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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 [link] , 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 ) sin θ . This distance is equivalent to a phase difference of ( 2 π D / λ N ) sin θ . The phasor diagram for the waves arriving at the point whose angular position is θ is shown in [link] . The amplitude of the phasor for each Huygens wavelet is Δ E 0 , the amplitude of the resultant phasor is E , and the phase difference between the wavelets from the first and the last sources is

ϕ = ( 2 π λ ) D sin θ .

With N , the phasor diagram approaches a circular arc of length N Δ E 0 and radius r . Since the length of the arc is N Δ E 0 for any ϕ , the radius r of the arc must decrease as ϕ 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.
(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 = ( 2 π / λ ) D sin θ . (b) The geometry of the phasor diagram.

The phasor diagram for ϕ = 0 (the center of the diffraction pattern) is shown in [link] (a) using N = 30 . In this case, the phasors are laid end to end in a straight line of length N Δ E 0 , the radius r goes to infinity, and the resultant has its maximum value E = N Δ E 0 . The intensity of the light can be obtained using the relation I = 1 2 c ε 0 E 2 from Electromagnetic Waves . The intensity of the maximum is then

I 0 = 1 2 c ε 0 ( N Δ E 0 ) 2 = 1 2 μ 0 c ( N Δ E 0 ) 2 ,

where ε 0 = 1 / μ 0 c 2 . 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 = 2 π rad for m = 1 and 4 π rad for m = 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.
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 = 3 π rad and have formed a resultant phasor of magnitude E 1 . The length of the arc formed by the phasors is N Δ E 0 . Since this corresponds to 1.5 rotations around a circle of diameter E 1 , we have

3 2 π E 1 = N Δ E 0 ,

so

E 1 = 2 N Δ E 0 3 π

and

I 1 = 1 2 μ 0 c E 1 2 = 4 ( N Δ E 0 ) 2 ( 9 π 2 ) ( 2 μ 0 c ) = 0.045 I 0 ,

where

I 0 = ( N Δ E 0 ) 2 2 μ 0 c .

In part (e), the phasors have rotated through ϕ = 5 π rad, corresponding to 2.5 rotations around a circle of diameter E 2 and arc length N Δ E 0 . This results in I 2 = 0.016 I 0 . The proof is left as an exercise for the student ( [link] ).

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Source:  OpenStax, University physics volume 3. OpenStax CNX. Nov 04, 2016 Download for free at http://cnx.org/content/col12067/1.4
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