show the plot of variation of intensity with a angle and state of reason for the reduction in intensity of secondary maxima compound to central maximum in case of diffraction of light due to single slit
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Answer:
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), 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 \left(D\text{/}N\right)\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta . This distance is equivalent to a phase difference of \left(2\pi D\text{/}\lambda N\right)\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta . The phasor diagram for the waves arriving at the point whose angular position is \theta is shown in (Figure). The amplitude of the phasor for each Huygens wavelet is \text{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
\varphi =\left(\frac{2\pi }{\lambda }\right)D\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta .
With N\to \infty, the phasor diagram approaches a circular arc of length N\text{Δ}{E}_{0} and radius r. Since the length of the arc is N\text{Δ}{E}_{0} for any \varphi, the radius r of the arc must decrease as \varphi increases (or equivalently, as the phasors form tighter spirals).
(a) Phasor diagram corresponding to the angular position \theta in the single-slit diffraction pattern. The phase difference between the wavelets from the first and last sources is \varphi =\left(2\pi \text{/}\lambda \right)D\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta. (b) The geometry of the phasor diagram.
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.
The phasor diagram for \varphi =0 (the center of the diffraction pattern) is shown in (Figure)(a) using N=30. In this case, the phasors are laid end to end in a straight line of length N\text{Δ}{E}_{0}, the radius r goes to infinity, and the resultant has its maximum value E=N\text{Δ}{E}_{0}. The intensity of the light can be obtained using the relation I=\frac{1}{2}c{\epsilon }_{0}{E}^{2} from Electromagnetic Waves. The intensity of the maximum is then
{I}_{0}=\frac{1}{2}c{\epsilon }_{0}{\left(N\text{Δ}{E}_{0}\right)}^{2}=\frac{1}{2{\mu }_{0}c}{\left(N\text{Δ}{E}_{0}\right)}^{2},
where {\epsilon }_{0}=1\text{/}{\mu }_{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 \varphi =2\pi rad for m=1 and 4\pi 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.