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Chapter 27: Problem 8

At most, how many bright fringes can be formed on either side of the central bright fringe when light of wavelength \(625 \mathrm{nm}\) falls on a double slit whose slit separation is \(3.76 \times 10^{-6} \mathrm{m} ?\)

Short Answer

Expert verified
There can be at most 6 bright fringes on either side of the central fringe.

Step by step solution

01

Set up the relevant equation

The formula to find the maximum number of bright fringes on either side of the central fringe in a double-slit interference pattern is given by the equation for constructive interference: \(d \sin \theta = m \lambda\). Here, \(d\) is the slit separation, \(\lambda\) is the wavelength of the light, and \(m\) is the order of the fringe. Since \(\sin \theta\) must be less than or equal to 1, the equation becomes \(m \leq \frac{d}{\lambda}\).
02

Substitute the given values

Substitute the slit separation \(d = 3.76 \times 10^{-6} \, \mathrm{m}\) and the wavelength \(\lambda = 625 \, \mathrm{nm} = 625 \times 10^{-9} \, \mathrm{m}\) into the equation for \(m\): \(m \leq \frac{3.76 \times 10^{-6}}{625 \times 10^{-9}}\).
03

Calculate the maximum value of m

Perform the division to find the maximum value of \(m\):\[m \leq \frac{3.76 \times 10^{-6}}{625 \times 10^{-9}} = 6.016.\]The integer value of \(m\) must be 6 or less, as \(m\) must be an integer.
04

Conclusion about the number of fringes

The maximum number of bright fringes on each side of the central bright fringe is 6. Therefore, there can be 6 bright fringes at most on either side of the central fringe.

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Key Concepts

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

Constructive Interference
In the phenomenon of double-slit interference, constructive interference plays a central role in the formation of bright fringes. When light waves pass through two closely spaced slits, they spread out and overlap on the other side. This results in an interference pattern of light and dark bands.

Constructive interference occurs when the peaks of two light waves align perfectly. This alignment amplifies the light, resulting in bright fringes. Mathematically, constructive interference is described by the equation:
  • \(d \sin \theta = m \lambda\)
Here:
  • \(d\) is the distance between the slits (slit separation)
  • \(\theta\) is the angle of deviation of the light
  • \(m\) represents the order of the fringe (an integer)
  • \(\lambda\) is the wavelength of the light
Bright fringes indicate constructive interference because the wave summation results in increased amplitude at those points.
By determining the conditions under which this occurs, we can calculate the number of bright fringes produced in a double-slit experiment.
Wavelength
The wavelength of light is a crucial factor in understanding double-slit interference. Wavelength (\(\lambda\)) is the distance between successive peaks of a wave. In the context of light, it determines the light's color and other properties. The double-slit experiment typically involves monochromatic light, meaning a single wavelength is used.

In calculations involving double-slit interference, knowing the wavelength allows us to predict the pattern of interference fringes that will form. The spacing and frequency of fringes are directly related to the wavelength used.
  • In our example, the wavelength \(625\, \text{nm}\) is used.
This particular value is crucial as it dictates how far apart the fringes will appear and how many can fit into a given space. Smaller wavelengths result in fringes that are closer together, while larger wavelengths spread the fringes apart.
By understanding the wavelength, we can begin to unravel the pattern that will emerge from the slits.
Slit Separation
Slit separation, denoted by \(d\), refers to the distance between the two slits in a double-slit interference setup. This distance dictates many aspects of the interference pattern, including the spacing of the bright and dark fringes.

In mathematical terms, the slit separation affects the angle \(\theta\) at which constructive interference (bright fringes) occurs, as seen in the equation \(d \sin \theta = m \lambda\). A larger slit separation results in fringes that are spaced farther apart.
  • When \(d\) is large, the angle \(\theta\) needed for constructive interference decreases.
  • Conversely, for small \(d\), the angle \(\theta\) increases.
In our specific example, the slit separation is \(3.76 \times 10^{-6} \, \text{m}\). This value is pivotal, as it influences how many bright fringes can appear on either side of the central bright fringe. By increasing the separation, you can consistently increase the number of detectable fringes, showcasing the delicate balance of these physical parameters in wave interference.
This understanding of slit separation allows physicists to finely tune experiments to observe particular interference patterns.

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Most popular questions from this chapter

Light of wavelength \(410 \mathrm{nm}\) (in vacuum) is incident on a diffraction grating that has a slit separation of \(1.2 \times 10^{-5} \mathrm{m} .\) The distance between the grating and the viewing screen is \(0.15 \mathrm{m} .\) A diffraction pattern is produced on the screen that consists of a central bright fringe and higher-order bright fringes (see the drawing). (a) Determine the distance \(y\) from the central bright fringe to the second-order bright fringe. (Hint: The diffraction angles are small enough that the approximation \(\tan \theta \approx \sin \theta\) can be used.) (b) If the entire apparatus is submerged in water \(\left(n_{\text {water }}=1.33\right),\) what is the distance \(y ?\)

You are standing in air and are looking at a flat piece of glass \((n=\) 1.52 ) on which there is a layer of transparent plastic \((n=1.61) .\) Light whose wavelength is 589 nm in vacuum is incident nearly perpendicularly on the coated glass and reflects into your eyes. The layer of plastic looks dark. Find the two smallest possible nonzero values for the thickness of the layer.

A tank of gasoline \((n=1.40)\) is open to the air \((n=1.00) .\) A thin film of liquid floats on the gasoline and has a refractive index that is between 1.00 and \(1.40 .\) Light that has a wavelength of \(625 \mathrm{nm}\) (in vacuum) shines perpendicularly down through the air onto this film, and in this light the film looks bright due to constructive interference. The thickness of the film is \(242 \mathrm{nm}\) and is the minimum nonzero thickness for which constructive interference can occur. What is the refractive index of the film?

An inkjet printer uses tiny dots of red, green, and blue ink to produce an image. Assume that the dot separation on the printed page is the same for all colors. At normal viewing distances, the eye does not resolve the individual dots, regardless of color, so that the image has a normal look. The wavelengths for red, green, and blue are \(\lambda_{\text {red }}=660 \mathrm{nm}, \lambda_{\text {green }}=550 \mathrm{nm},\) and \(\lambda_{\text {blue }}=470 \mathrm{nm} .\) The diameter of the pupil through which light enters the eye is \(2.0 \mathrm{mm}\). For a viewing distance of \(0.40 \mathrm{m},\) what is the maximum allowable dot separation?

Astronomers have discovered a planetary system orbiting the star Upsilon Andromedae, which is at a distance of \(4.2 \times 10^{17} \mathrm{m}\) from the earth. One planet is believed to be located at a distance of \(1.2 \times 10^{11} \mathrm{m}\) from the star. Using visible light with a vacuum wavelength of \(550 \mathrm{nm},\) what is the minimum necessary aperture diameter that a telescope must have so that it can resolve the planet and the star?
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