m58544 Intensity in Single-Slit Diffraction 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 173c9f53-3485-457c-b14d-5f2dfad407da 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 , with each source separated by a distance a/N from its adjacent neighbors, the path difference between waves from adjacent sources reaching the arbitrary point P on the screen is (a/N)sinθ. This distance is equivalent to a phase difference of (2πa/λN)sinθ. The phasor diagram for the waves arriving at the point whose angular position is θ is shown in . The amplitude of the phasor for each Huygens wavelet is ΔE0, the amplitude of the resultant phasor is E, and the phase difference between the wavelets from the first and the last sources isϕ=(2πλ)asinθ. With N→∞, the phasor diagram approaches a circular arc of length NΔE0 and radius r. Since the length of the arc is NΔE0 for any ϕ, the radius r of the arc must decrease as ϕ increases (or equivalently, as the phasors form tighter spirals).

(a) Phasor diagram corresponding to the angular position θ in the single-slit diffraction pattern. The phase difference between the wavelets from the first and last sources is ϕ=(2π/λ)asinθ. (b) The geometry of the phasor diagram.

The phasor diagram for ϕ=0 (the center of the diffraction pattern) is shown in (a) using N=30. In this case, the phasors are laid end to end in a straight line of length NΔE0, the radius r goes to infinity, and the resultant has its maximum value E=NΔE0. The intensity of the light can be obtained using the relation I=12cε0E2 from Electromagnetic Waves. The intensity of the maximum is thenI0=12cε0(NΔE0)2=12μ0c(NΔE0)2, where ε0=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=2π rad for m=1 and 4π rad for m=2.

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=3π rad and have formed a resultant phasor of magnitude E1. The length of the arc formed by the phasors is NΔE0. Since this corresponds to 1.5 rotations around a circle of diameter E1, we have 32πE1≈NΔE0,so E1=2NΔE03π and I1=12μ0cE12=4(NΔE0)2(9π2)(2μ0c)≈0.045I0,where I0=(NΔE0)22μ0c. In part (e), the phasors have rotated through ϕ=5π rad, corresponding to 2.5 rotations around a circle of diameter E2 and arc length NΔE0. This results in I2≈0.016I0. The proof is left as an exercise for the student ().These two maxima actually correspond to values of ϕ slightly less than 3π rad and 5π rad. Since the total length of the arc of the phasor diagram is always NΔE0, the radius of the arc decreases as ϕ increases. As a result, E1 and E2 turn out to be slightly larger for arcs that have not quite curled through 3π rad and 5π rad, respectively. The exact values of ϕ for the maxima are investigated in . In solving that problem, you will find that they are less than, but very close to, ϕ=3π,5π,7π,…rad.To calculate the intensity at an arbitrary point P on the screen, we return to the phasor diagram of . Since the arc subtends an angle ϕ at the center of the circle, NΔE0=rϕ and 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. Now defining β=ϕ2=πasinθλ we obtain E=NΔE0sinββ This equation relates the amplitude of the resultant field at any point in the diffraction pattern to the amplitude NΔE0 at the central maximum. The intensity is proportional to the square of the amplitude, so I=I0(sinββ)2 where I0=(NΔE0)2/2μ0c is the intensity at the center of the pattern. For the central maximum, ϕ=0, β is also zero and we see from l’Hôpital’s rule that limβ→0(sinβ/β)=1, so that limϕ→0I=I0. For the next maximum, ϕ=3π rad, we have β=3π/2 rad and when substituted into , it yields I1=I0(sin3π/23π/2)2≈0.045I0,in agreement with what we found earlier in this section using the diameters and circumferences of phasor diagrams. Substituting ϕ=5π rad into yields a similar result for I2. A plot of is shown in 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, which occurs when β=mπ rad. This corresponds to πasinθλ=mπ, or asinθ=mλ, which is .

(a) The calculated intensity distribution of a single-slit diffraction pattern. (b) The actual diffraction pattern.

Light of wavelength 550 nm passes through a slit of width 2.00μm and produces a diffraction pattern similar to that shown in . (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. The minima are given by , asinθ=mλ. The first two minima are for m=1 and m=2. and can be used to determine the intensity once the angle has been worked out. Solving for θ gives us θm=sin−1(mλ/a), so that 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°. The halfway point between θ1 and θ2 is θ=(θ1+θ2)/2=(16.0°+33.4°)/2=24.7°. gives β=πasinθλ=π(2.00×10−6m)sin(24.7°)(550×10−9m)=1.52πor4.77rad. From , we can calculate IIo=(sinββ)2=(sin(4.77)4.77)2=(−0.99854.77)2=0.044. This position, halfway between two minima, is very close to the location of the maximum, expected near β=3π/2,or1.5π. For the experiment in , at what angle from the center is the third maximum and what is its intensity relative to the central maximum? 74.3°, 0.0083I0 If the slit width a is varied, the intensity distribution changes, as illustrated in . The central peak is distributed over the region from sinθ=−λ/a to sinθ=+λ/a. For small θ, this corresponds to an angular width Δθ≈2λ/a. Hence, an increase in the slit width results in a decrease in the width of the central peak. For a slit with a≫λ, the central peak is very sharp, whereas if a≈λ, it becomes quite broad.

Single-slit diffraction patterns for various slit widths. As the slit width a increases from a=λto5λ and then to 10λ, the width of the central peak decreases as the angles for the first minima decrease as predicted by .

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.

The intensity pattern for diffraction due to a single slit can be calculated using phasors as I=I0(sinββ)2, where β=ϕ2=πasinθλ, a is the slit width, λ is the wavelength, and θ is the angle from the central peak.

In , the parameter β looks like an angle but is not an angle that you can measure with a protractor in the physical world. Explain what β represents. The parameter β=ϕ/2 is the arc angle shown in the phasor diagram in . The phase difference between the first and last Huygens wavelet across the single slit is 2β and is related to the curvature of the arc that forms the resultant phasor that determines the light intensity.

A single slit of width 3.0μm is illuminated by a sodium yellow light of wavelength 589 nm. Find the intensity at a 15° angle to the axis in terms of the intensity of the central maximum. A single slit of width 0.1 mm is illuminated by a mercury light of wavelength 576 nm. Find the intensity at a 10° angle to the axis in terms of the intensity of the central maximum. I/I0=2.2×10−5 The width of the central peak in a single-slit diffraction pattern is 5.0 mm. The wavelength of the light is 600 nm, and the screen is 2.0 m from the slit. (a) What is the width of the slit? (b) Determine the ratio of the intensity at 4.5 mm from the center of the pattern to the intensity at the center. Consider the single-slit diffraction pattern for λ=600nm, a=0.025mm, and x=2.0m. Find the intensity in terms of Io at θ=0.5°, 1.0°, 1.5°, 3.0°, and 10.0°. 0.63I0,0.11I0,0.0067I0,0.0062I0,0.00088I0

width of the central peak angle between the minimum for m=1 and m=−1