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In a Young's double-slit experiment, two rays of monochromatic light emerge from the slits and meet at a point on a distant screen, as in Figure 27.6\(a .\) The point on the screen where these two rays meet is the eighth-order bright fringe. The difference in the distances that the two rays travel is \(4.57 \times 10^{-6} \mathrm{m} .\) What is the wavelength (in \(\mathrm{nm} )\) of the monochromatic light?

Short Answer

Expert verified
The wavelength of the monochromatic light is approximately 571 nm.

Step by step solution

01

Understand the Given Data

We are given that the two rays meet at the eighth-order bright fringe. This implies that the path difference between the two rays corresponds to the eighth-order, or 8 times the wavelength (8λ). We are also given the path difference between the two rays is 4.57 x 10^-6 meters.
02

Use the Formula for Path Difference

In Young's double-slit experiment, the path difference for the m-th bright fringe is given by the equation \[\Delta d = m \lambda\]where \(m\) is the order of the fringe and \(\lambda\) is the wavelength. For the eighth-order bright fringe, \(m = 8\), so the equation becomes\[ 4.57 \times 10^{-6} = 8 \lambda\]
03

Solve for Wavelength

Rearrange the equation to find the wavelength as follows:\[\lambda = \frac{4.57 \times 10^{-6}}{8}\]Calculate the wavelength:\[\lambda = 5.7125 \times 10^{-7} \text{ m}\]
04

Convert to Nanometers

Convert the wavelength from meters to nanometers by using the conversion factor: 1 m = 10^9 nm. Multiply the wavelength by this factor to obtain it in nanometers:\[\lambda = 5.7125 \times 10^{-7} \times 10^{9} = 571.25 \text{ nm}\]
05

Round the Result

For simplicity and standard representation, we round the wavelength to the nearest integer:\[\lambda \approx 571 \text{ nm}\]

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Key Concepts

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

Wavelength Calculation
Wavelength calculation is a key concept in understanding how light behaves in different conditions, especially in experiments like Young's double-slit experiment. In the given problem, we determined the wavelength by using the known path difference and the order of the bright fringe. The path difference is given as the amount by which one wave path differs from another.
To calculate the wavelength (\(\lambda\)), we rely on the formula:
  • \(\Delta d = m \lambda\)
Here, \(\Delta d\) represents the path difference, \(m\) is the fringe order, and \(\lambda\) is the wavelength.
In this example, we solve for \(\lambda\) using:\[\lambda = \frac{4.57 \times 10^{-6}}{8}\] This shows how we can find a precise measurement of a light's wavelength using simple math and given data. By converting meters to nanometers, we further simplify understanding, as wavelengths are typically expressed in nanometers.
Bright Fringe
In Young's double-slit experiment, bright fringes represent points on the screen where constructive interference occurs. This happens when two light waves meet in phase, meaning their crests and troughs align perfectly.
These points become intensely bright due to the addition of the wave amplitudes.
  • The order of a bright fringe is denoted by \(m\), which indicates how many wavelengths fit into the path difference.
  • In our example, the eighth-order bright fringe means the path difference equals eight times the wavelength: \(\Delta d = 8\lambda\).
The position of each bright fringe is critical as it reveals the stepwise increase in wavelengths matching the path difference, allowing us to visually measure the wavelength of the light used.
Path Difference
Path difference is a fundamental concept in interference patterns involving waves, such as light waves in the double-slit experiment. It refers to the difference in distance traveled by two rays of light from their slits to a point on the screen.
This distance difference can affect how the waves interact at their meeting point.
  • Constructive interference, producing a bright fringe, occurs when the path difference is a whole number multiple of the wavelength, \(m\lambda\).
  • This concept is crucial for identifying the appearance and location of bright and dark fringes on the screen.
In the exercise, the path difference is given as \(4.57 \times 10^{-6} \, \mathrm{m}\), which tells us how much longer one path is than the other, directly permitting the calculation of the light's wavelength.
Monochromatic Light
Monochromatic light refers to light of a single wavelength or frequency. In experiments like Young's double-slit, using monochromatic light ensures clear and consistent interference patterns.
This is because each wave crest from the slits will meet its corresponding crest consistently, generating distinct bright and dark fringes.
  • This property highlights the pure form of light of one color, such as laser light or light filtered to a specific wavelength.
  • It's the simplicity of monochromatic light that makes it excellent for precision experiments in physics.
In this exercise, dealing with monochromatic light helped us determine the exact wavelength involved by examining the interference pattern produced.

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Most popular questions from this chapter

The distance between adjacent slits of a certain diffraction grating is \(1.250 \times 10^{-5} \mathrm{m}\) . The grating is illuminated by monochromatic light with a wavelength of \(656.0 \mathrm{nm},\) and is then heated so that its temperature increases by 100.0 \(\mathrm{C}^{\circ} .\) Determine the change in the angle of the seventh order principal maximum that occurs as a result of thermal expansion of the grating. The cofficient of linear expansion for the diffaction grating is \(1.30 \times 10^{-4}\left(\mathrm{C}^{9}\right)^{-1} .\) Be sure to include the proper algebraic sign with your answer: \(+\) if the angle increases, - if the angle decreases.

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?

A diffraction grating has 2604 lines per centimeter, and it produces a principal maximum at \(\theta=30.0^{\circ} .\) The grating is used with light that contains all wavelengths between 410 and 660 \(\mathrm{nm}\) . What is (are) the wavelength(s) of the incident light that could have produced this maximum?

Light of wavelength 410 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 \(y ?\)

A diffraction pattern forms when light passes through a single slit. The wavelength of the light is 675 nm. Determine the angle that locates the first dark fringe when the width of the slit is (a) \(1.8 \times 10^{-4} \mathrm{m}\) and \(\quad(\mathbf{b}) 1.8 \times 10^{-6} \mathrm{m}\)

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