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University Physics Volume 3

4.2 Intensity in Single-Slit Diffraction

University Physics Volume 34.2 Intensity in Single-Slit Diffraction
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  1. Preface
  2. Unit 1. Optics
    1. 1 The Nature of Light
      1. Introduction
      2. 1.1 The Propagation of Light
      3. 1.2 The Law of Reflection
      4. 1.3 Refraction
      5. 1.4 Total Internal Reflection
      6. 1.5 Dispersion
      7. 1.6 Huygens’s Principle
      8. 1.7 Polarization
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    2. 2 Geometric Optics and Image Formation
      1. Introduction
      2. 2.1 Images Formed by Plane Mirrors
      3. 2.2 Spherical Mirrors
      4. 2.3 Images Formed by Refraction
      5. 2.4 Thin Lenses
      6. 2.5 The Eye
      7. 2.6 The Camera
      8. 2.7 The Simple Magnifier
      9. 2.8 Microscopes and Telescopes
      10. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
    3. 3 Interference
      1. Introduction
      2. 3.1 Young's Double-Slit Interference
      3. 3.2 Mathematics of Interference
      4. 3.3 Multiple-Slit Interference
      5. 3.4 Interference in Thin Films
      6. 3.5 The Michelson Interferometer
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    4. 4 Diffraction
      1. Introduction
      2. 4.1 Single-Slit Diffraction
      3. 4.2 Intensity in Single-Slit Diffraction
      4. 4.3 Double-Slit Diffraction
      5. 4.4 Diffraction Gratings
      6. 4.5 Circular Apertures and Resolution
      7. 4.6 X-Ray Diffraction
      8. 4.7 Holography
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  3. Unit 2. Modern Physics
    1. 5 Relativity
      1. Introduction
      2. 5.1 Invariance of Physical Laws
      3. 5.2 Relativity of Simultaneity
      4. 5.3 Time Dilation
      5. 5.4 Length Contraction
      6. 5.5 The Lorentz Transformation
      7. 5.6 Relativistic Velocity Transformation
      8. 5.7 Doppler Effect for Light
      9. 5.8 Relativistic Momentum
      10. 5.9 Relativistic Energy
      11. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
    2. 6 Photons and Matter Waves
      1. Introduction
      2. 6.1 Blackbody Radiation
      3. 6.2 Photoelectric Effect
      4. 6.3 The Compton Effect
      5. 6.4 Bohr’s Model of the Hydrogen Atom
      6. 6.5 De Broglie’s Matter Waves
      7. 6.6 Wave-Particle Duality
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
    3. 7 Quantum Mechanics
      1. Introduction
      2. 7.1 Wave Functions
      3. 7.2 The Heisenberg Uncertainty Principle
      4. 7.3 The Schrӧdinger Equation
      5. 7.4 The Quantum Particle in a Box
      6. 7.5 The Quantum Harmonic Oscillator
      7. 7.6 The Quantum Tunneling of Particles through Potential Barriers
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    4. 8 Atomic Structure
      1. Introduction
      2. 8.1 The Hydrogen Atom
      3. 8.2 Orbital Magnetic Dipole Moment of the Electron
      4. 8.3 Electron Spin
      5. 8.4 The Exclusion Principle and the Periodic Table
      6. 8.5 Atomic Spectra and X-rays
      7. 8.6 Lasers
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
    5. 9 Condensed Matter Physics
      1. Introduction
      2. 9.1 Types of Molecular Bonds
      3. 9.2 Molecular Spectra
      4. 9.3 Bonding in Crystalline Solids
      5. 9.4 Free Electron Model of Metals
      6. 9.5 Band Theory of Solids
      7. 9.6 Semiconductors and Doping
      8. 9.7 Semiconductor Devices
      9. 9.8 Superconductivity
      10. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    6. 10 Nuclear Physics
      1. Introduction
      2. 10.1 Properties of Nuclei
      3. 10.2 Nuclear Binding Energy
      4. 10.3 Radioactive Decay
      5. 10.4 Nuclear Reactions
      6. 10.5 Fission
      7. 10.6 Nuclear Fusion
      8. 10.7 Medical Applications and Biological Effects of Nuclear Radiation
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    7. 11 Particle Physics and Cosmology
      1. Introduction
      2. 11.1 Introduction to Particle Physics
      3. 11.2 Particle Conservation Laws
      4. 11.3 Quarks
      5. 11.4 Particle Accelerators and Detectors
      6. 11.5 The Standard Model
      7. 11.6 The Big Bang
      8. 11.7 Evolution of the Early Universe
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  4. A | Units
  5. B | Conversion Factors
  6. C | Fundamental Constants
  7. D | Astronomical Data
  8. E | Mathematical Formulas
  9. F | Chemistry
  10. G | The Greek Alphabet
  11. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
    8. Chapter 8
    9. Chapter 9
    10. Chapter 10
    11. Chapter 11
  12. Index

