Lesson 3.6 · 3. Electromagnetism

Geometric Optics

In the previous lesson, we explored the wave nature of light in detail: interference, diffraction, and coherence. But in many practical situations, we do not need to track every oscillation and phase. When light interacts with objects much larger than its wavelength, the wave effects become negligible, and light can be described as traveling along rays, straight lines that indicate the direction of energy flow. This is the domain of geometric optics, the oldest branch of optics and still the foundation of lens and mirror design, imaging systems, and our everyday understanding of how we see.

The ray approximation

When the wavelength $\lambda$ is much smaller than all relevant dimensions (aperture sizes, object sizes, distances), diffraction effects are negligible and light propagates along straight lines called rays. Geometric optics is the $\lambda \to 0$ limit of wave optics. It is an enormously useful approximation: virtually all optical instrument design (cameras, telescopes, microscopes, eyeglasses) is based on it.

Fermat's Principle

All of geometric optics can be derived from a single, elegant statement:

Fermat's Principle of Least Time

Light travels between two points along the path that takes the least time (more precisely, the path for which the travel time is stationary with respect to small variations).

$$\delta \int_A^B \frac{ds}{v} = \delta \int_A^B \frac{n(s)}{c} \, ds = 0$$

where $n = c/v$ is the refractive index and $ds$ is an infinitesimal arc length element.

Fermat's Principle IS the Action Principle

In Lesson 1 (Principle of Least Action), we learned that nature chooses the path that makes the action $S = \int L \, dt$ stationary. Fermat's principle is exactly the same idea applied to light: the "action" is the optical path length $\int n \, ds$, and light "chooses" the path that makes this quantity stationary. This is not a coincidence. In the quantum description (Lesson 11, Path Integrals), light explores all paths simultaneously, and the classical ray emerges as the path of stationary phase, exactly as the classical trajectory emerges from Feynman's sum over histories. The deep unity of physics reveals itself: mechanics and optics are governed by the same variational principle.

Let us see how Fermat's principle yields the laws of reflection and refraction.

Reflection

The Law of Reflection from Fermat's Principle

Consider light traveling from point A to point B via reflection off a flat mirror. We want to find the reflection point P that minimizes the total travel time.

Step 1: Set up coordinates. Let the mirror lie along the $x$-axis. Let A be at $(0, h_1)$ and B at $(d, h_2)$, where $h_1, h_2 > 0$. The reflection point P is at $(x, 0)$.

Step 2: Total path length. In a uniform medium (constant $n$), minimizing time is equivalent to minimizing distance: $$L(x) = \sqrt{x^2 + h_1^2} + \sqrt{(d-x)^2 + h_2^2}$$

Step 3: Minimize. Setting $dL/dx = 0$: $$\frac{x}{\sqrt{x^2 + h_1^2}} - \frac{d-x}{\sqrt{(d-x)^2 + h_2^2}} = 0$$ The first term is $\sin\theta_i$ (the sine of the angle of incidence) and the second is $\sin\theta_r$ (the sine of the angle of reflection).

Step 4: Conclusion. $$\sin\theta_i = \sin\theta_r \implies \boxed{\theta_i = \theta_r}$$ The angle of incidence equals the angle of reflection. This ancient law, known empirically for millennia, follows inevitably from Fermat's principle.

The law of reflection: the angle of incidence θᵢ equals the angle of reflection θᵣ, both measured from the normal to the surface.

Refraction: Snell's Law

Deriving Snell's Law from Fermat's Principle

Now consider light traveling from point A in medium 1 (refractive index $n_1$) to point B in medium 2 (refractive index $n_2$), crossing a flat interface.

Step 1: Setup. Let the interface be at $y = 0$. Point A is at $(0, h_1)$ in medium 1, point B is at $(d, -h_2)$ in medium 2. The crossing point is at $(x, 0)$.

