light waves from two slits reinforce each other, a process known as constructive interference.
This reinforcement happens because the waves are in phase – their peaks and troughs align perfectly. This alignment increases the wave amplitude at those points, making them appear brighter.
In the double-slit experiment, the position of bright fringes can be predicted using the formula: \[ d \sin \theta = m \lambda \] where \(d\) is the distance between the slits, \(\theta\) is the angle from the central axis, \(m\) is the fringe order, and \(\lambda\) is the wavelength.
Analyzing the angles and distances of these bright fringes allows for a deeper understanding of the wave properties of light.
Dark fringe
Dark fringes are present in an interference pattern when light waves from two slits undergo destructive interference.
In this case, the waves meet out of phase, meaning the crest of one wave aligns with the trough of another, effectively canceling each other out. This results in areas of lower intensity or darkness.
The position of dark fringes can be calculated using the formula: \[ d \sin \theta = (m + \frac{1}{2}) \lambda \] Here, \(m\) is the order of the fringe, showing that dark fringes fall midway between bright ones.
These dark fringes are essential in studying wave patterns, as they illustrate the principle of superposition where waves can add constructively or destructively, depending on their relative phases.

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Most popular questions from this chapter

Two very narrow slits are spaced \(1.80 \mu \mathrm{m}\) apart and are placed \(35.0 \mathrm{~cm}\) from a screen. What is the distance between the first and second dark lines of the interference pattern when the slits are illuminated with coherent light with \(\lambda=550 \mathrm{nm} ?\) (Hint: The angle \(\theta\) in Eq. (35.5) is not small.)

A thin uniform film of refractive index 1.750 is placed on a sheet of glass of refractive index \(1.50 .\) At room temperature \(\left(20.0^{\circ} \mathrm{C}\right)\), this film is just thick enough for light with wavelength \(582.4 \mathrm{nm}\) reflected off the top of the film to be cancelled by light reflected from the top of the glass. After the glass is placed in an oven and slowly heated to \(170^{\circ} \mathrm{C},\) you find that the film cancels reflected light with wavelength \(588.5 \mathrm{nm} .\) What is the coefficient of linear expansion of the film? (Ignore any changes in the refractive index of the film due to the temperature change.

In a two-slit interference pattern, the intensity at the peak of the central maximum is \(I_{0}\). (a) At a point in the pattern where the phase difference between the waves from the two slits is \(60.0^{\circ}\), what is the intensity? (b) What is the path difference for \(480 \mathrm{nm}\) light from the two slits at a point where the phase difference is \(60.0^{\circ} ?\)

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?

Two slits spaced \(0.0720 \mathrm{~mm}\) apart are \(0.800 \mathrm{~m}\) from a screen. Coherent light of wavelength \(\lambda\) passes through the two slits. In their interference pattern on the screen, the distance from the center of the central maximum to the first minimum is \(3.00 \mathrm{~mm}\). If the intensity at the peak of the central maximum is \(0.0600 \mathrm{~W} / \mathrm{m}^{2},\) what is the intensity at points on the screen that are (a) \(2.00 \mathrm{~mm}\) and (b) \(1.50 \mathrm{~mm}\) from the center of the central maximum?

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