Learning Objectives

By the end of this section, you will be able to:
  • Calculate the intensity relative to the central maximum of the single-slit diffraction peaks
  • Calculate the intensity relative to the central maximum of an arbitrary point on the screen

To calculate the intensity of the diffraction pattern, we follow the phasor method used for calculations with ac circuits in Alternating-Current Circuits. If we consider that there are N Huygens sources across the slit shown in Figure 4.4, with each source separated by a distance D/N from its adjacent neighbors, the path difference between waves from adjacent sources reaching the arbitrary point P on the screen is (D/N)sinθ.(D/N)sinθ. This distance is equivalent to a phase difference of (2πD/λN)sinθ.(2πD/λN)sinθ. The phasor diagram for the waves arriving at the point whose angular position is θθ is shown in Figure 4.7. The amplitude of the phasor for each Huygens wavelet is ΔE0,ΔE0, the amplitude of the resultant phasor is E, and the phase difference between the wavelets from the first and the last sources is

ϕ=(2πλ)Dsinθ.ϕ=(2πλ)Dsinθ.

With NN, the phasor diagram approaches a circular arc of length NΔE0NΔE0 and radius r. Since the length of the arc is NΔE0NΔE0 for any ϕϕ, the radius r of the arc must decrease as ϕϕ increases (or equivalently, as the phasors form tighter spirals).

Figure a shows an arc with phasors labeled delta E subscript 0. This subtends an angle at the center of the circle, through two lines labeled r. This angle is bisected and each half is labeled phi by 2. The endpoints of the arc are connected by an arrow labeled E. The tangent at one endpoint of the arc is horizontal. The tangent at the other endpoint of the arc makes an angle phi with the horizontal. Figure b shows the arc and the angle phi subtended by it. A dotted line extends from one endpoint of the arc to the opposite line r. It is perpendicular to r. It makes an angle phi with the arc and an angle 90 minus phi with the adjacent line r.
Figure 4.7 (a) Phasor diagram corresponding to the angular position θθ in the single-slit diffraction pattern. The phase difference between the wavelets from the first and last sources is ϕ=(2π/λ)Dsinθϕ=(2π/λ)Dsinθ. (b) The geometry of the phasor diagram.

The phasor diagram for ϕ=0ϕ=0 (the center of the diffraction pattern) is shown in Figure 4.8(a) using N=30N=30. In this case, the phasors are laid end to end in a straight line of length NΔE0,NΔE0, the radius r goes to infinity, and the resultant has its maximum value E=NΔE0.E=NΔE0. The intensity of the light can be obtained using the relation I=12cε0E2I=12cε0E2 from Electromagnetic Waves. The intensity of the maximum is then

I0=12cε0(NΔE0)2=12μ0c(NΔE0)2,I0=12cε0(NΔE0)2=12μ0c(NΔE0)2,

where ε0=1/μ0c2ε0=1/μ0c2. The phasor diagrams for the first two zeros of the diffraction pattern are shown in parts (b) and (d) of the figure. In both cases, the phasors add to zero, after rotating through ϕ=2πϕ=2π rad for m=1m=1 and 4π4π rad for m=2m=2.

