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Formula for single slit diffraction minima
Formula for single slit diffraction minima













As a result, E 1 E 1 and E 2 E 2 turn out to be slightly larger for arcs that have not quite curled through 3 π 3 π rad and 5 π 5 π rad, respectively. Since the total length of the arc of the phasor diagram is always N Δ E 0, N Δ E 0, the radius of the arc decreases as ϕ ϕ increases. These two maxima actually correspond to values of ϕ ϕ slightly less than 3 π 3 π rad and 5 π 5 π rad. The proof is left as an exercise for the student ( Exercise 4.119).

formula for single slit diffraction minima

In part (e), the phasors have rotated through ϕ = 5 π ϕ = 5 π rad, corresponding to 2.5 rotations around a circle of diameter E 2 E 2 and arc length N Δ E 0. The amplitude of the phasor for each Huygens wavelet is Δ E 0, 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 The phasor diagram for the waves arriving at the point whose angular position is θ θ is shown in Figure 4.7. This distance is equivalent to a phase difference of ( 2 π a / λ N ) sin θ. If we consider that there are N Huygens sources across the slit shown in Figure 4.4, 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 ) sin θ.

formula for single slit diffraction minima

To calculate the intensity of the diffraction pattern, we follow the phasor method used for calculations with ac circuits in Alternating-Current Circuits.

formula for single slit diffraction minima

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















    Formula for single slit diffraction minima