Suppose you point a telescope at two nearby stars in a galaxy far, far away. If you use the best possible detector, will you be able to distinguish separate images for the two stars, or will they blur into a single blob of light? A similar question could be asked of a microscope. Can two microscopic objects, very close together, be distinguished if sufficient magnification is used? Or is there some size limit at which their images will blur together and never be separated? These are important questions about the resolution of optical instruments.

Because of diffraction, the image of a distant star or a distant headlight is not a point but a circular diffraction pattern. Our question, then, really is: How close together can two diffraction patterns be before you can no longer distinguish them? One of the major scientists of the 19th century, Lord Rayleigh, studied this problem and suggested a reasonable rule that today is called Rayleigh’s criterion.

FIGURE 35.25 shows two distant point sources being imaged by a lens of diameter $D$. The angular separation between the objects, as seen from the lens, is $\alpha$. Rayleigh’s criterion states that

• The two objects are resolvable if $\alpha >{\theta }_{\text{min}}$, where ${\theta }_{\text{min}}={\theta }_1=1.22\text{λ}/D$ is the angle of the first dark fringe in the circular diffraction pattern.

• The two objects are not resolvable if $\alpha <{\theta }_{\text{min}}$ because their diffraction patterns are too overlapped.

• The two objects are marginally resolvable if $\alpha ={\theta }_{\text{min}}$. The central maximum of one image falls exactly on top of the first dark fringe of the other image. This is the situation shown in the figure.

FIGURE 35.25 Two images that are marginally resolved.

FIGURE 35.26 shows enlarged photographs of the images of two point sources. The images are circular diffraction patterns, not points. The two images are close but distinct where the objects are separated by $\alpha >{\theta }_{\text{min}}$. Two objects really were recorded in the photo at the bottom, but their separation is $\alpha <{\theta }_{\text{min}}$ and their images have blended together. In the middle photo, with $\alpha ={\theta }_{\text{min}}$, you can see that the two images are just barely resolved.

FIGURE 35.26 Enlarged photographs of the images of two point sources.

The angle

is called the angular resolution of a lens. The angular resolution of a telescope depends on the diameter of the objective lens (or the primary mirror) and the wavelength of the light; magnification is not a factor. Two images will remain overlapped and unresolved no matter what the magnification if their angular separation is less than ${\theta }_{\text{min}}$. For visible light, where $\text{λ}$ is pretty much fixed, the only parameter over which the astronomer has any control is the diameter of the lens or mirror of the telescope. The urge to build ever-larger telescopes is motivated, in part, by a desire to improve the angular resolution. (Another motivation is to increase the light-gathering power so as to see objects farther away.)

The performance of a microscope is also limited by the diffraction of light passing through the objective lens. Just as light cannot be focused to a spot smaller than about a wavelength, the most perfect microscope cannot resolve the features of objects separated by less than one wavelength, or roughly 500 nm. (Some clever tricks with the phase of the light can improve the resolution to about 200 nm, but diffraction is still the limiting factor.) Because atoms are approximately 0.1 nm in diameter, vastly smaller than the wavelength of visible or even ultraviolet light, there is no hope of ever seeing atoms with an optical microscope. This limitation is not simply a matter of needing a better design or more precise components; it is a fundamental limit set by the wave nature of the light with which we see.

CHAPTER 35 CHALLENGE EXAMPLE Visual acuity

The normal human eye has maximum visual acuity with a pupil diameter of about 3 mm. For larger pupils, acuity decreases due to increasing aberrations; for smaller pupils, acuity decreases due to increasing diffraction. If your pupil diameter is 2.0 mm, as it would be in bright light, what is the smallest-diameter circle that you should be able to see as a circle, rather than just an unresolved blob, on an eye chart at the standard distance of 20 ft? The index of refraction inside the eye is 1.34.

MODEL Assume that a 2.0-mm-diameter pupil is diffraction limited. Then the angular resolution is given by Rayleigh’s criterion. Diffraction increases with wavelength, so the eye’s acuity will be affected more by longer wavelengths than by shorter wavelengths. Consequently, assume that the light’s wavelength in air is 600 nm.

VISUALIZE Let the diameter of the circle be $d$. FIGURE 35.27 shows the circle at distance $s=20\text{ft}=6.1\text{m}$. “Seeing the circle,” shown edge-on, requires resolving the top and bottom lines as distinct.

FIGURE 35.27 Viewing a circle of diameter $d$.

SOLVE The angular separation of the top and bottom lines of the circle is $\alpha =d/s$. Rayleigh’s criterion says that a perfect lens with aperture $D$ can just barely resolve these two lines if

$$\alpha ={\displaystyle \frac{d}{s}}={\theta }_{\text{min}}={\displaystyle \frac{1.22{\text{λ}}_{\text{eye}}}{D}}={\displaystyle \frac{1.22{\text{λ}}_{\text{air}}}{n_{\text{eye}}D}}$$

The diffraction takes place inside the eye, where the wavelength is shortened to ${\text{λ}}_{\text{eye}}={\text{λ}}_{\text{air}}/n_{\text{eye}}$. Thus the circle diameter that can barely be resolved with perfect vision is

$$d={\displaystyle \frac{1.22{\text{λ}}_{\text{air}}s}{n_{\text{eye}}D}}={\displaystyle \frac{1.22(600\times {10}^{-9}\text{m})\left(6.1\text{m}\right)}{\left(1.34\right)\left(0.0020\text{m}\right)}}\approx 2\text{mm}$$

That’s about the height of a capital O in a printed book, so in principle you should—in very bright light—just barely be able to recognize it as an O at 20 feet.

REVIEW On an eye chart, the O on the line for 20/20 vision—the standard of excellent vision—is about 7 mm tall, so the calculated 2 mm, although in the right range, is a bit too small. There are two reasons. First, although aberrations of the eye are reduced with a smaller pupil, they haven’t vanished. And second, for a 2-mm-tall object at 20 ft, the size of the image on the retina is barely larger than the spacing between the cone cells, so the resolution of the “detector” is also a factor. Your eye is a very good optical instrument, but not perfect.