4.2 Intensity in single-slit diffraction  (Page 2/3)

These two maxima actually correspond to values of $\varphi$ slightly less than $3\pi$ rad and $5\pi$ rad. Since the total length of the arc of the phasor diagram is always $N\text{Δ}{E}_0,$ the radius of the arc decreases as $\varphi$ increases. As a result, $E_1$ and $E_2$ turn out to be slightly larger for arcs that have not quite curled through $3\pi$ rad and $5\pi$ rad, respectively. The exact values of $\varphi$ for the maxima are investigated in [link] . In solving that problem, you will find that they are less than, but very close to, $\varphi =3\pi ,5\pi ,7\pi ,\text{…}\text{rad}.$

To calculate the intensity at an arbitrary point P on the screen, we return to the phasor diagram of [link] . Since the arc subtends an angle $\varphi$ at the center of the circle,

and

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

Now defining

we obtain

This equation relates the amplitude of the resultant field at any point in the diffraction pattern to the amplitude $N\text{Δ}{E}_0$ at the central maximum. The intensity is proportional to the square of the amplitude, so

where $I_0={\left(N\text{Δ}{E}_0\right)}^2\text{/}2{\mu }_{0}c$ is the intensity at the center of the pattern.

For the central maximum, $\varphi =0$ , $\beta$ is also zero and we see from l’Hôpital’s rule that ${\text{lim}}_{\beta \to 0}\left(\text{sin}\beta \text{/}\beta \right)=1,$ so that ${\text{lim}}_{\varphi \to 0}I=I_0.$ For the next maximum, $\varphi =3\pi$ rad, we have $\beta =3\pi \text{/}2$ rad and when substituted into [link] , it yields

in agreement with what we found earlier in this section using the diameters and circumferences of phasor diagrams. Substituting $\varphi =5\pi$ rad into [link] yields a similar result for $I_2$ .

A plot of [link] is shown in [link] 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 $\text{sin}\beta =0,$ which occurs when $\beta =m\pi$ rad. This corresponds to

or

which is [link] .

If the slit width D is varied, the intensity distribution changes, as illustrated in [link] . The central peak is distributed over the region from $\text{sin}\theta =\text{−}\lambda \text{/}D$ to $\text{sin}\theta =+\lambda \text{/}D$ . For small $\theta$ , this corresponds to an angular width $\text{Δ}\theta \approx 2\lambda \text{/}D.$ Hence, an increase in the slit width results in a decrease in the width of the central peak    . For a slit with $D\gg \lambda ,$ the central peak is very sharp, whereas if $D\approx \lambda$ , it becomes quite broad.

Summary

• The intensity pattern for diffraction due to a single slit can be calculated using phasors as
  
  $I=I_0{\left(\frac{\text{sin}\beta }{\beta }\right)}^2,$
  
  where $\beta =\frac{\varphi }{2}=\frac{\pi D\text{sin}\theta }{\lambda }$ , D is the slit width, $\lambda$ is the wavelength, and $\theta$ is the angle from the central peak.

Conceptual questions

Problems