Figure a shows 30 phasors in a line of length N delta E subscript 0. The length of a phasor is delta E subscript 0. Figure b shows a circle with phasors pointing in the anticlockwise direction. This is labeled m equal to 1, E equal to 0. Figure c shows phasors along a circle. They start from the bottom and go one and a half times around the circle in the anticlockwise direction. An arrow from the starting point to the ending point is labeled E1. It forms a diameter of the circle. Figure c is labeled 3 by 2 pi E1 equal to N delta E0. Figure d shows phasors along a circle. They start from the bottom and go twice around the circle in the anticlockwise direction. The figure is labeled m equal to 2, E equal to 0. Figure e shows phasors along a circle. They start from the bottom and go two and a half times around the circle in the anticlockwise direction. An arrow from the starting point to the ending point is labeled E2. It forms a diameter of the circle. Figure c is labeled 5 by 2 pi E2 equal to N delta E0.
Figure 4.8 Phasor diagrams (with 30 phasors) for various points on the single-slit diffraction pattern. Multiple rotations around a given circle have been separated slightly so that the phasors can be seen. (a) Central maximum, (b) first minimum, (c) first maximum beyond central maximum, (d) second minimum, and (e) second maximum beyond central maximum.

The next two maxima beyond the central maxima are represented by the phasor diagrams of parts (c) and (e). In part (c), the phasors have rotated through ϕ=3πϕ=3π rad and have formed a resultant phasor of magnitude E1E1. The length of the arc formed by the phasors is NΔE0.NΔE0. Since this corresponds to 1.5 rotations around a circle of diameter E1E1, we have

32πE1NΔE0,32πE1NΔE0,

so

E1=2NΔE03πE1=2NΔE03π

and

I1=12μ0cE12=4(NΔE0)2(9π2)(2μ0c)0.045I0,I1=12μ0cE12=4(NΔE0)2(9π2)(2μ0c)0.045I0,

where

I0=(NΔE0)22μ0c.I0=(NΔE0)22μ0c.

In part (e), the phasors have rotated through ϕ=5πϕ=5π rad, corresponding to 2.5 rotations around a circle of diameter E2E2 and arc length NΔE0.NΔE0. This results in I20.016I0I20.016I0. The proof is left as an exercise for the student (Exercise 4.119).

These two maxima actually correspond to values of ϕϕ slightly less than 3π3π rad and 5π5π rad. Since the total length of the arc of the phasor diagram is always NΔE0,NΔE0, the radius of the arc decreases as ϕϕ increases. As a result, E1E1 and E2E2 turn out to be slightly larger for arcs that have not quite curled through 3π3π rad and 5π5π rad, respectively. The exact values of ϕϕ for the maxima are investigated in Exercise 4.120. In solving that problem, you will find that they are less than, but very close to, ϕ=3π,5π,7π,rad.ϕ=3π,5π,7π,rad.

To calculate the intensity at an arbitrary point P on the screen, we return to the phasor diagram of Figure 4.7. Since the arc subtends an angle ϕϕ at the center of the circle,

NΔE0=rϕNΔE0=rϕ

and

sin(ϕ2)=E2r.sin(ϕ2)=E2r.

where E is the amplitude of the resultant field. Solving the second equation for E and then substituting r from the first equation, we find

E=2rsinϕ2=2NΔEoϕsinϕ2.E=2rsinϕ2=2NΔEoϕsinϕ2.

Now defining

β=ϕ2=πDsinθλβ=ϕ2=πDsinθλ
(4.2)

we obtain

E=NΔE0sinββE=NΔE0sinββ
(4.3)

This equation relates the amplitude of the resultant field at any point in the diffraction pattern to the amplitude NΔE0NΔE0 at the central maximum. The intensity is proportional to the square of the amplitude, so

I=I0(sinββ)2I=I0(sinββ)2
(4.4)

where I0=(NΔE0)2/2μ0cI0=(NΔE0)2/2μ0c is the intensity at the center of the pattern.

For the central maximum, ϕ=0ϕ=0, ββ is also zero and we see from l’Hôpital’s rule that limβ0(sinβ/β)=1,limβ0(sinβ/β)=1, so that limϕ0I=I0.limϕ0I=I0. For the next maximum, ϕ=3πϕ=3π rad, we have β=3π/2β=3π/2 rad and when substituted into Equation 4.4, it yields

I1=I0(sin3π/23π/2)20.045I0,I1=I0(sin3π/23π/2)20.045I0,

in agreement with what we found earlier in this section using the diameters and circumferences of phasor diagrams. Substituting ϕ=5πϕ=5π rad into Equation 4.4 yields a similar result for I2I2.

A plot of Equation 4.4 is shown in Figure 4.9 and directly below it is a photograph of an actual diffraction pattern. Notice that the central peak is much brighter than the others, and that the zeros of the pattern are located at those points where sinβ=0,sinβ=0, which occurs when β=mπβ=mπ rad. This corresponds to

πDsinθλ=mπ,πDsinθλ=mπ,

or

Dsinθ=mλ,Dsinθ=mλ,

which is Equation 4.1.

