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Chapter 35: Problem 7

Young's experiment is performed with light from excited helium atoms \((\lambda=502 \mathrm{nm}) .\) Fringes are measured carefully on a screen \(1.20 \mathrm{~m}\) away from the double slit, and the center of the 20 th fringe \((\) not counting the central bright fringe) is found to be \(10.6 \mathrm{~mm}\) from the center of the central bright fringe. What is the separation of the two slits?

Short Answer

Expert verified
By substituting the given values into the formula, the separation of the two slits in the Young's experiment setup can be calculated.

Step by step solution

01

Convert all measurements to meters

First, all the measurements need to be in the same unit, preferably meters. So, convert the wavelength \(\lambda\) to \(502 \times 10^{-9} m\), the distance to the screen \(L\) remains as \(1.20 m\), and the displacement \(d\) to \(10.6 \times 10^{-3} m\).
02

Calculate the fringe spacing

Calculate the fringe spacing or the distance between successive fringes (\(\Delta y\)). Considering we have the displacement from the 0th to the 20th fringe, the fringe spacing can be found by dividing the total displacement by the number of fringe intervals, i.e. 20. Thus, \(\Delta y = d / 20\)
03

Solve for slit separation

Rewrite the formula for fringe spacing to solve for the slit separation \(d\). This gives: \(d = \lambda \cdot (L / \Delta y)\). Substitute the values for the wavelength, screen-to-slit distance, and the fringe spacing in the equation and compute the value for the slit separation.

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Key Concepts

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

Wave Interference
When discussing Young's double-slit experiment, it is crucial to understand the concept of wave interference. This phenomenon occurs when two or more waves meet and combine to form a new wave pattern. In Young's experiment, light waves pass through two slits, creating two sets of waves that overlap.

Here are some important notes about wave interference:
  • Constructive Interference: This happens when two waves combine to form a wave with a larger amplitude. This results in a bright fringe on the screen.
  • Destructive Interference: This happens when two waves combine to form a wave with a smaller amplitude, resulting in a dark fringe.
  • Coherence: For interference to be observable, the waves must be coherent, meaning they have a constant phase difference.
Wave interference creates a pattern of alternating bright and dark fringes on the screen, which is central to the analysis in Young鈥檚 experiment. Observing these patterns provides valuable insights into the properties of the waves involved.
Fringe Spacing
Fringe spacing, indicated by the symbol \(\Delta y\), represents the distance between consecutive bright (or dark) interference fringes observed on the screen. Calculating fringe spacing helps to elaborate on the behavior of light in the double-slit experiment.

In the given exercise, fringe spacing is determined by dividing the total measured distance between the interference patterns by the number of observed fringes or intervals. For our case, the center of the 20th fringe is 10.6 mm from the central bright fringe.

Thus, the fringe spacing formula becomes:
  • \(\Delta y = \frac{10.6 \times 10^{-3} \text{ m}}{20}\)
Understanding fringe spacing allows students to grasp how strikingly different wave interactions can form predictable and measurable patterns, an essential aspect of wave mechanics.
Wavelength Calculation
A core part of Young's double-slit experiment is the calculation of the wavelength involved. In our exercise, you are provided with the wavelength of light from helium atoms, which is 502 nm.

To solve the problem, we must apply the relationship between wavelength, fringe spacing, and slit separation. The formula is derived from:
  • The wavelength of light \(\lambda\)
  • The distance between the screen and the slits (\(L = 1.20 \,\text{m}\))
  • The fringe spacing \(\Delta y\)
The formula for the slit separation \(d\) is:
  • \(d = \lambda \cdot (L / \Delta y)\)
By substituting the values found for \(\Delta y\), \(\lambda\) as \(502 \times 10^{-9} \,\text{m}\), and \(L\), you can calculate the separation of the slits. This calculation is critical to bridging the physical dimensions of the experiment with the theoretical predictions from wave interference.

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Most popular questions from this chapter

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?

The professor returns the apparatus to the original setting. She then adjusts the speakers again. All of the students who had heard nothing originally now hear a loud tone, while you and the others who had originally heard the loud tone hear nothing. What did the professor do? (a) She turned off the oscillator. (b) She turned down the volume of the speakers. (c) She changed the phase relationship of the speakers. (d) She disconnected one speaker.

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.)

Figure \(\mathbf{P} 35.56\) shows an interferometer known as Fresnel's biprism. The magnitude of the prism angle \(A\) is extremely small. (a) If \(S_{0}\) is a very narrow source slit, show that the separation of the two virtual coherent sources \(S_{1}\) and \(S_{2}\) is given by \(d=2 a A(n-1)\), where \(n\) is the index of refraction of the material of the prism. (b) Calculate the spacing of the fringes of green light with wavelength \(500 \mathrm{nm}\) on a screen \(2.00 \mathrm{~m}\) from the biprism. Take \(a=0.200 \mathrm{~m}\) \(A=3.50 \mathrm{mrad},\) and \(n=1.50\)

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 \(540.6 \mathrm{nm} ?\)
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