Step 2: Optical path length. The time is: $$T(x) = \frac{n_1}{c}\sqrt{x^2 + h_1^2} + \frac{n_2}{c}\sqrt{(d-x)^2 + h_2^2}$$

Step 3: Minimize. Setting $dT/dx = 0$: $$n_1 \frac{x}{\sqrt{x^2 + h_1^2}} = n_2 \frac{d-x}{\sqrt{(d-x)^2 + h_2^2}}$$

Step 4: Snell's law. Recognizing the geometric factors as sines of the angles: $$\boxed{n_1 \sin\theta_1 = n_2 \sin\theta_2}$$ This is Snell's law of refraction, derived purely from the principle that light minimizes its travel time.

Total internal reflection

When light travels from a denser medium to a less dense one ($n_1 > n_2$), Snell's law gives $\sin\theta_2 = (n_1/n_2)\sin\theta_1$. If $\sin\theta_1 > n_2/n_1$, there is no real solution for $\theta_2$: the light is completely reflected back into the denser medium. The critical angle is:

$$\theta_c = \arcsin\left(\frac{n_2}{n_1}\right)$$

Total internal reflection is the principle behind optical fibers (explored in Lesson 67) and is used in prisms, binoculars, and many other optical devices.

Lenses

The Thin Lens

A lens is a piece of transparent material (usually glass, $n \approx 1.5$) with curved surfaces that refracts light. A thin lens is one whose thickness is negligible compared to the radii of curvature and the focal length.

The key property of a converging thin lens: all rays parallel to the optical axis, after passing through the lens, converge to a single point called the focal point F, located at a distance $f$ (the focal length) behind the lens.

The Thin Lens Equation

For a thin lens with focal length $f$, an object at distance $s$ from the lens produces an image at distance $s'$ given by:

$$\frac{1}{s} + \frac{1}{s'} = \frac{1}{f}$$

The magnification is $M = -s'/s$. A negative magnification means the image is inverted. This equation applies to both converging ($f > 0$) and diverging ($f < 0$) lenses, with the sign convention that real objects and images have positive distances on their respective sides.

Ray diagram for a converging thin lens. Three principal rays locate the image: (1) parallel to axis, refracts through F'; (2) through the center, undeviated; (3) through F, refracts parallel. The image is real, inverted, and magnified when s is between f and 2f.

The Lensmaker's Equation

The focal length of a thin lens depends on its shape and the refractive index of its material. The lensmaker's equation relates these:

$$\frac{1}{f} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

where $R_1$ and $R_2$ are the radii of curvature of the two surfaces (positive if the center of curvature is on the transmission side). For a symmetric biconvex lens with equal radii $R$, this gives $1/f = 2(n-1)/R$.

Mirrors

Concave Mirrors

A concave (converging) mirror has a reflecting surface that curves inward. For a spherical mirror with radius of curvature $R$, the focal length is:

$$f = \frac{R}{2}$$

The mirror equation has the same form as the thin lens equation:

$$\frac{1}{s} + \frac{1}{s'} = \frac{1}{f} = \frac{2}{R}$$

Concave mirrors focus parallel rays to a focal point and are used in reflecting telescopes, car headlights, and solar concentrators.

Convex Mirrors

A convex (diverging) mirror curves outward. Parallel rays diverge after reflection, appearing to come from a virtual focal point behind the mirror. The focal length is negative: $f = -R/2$. Convex mirrors always produce virtual, upright, reduced images, which is why they are used as wide-angle rear-view mirrors (with the warning: "objects in mirror are closer than they appear").

Left: a concave mirror focuses parallel rays to a real focal point F. Right: a convex mirror diverges rays; extending them backward, they appear to come from a virtual focal point F behind the mirror.

Optical Instruments

The Telescope

A simple refracting telescope consists of two converging lenses: a large objective (focal length $f_o$) that collects light and forms a real image, and a smaller eyepiece (focal length $f_e$) through which the observer views a magnified image.

The angular magnification of a telescope is:

$$M = -\frac{f_o}{f_e}$$

A large objective focal length and a small eyepiece focal length give high magnification. The negative sign indicates the image is inverted. The objective's aperture determines both the light-gathering power (proportional to area) and the angular resolution (limited by diffraction: $\theta_{\min} \approx 1.22\lambda/D$, where $D$ is the aperture diameter).

Why are large telescopes always reflectors, not refractors?

A lens must be supported at its edges, so large lenses sag under their own weight, distorting the image. The largest refracting telescope ever built (Yerkes Observatory, 1897) has a 1-meter lens and was never surpassed. A mirror, by contrast, can be supported across its entire back surface. Modern telescopes use mirrors up to 10 meters in diameter (and segmented mirrors even larger). Additionally, mirrors have no chromatic aberration (see below), and a single reflecting surface introduces no dispersion.

The Microscope

A compound microscope also uses two lenses, but the geometry differs. A short-focal-length objective creates a magnified real image of a nearby object, and the eyepiece further magnifies this intermediate image.

The total magnification is:

$$M = M_o \times M_e = -\frac{L}{f_o} \times \frac{25\text{ cm}}{f_e}$$

where $L$ is the tube length (distance between the focal points of the two lenses) and $25$ cm is the conventional near-point distance of the human eye. The resolution of a microscope is limited by diffraction to approximately $\lambda/(2 \cdot \text{NA})$, where NA is the numerical aperture.

Aberrations

Real lenses and mirrors are not perfect. Deviations from ideal imaging are called aberrations.

Spherical Aberration

Spherical aberration

A spherical mirror or lens does not focus all parallel rays to the same point. Rays far from the optical axis (marginal rays) are focused closer to the mirror than rays near the axis (paraxial rays). This causes a blurred image. The effect can be reduced by using parabolic mirrors (which have exactly the right curvature to eliminate the aberration for on-axis objects) or by using only the central portion of the lens (at the cost of light-gathering power).

Chromatic Aberration

Chromatic aberration

The refractive index $n$ of glass depends on wavelength (dispersion). Blue light ($n$ larger) is refracted more strongly than red light ($n$ smaller), so a simple lens focuses different colors at different distances. This produces colored halos around images. The standard remedy is an achromatic doublet: two lenses made of different glasses (e.g., crown and flint) cemented together, designed so that their dispersions cancel while maintaining a net focusing power.

Mirrors are completely free of chromatic aberration, since reflection does not depend on wavelength. This is another reason why large telescopes use mirrors.

Other aberrations include coma (off-axis points imaged as comet-shaped blurs), astigmatism (different focal lengths in different planes), field curvature (the image lies on a curved surface rather than a flat one), and distortion (straight lines in the object appear curved in the image). Optical design is the art of balancing these aberrations using combinations of lenses and mirrors.

Dispersion and Prisms

Since the refractive index depends on wavelength, $n = n(\lambda)$, a prism separates white light into its component colors. The angular spread between red and violet light after passing through a prism is called the angular dispersion.

For most transparent materials in the visible range, the refractive index is well described by the Cauchy equation:

$$n(\lambda) = A + \frac{B}{\lambda^2} + \frac{C}{\lambda^4} + \cdots$$

The higher-order terms are usually negligible. The coefficient $B > 0$ means $n$ decreases with wavelength: blue light is bent more than red (normal dispersion).

From Prisms to Quantum Mechanics

Newton's famous prism experiment (1666) showed that white light is a mixture of colors, but it raised a question: why does glass bend blue light more than red? The answer had to wait for quantum mechanics and the atomic theory of matter (Lesson 7). The refractive index arises from the interaction of light with atomic electrons. The resonance frequencies of these electrons lie in the ultraviolet, and the response function naturally produces a refractive index that decreases with wavelength in the visible range (far from resonance).

Exercises

Exercise 1: Image Formation

An object is placed 30 cm in front of a converging lens with focal length 20 cm. (a) Where is the image? (b) Is it real or virtual? (c) What is the magnification? (d) If the object is 5 cm tall, how tall is the image?

(a) Using the thin lens equation: $$\frac{1}{s'} = \frac{1}{f} - \frac{1}{s} = \frac{1}{20} - \frac{1}{30} = \frac{3 - 2}{60} = \frac{1}{60}$$ So $s' = 60$ cm.

(b) Since $s' > 0$, the image is real (on the opposite side of the lens from the object).

(c) $M = -s'/s = -60/30 = -2$. The image is inverted and magnified by a factor of 2.

(d) Image height: $h' = M \times h = -2 \times 5 = -10$ cm (10 cm tall, inverted).

Exercise 2: Total Internal Reflection

A light ray travels inside a glass fiber ($n = 1.50$) surrounded by a cladding ($n = 1.46$). (a) What is the critical angle for total internal reflection? (b) What is the maximum angle of incidence (measured from the fiber axis) at the entrance face of the fiber for light to be guided?

(a) Critical angle: $$\theta_c = \arcsin\left(\frac{n_2}{n_1}\right) = \arcsin\left(\frac{1.46}{1.50}\right) = \arcsin(0.973) = 76.7°$$

(b) At the entrance face, the ray refracts into the fiber. For it to hit the core-cladding interface at the critical angle, the angle inside the fiber (from the axis) must be $90° - 76.7° = 13.3°$. By Snell's law at the entrance: $$\sin\theta_{\max} = n_1 \sin(13.3°) = 1.50 \times 0.230 = 0.345$$ $$\theta_{\max} = 20.2°$$ This defines the numerical aperture: $\text{NA} = \sin\theta_{\max} = \sqrt{n_1^2 - n_2^2} = \sqrt{1.50^2 - 1.46^2} = 0.344$.

Exercise 3: Telescope Design

A refracting telescope has an objective with focal length $f_o = 1200$ mm and an eyepiece with $f_e = 25$ mm. (a) What is the angular magnification? (b) If you wanted to resolve two stars separated by 1 arcsecond, what minimum objective diameter is needed (at $\lambda = 550$ nm)?

(a) Angular magnification: $|M| = f_o / f_e = 1200/25 = 48\times$.

(b) The diffraction-limited angular resolution is: $$\theta_{\min} = 1.22 \frac{\lambda}{D}$$ Setting $\theta_{\min} = 1'' = 4.85 \times 10^{-6}$ rad: $$D = 1.22 \frac{\lambda}{\theta_{\min}} = 1.22 \times \frac{550 \times 10^{-9}}{4.85 \times 10^{-6}} = 0.138 \text{ m} \approx 14 \text{ cm}$$

Exercise 4: Fermat's Principle Applied

A lifeguard at point A on the beach must reach a drowning swimmer at point B in the water. The lifeguard runs at speed $v_1$ on sand and swims at speed $v_2 < v_1$ in water. Using Fermat's principle (replace "time" for "optical path length"), show that the optimal path obeys Snell's law: $\sin\theta_1 / v_1 = \sin\theta_2 / v_2$, where $\theta_1$ and $\theta_2$ are the angles from the normal to the shoreline.

Solution: The total time is $T = \sqrt{x^2 + a^2}/v_1 + \sqrt{(d-x)^2 + b^2}/v_2$. Minimizing $T$ with respect to the entry point $x$: $$\frac{dT}{dx} = \frac{x}{v_1\sqrt{x^2 + a^2}} - \frac{d-x}{v_2\sqrt{(d-x)^2 + b^2}} = 0$$ $$\frac{\sin\theta_1}{v_1} = \frac{\sin\theta_2}{v_2}$$ This is identical to Snell's law with $n \propto 1/v$. The lifeguard should not run straight to the swimmer but should take a path with a sharper angle in the water, spending more time on the fast medium (sand). Nature does the same with light.

Key Takeaways

Geometric optics is the $\lambda \to 0$ limit of wave optics, valid when all relevant dimensions are much larger than the wavelength.
Fermat's principle (light takes the path of stationary time) is the optical analog of the principle of least action, unifying mechanics and optics at the deepest level.
Snell's law $n_1\sin\theta_1 = n_2\sin\theta_2$ and the law of reflection $\theta_i = \theta_r$ both follow from Fermat's principle.
The thin lens equation $1/s + 1/s' = 1/f$ and the mirror equation govern image formation in all simple optical systems.
Optical instruments (telescopes, microscopes) combine lenses to achieve magnification, with ultimate resolution limited by diffraction.
Aberrations (spherical, chromatic, coma, astigmatism) limit real optical systems and drive the art of lens design.

← 3.5 Wave Optics 3.7 Modern Optics →