Figure a shows a graph of I by I0 versus beta. There is a crest at the center of the graph at beta equal to 0. The y-value of this is 1. The graph has ripples on both sides of this which grow smaller as you go outwards. The graph has zeroes at minus 3 pi, minus 2 pi, minus pi, pi, 2 pi, 3 pi. Figure b shows a strip with alternating light and dark regions. The central portion is brightest.
Figure 4.9 (a) The calculated intensity distribution of a single-slit diffraction pattern. (b) The actual diffraction pattern.

Example 4.2

Intensity in Single-Slit Diffraction Light of wavelength 550 nm passes through a slit of width 2.00μm2.00μm and produces a diffraction pattern similar to that shown in Figure 4.9. (a) Find the locations of the first two minima in terms of the angle from the central maximum and (b) determine the intensity relative to the central maximum at a point halfway between these two minima.

Strategy The minima are given by Equation 4.1, Dsinθ=mλDsinθ=mλ. The first two minima are for m=1m=1 and m=2.m=2. Equation 4.4 and Equation 4.2 can be used to determine the intensity once the angle has been worked out.

Solution

  1. Solving Equation 4.1 for θθ gives us θm=sin−1(mλ/D),θm=sin−1(mλ/D), so that
    θ1=sin−1((+1)(550×10−9m)2.00×10−6m)=+16.0°θ1=sin−1((+1)(550×10−9m)2.00×10−6m)=+16.0°

    and
    θ2=sin−1((+2)(550×10−9m)2.00×10−6m)=+33.4°.θ2=sin−1((+2)(550×10−9m)2.00×10−6m)=+33.4°.
  2. The halfway point between θ1θ1 and θ2θ2 is
    θ=(θ1+θ2)/2=(16.0°+33.4°)/2=24.7°.θ=(θ1+θ2)/2=(16.0°+33.4°)/2=24.7°.

Equation 4.2 gives

β=πDsinθλ=π(2.00×10−6m)sin(24.7°)(550×10−9m)=1.52πor4.77rad.β=πDsinθλ=π(2.00×10−6m)sin(24.7°)(550×10−9m)=1.52πor4.77rad.

From Equation 4.4, we can calculate

IIo=(sinββ)2=(sin(4.77)4.77)2=(−0.99854.77)2=0.044.IIo=(sinββ)2=(sin(4.77)4.77)2=(−0.99854.77)2=0.044.

Significance This position, halfway between two minima, is very close to the location of the maximum, expected near β=3π/2,or1.5πβ=3π/2,or1.5π.

Check Your Understanding 4.2

For the experiment in Example 4.2, at what angle from the center is the third maximum and what is its intensity relative to the central maximum?

If the slit width D is varied, the intensity distribution changes, as illustrated in Figure 4.10. The central peak is distributed over the region from sinθ=λ/Dsinθ=λ/D to sinθ=+λ/Dsinθ=+λ/D. For small θθ, this corresponds to an angular width Δθ2λ/D.Δθ2λ/D. Hence, an increase in the slit width results in a decrease in the width of the central peak. For a slit with Dλ,Dλ, the central peak is very sharp, whereas if DλDλ, it becomes quite broad.

Figures a through c show graphs of I by I0 versus theta in degrees. Each has a wave crest with y value 1 at x=0. Figure a, labeled D equal to lambda has a broad arc. Figure b, labeled D equal to 5 lambda has a narrower crest. It has zeroes roughly between 10 and 15 and between minus 10 and minus 15. Figure c, labeled D equal to 10 lambda has a narrow crest. It has zeroes at plus and minus 5, roughly between 10 and 15 and between minus 10 and minus 15.
Figure 4.10 Single-slit diffraction patterns for various slit widths. As the slit width D increases from D=λto5λD=λto5λ and then to 10λ10λ, the width of the central peak decreases as the angles for the first minima decrease as predicted by Equation 4.1.

Interactive

A diffraction experiment in optics can require a lot of preparation but this simulation by Andrew Duffy offers not only a quick set up but also the ability to change the slit width instantly. Run the simulation and select “Single slit.” You can adjust the slit width and see the effect on the diffraction pattern on a screen and as a graph.

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