Diffraction occurs if a wave passes through an opening of any shape. Diffraction by a single slit establishes the basic ideas of diffraction, but a common situation of practical importance is diffraction of a wave by a circular aperture. Circular diffraction is mathematically more complex than diffraction from a slit, and we will present results without derivation.
Consider some examples. A loudspeaker cone generates sound by the rapid oscillation of a diaphragm, but the sound wave must pass through the circular aperture defined by the outer edge of the speaker cone before it travels into the room beyond. This is diffraction by a circular aperture. Telescopes and microscopes are the reverse. Light waves from outside need to enter the instrument. To do so, they must pass through a circular lens. In fact, the performance limit of optical instruments is determined by the diffraction of the circular openings through which the waves must pass. This is an issue we’ll look at in Chapter 35.
FIGURE 33.21 shows a circular aperture of diameter . Light waves passing through this aperture spread out to generate a circular diffraction pattern. You should compare this to Figure 33.11 for a single slit to note the similarities and differences. The diffraction pattern still has a central maximum, now circular, and it is surrounded by a series of secondary bright fringes.
Angle locates the first minimum in the intensity, where there is perfect destructive interference. A mathematical analysis of circular diffraction finds
where is the diameter of the circular opening. The last step in Equation 33.30 uses the small-angle approximation, which is almost always valid for the diffraction of light but usually is not valid for the diffraction of sound waves.
Within the small-angle approximation, the width of the central maximum is
The diameter of the diffraction pattern increases with distance , showing that light spreads out behind a circular aperture, but it decreases if the size of the aperture is increased.
Light from a helium-neon laser passes through a 0.50-mm-diameter hole. How far away should a viewing screen be placed to observe a diffraction pattern whose central maximum is 3.0 mm in diameter?
SOLVE Equation 33.31 gives us the appropriate screen distance: