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Coherent light of frequency \(f\) travels in air and is incident on two narrow slits. The interference pattern is observed on a distant screen that is directly opposite the slits. The frequency of light \(f\) can be varied. For \(f=5.60 \times 10^{12} \mathrm{~Hz}\) there is an interference maximum for \(\theta=60.0^{\circ} .\) The next higher frequency for which there is an interference maximum at this angle is \(7.47 \times 10^{12} \mathrm{~Hz}\). What is the separation \(d\) between the two slits?

Short Answer

Expert verified
The separation between the two slits is approximately \(6.87 × 10^{-5} m\).

Step by step solution

01

Find the wavelengths of the light at both frequencies

The speed of light \(c\) is the product of its frequency \(f\) and its wavelength \(\lambda\). So \(\lambda = \frac{c}{f}\). For the two given frequencies of \(5.60 × 10^{12} Hz\) and \(7.47 × 10^{12} Hz\) and with \(c = 3.00 × 10^{8} m/s\) you will find the corresponding wavelengths as \(\lambda_{1} = \frac{c}{f_{1}} = \frac{3.00 × 10^{8} m/s}{5.60 × 10^{12} Hz} = 5.36 × 10^{-5} m\) and \(\lambda_{2} = \frac{c}{f_{2}} = \frac{3.00 × 10^{8} m/s}{7.47 × 10^{12} Hz} = 4.01 × 10^{-5} m\).
02

Use the interference condition to find the slit separation.

For the same angle \(\theta\) and order \(m\), the condition for an interference maximum can be written as \(d \sin \theta = m \lambda_{1}\) and \(d \sin \theta = m \lambda_{2}\). These imply that \(m \lambda_{1} = m \lambda_{2}\) which simplifies to \(\lambda_{1} = \lambda_{2}\). This is an inconsistency, but it is resolved if we realize they represent different order maxima, thus \(m_{1} \lambda_{1} = m_{2} \lambda_{2}\). Given that the frequencies are in a \(5.60:7.47\) ratio it’s logical to assume that the orders are in a \(7:6\) ratio. So, we can write \(m_{1} = 7\) and \(m_{2} = 6\). Substitute these into the earlier equations to solve for \(d\) and get \(d = \frac{m_{1} \lambda_{1}}{\sin \theta} = \frac{7 \times 5.36 × 10^{-5} m}{\sin 60.0^\circ} ≈ 6.87 × 10^{-5} m\).
03

Write the final answer

The separation between the slits is approximately \(6.87 × 10^{-5} m\). This makes sense as typical slit separations in diffraction grating are of the order of \(10^{-5} m\).

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Key Concepts

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

Understanding Light Interference
Light interference is a fascinating phenomenon that occurs when two or more light waves converge and combine with each other. This results in the formation of a new wave pattern, which can either amplify or diminish the light intensity in certain areas. Interference patterns are created when the waves are coherent, meaning they have the same frequency and a stable phase relationship.

When light passes through two narrow slits, as in the exercise, it behaves like waves spreading out from each slit. These waves overlap and interfere with each other, creating a pattern of bright and dark fringes on a screen. Bright fringes, known as interference maxima, occur where the waves enhance each other, while dark fringes appear where they cancel out.

The exercise illustrates how varying the frequency of light affects the position of these interference maxima. By calculating the wavelengths for two different frequencies and using the interference condition, one can determine the separation between the slits, as this distance dictates the specific pattern of light and dark bands observed.
Decoding the Double Slit Experiment
The double slit experiment is a classic demonstration of the wave-like behavior of light and is central to the field of quantum mechanics. It reveals the interference pattern, which is one of the most direct pieces of evidence that light can behave as a wave.

In the experiment, coherent light shines through two closely spaced slits, producing a pattern of bright and dark fringes on a screen behind the slits. From the experiment in the exercise, we see that when light of a certain frequency causes an interference maximum at a set angle, a different frequency of light can produce a maximum at the same angle—indicating that the interference pattern is a function of both the frequency of light and the slit separation.

Through careful measurements and calculations, such as those presented in the problem solution, the separation between the slits can be determined. The logic applied in the example uses the assumption about the ratio of the maxima's orders derived from the ratio of frequencies to decode the underlying pattern and conclude the sought value of the slit separation.
Exploring Wave-Particle Duality
Wave-particle duality is a cornerstone concept of quantum mechanics proposing that every particle or quantum entity can exhibit both wave-like and particle-like properties. In the context of light, this concept was proposed to reconcile the particle theory of light, which includes photons, with its wave-like behavior, such as diffraction and interference.

Even though the exercise focuses primarily on the wave aspect of light to explain the interference pattern, the underlying principle of wave-particle duality is also implicitly at play. The fact that light can generate an interference pattern demonstrates its wave-like property; meanwhile, its particle aspect is acknowledged when considering that light is made up of photons—discrete packets of energy.

Realizing that matter and energy can switch between particle-like and wave-like behaviors depending on the conditions and measurements is pivotal in understanding the fundamentals of quantum phenomena. This is also essential in grasping the full scope of the double slit experiment, where not just waves of light but individual photons can produce interference patterns, a result that's deeply intertwined with the nature of wave-particle duality.

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Most popular questions from this chapter

White light reflects at normal incidence from the top and bottom surfaces of a glass plate \((n=1.52) .\) There is air above and below the plate. Constructive interference is observed for light whose wavelength in air is \(477.0 \mathrm{nm}\). What is the thickness of the plate if the next longer wavelength for which there is constructive interference is \(540.6 \mathrm{nm} ?\)

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 typical red laser pointer has a wavelength of \(650 \mathrm{nm}\). Suppose we wanted to test the wave nature of light by carefully cutting two parallel slits in a dark plastic sheet. We would then shine the laser through the slits onto a wall located \(1 \mathrm{~m}\) beyond the sheet. (a) Determine the slit spacing needed so that the bright spots on the wall would be discernible with \(1 \mathrm{~cm}\) spacing. (b) Is this a feasible "home experiment"? Is it possible to cut slits with that separation using typical household tools? (c) Suppose we wanted to test the wave nature of sound in a similar manner, by placing two small speakers driven in-phase near each other, separated by \(40 \mathrm{~cm}\), both facing a wall \(2 \mathrm{~m}\) distant. If we used a \(1.0 \mathrm{kHz}\) tone, determine the distance between points along the wall that would exhibit enhanced sound. (d) Suppose we wanted the soundenhanced points to be separated by only \(1.75 \mathrm{~m}\) to render this experiment feasible. Estimate an audible frequency \(f\) and use it to determine a speaker separation distance \(d\) that would accomplish this. (e) Is this a feasible "home experiment"?

The index of refraction of a glass rod is 1.48 at \(T=20.0^{\circ} \mathrm{C}\) and varies linearly with temperature, with a coefficient of \(2.50 \times 10^{-5} / \mathrm{C}^{\circ} .\) The coefficient of linear expansion of the glass is \(5.00 \times 10^{-6} / \mathrm{C}^{\circ} .\) At \(20.0^{\circ} \mathrm{C}\) the length of the rod is \(3.00 \mathrm{~cm}\) A Michelson interferometer has this glass rod in one arm, and the rod is being heated so that its temperature increases at a rate of \(5.00 \mathrm{C}^{\circ} / \mathrm{min} .\) The light source has wavelength \(\lambda=589 \mathrm{nm},\) and the rod initially is at \(T=20.0^{\circ} \mathrm{C}\). How many fringes cross the field of view each minute?

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.

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