31 May 2025

Wavefronts, Coherence, Fermat's Principle, and Cardinal Points

Electromagnetic light, wavefront propagation, Huygens construction, coherence, reflection and refraction, mirror and lens formulae, and cardinal points.

waves-and-optics electromagnetic-light wavefront coherence fermat-principle cardinal-points

Electromagnetic nature of light

In charge-free vacuum, Maxwell’s equations include

\[\boldsymbol\nabla\cdot\mathbf E=0, \quad \boldsymbol\nabla\cdot\mathbf B=0, \quad \boldsymbol\nabla\times\mathbf E=-\frac{\partial\mathbf B}{\partial t}, \quad \boldsymbol\nabla\times\mathbf B=\mu_0\epsilon_0 \frac{\partial\mathbf E}{\partial t}.\]

Take the curl of Faraday’s law and use the vector identity

\[\boldsymbol\nabla\times(\boldsymbol\nabla\times\mathbf E) =\boldsymbol\nabla(\boldsymbol\nabla\cdot\mathbf E)-\nabla^2\mathbf E.\]

It follows that

\[-\nabla^2\mathbf E =-\mu_0\epsilon_0\frac{\partial^2\mathbf E}{\partial t^2}.\]

The same step applied to $\mathbf B$ gives

\[\boxed{\nabla^2\mathbf E=\frac1{c^2}\frac{\partial^2\mathbf E}{\partial t^2}}, \qquad \boxed{\nabla^2\mathbf B=\frac1{c^2}\frac{\partial^2\mathbf B}{\partial t^2}},\]

with

\[\boxed{c=\frac1{\sqrt{\mu_0\epsilon_0}}}.\]

For a plane wave, Maxwell’s curl equations require

\[\boxed{\mathbf E\perp\mathbf B\perp\mathbf k}, \qquad \boxed{B_0=\frac{E_0}{c}}.\]

The energy flow is along $\mathbf E\times\mathbf B$, the direction of propagation.

Wavefronts, Huygens’s principle, and coherence

A wavefront is a surface of constant phase. Rays are normal to wavefronts in an isotropic medium. Plane waves have parallel plane wavefronts; point sources have concentric spherical wavefronts. Successive fronts whose phases differ by $2\pi$ are separated along a ray by one wavelength $\lambda=v/f$.

Huygens’s principle states that every point of a wavefront acts as a source of secondary wavelets. After time $\Delta t$, each wavelet has advanced by $v\Delta t$, and their forward envelope is the new wavefront. At a boundary, the construction uses speed $v_1$ before the boundary and $v_2$ after it. Its geometry gives

\[\frac{\sin i}{\sin r}=\frac{v_1}{v_2}=\frac{n_2}{n_1},\]

which is Snell’s law.

Temporal coherence is phase correlation at one point at different times. A source of frequency width $\Delta\nu$ has an order-of-magnitude coherence time and length

\[\tau_c\sim\frac1{\Delta\nu}, \qquad \boxed{\ell_c=v\tau_c}.\]

Stable interference requires an optical-path difference no greater than the source’s coherence length. Spatial coherence is phase correlation at different points of the same wavefront; it determines whether separated apertures can act as mutually coherent sources.

Fermat’s principle and the ray laws

For refractive index $n$, the optical path between $A$ and $B$ is

\[\boxed{\mathcal L=\int_A^B n\,ds}.\]

Fermat’s principle states that the physical ray makes the first variation stationary:

\[\boxed{\delta\mathcal L=0}.\]

Let a ray cross a plane interface at a variable horizontal coordinate $x$. If the fixed endpoints are at perpendicular distances $a$ and $b$, with horizontal separation $d$, then

\[\mathcal L(x)=n_1\sqrt{a^2+x^2} +n_2\sqrt{b^2+(d-x)^2}.\]

Stationarity gives

\[0=\frac{d\mathcal L}{dx} =n_1\frac{x}{\sqrt{a^2+x^2}} -n_2\frac{d-x}{\sqrt{b^2+(d-x)^2}}.\]

The two ratios are $\sin i$ and $\sin r$, so

\[\boxed{n_1\sin i=n_2\sin r}.\]

The incident ray, refracted ray, and interface normal are coplanar. For reflection, both segments lie in the same medium; the same variation gives $\sin i=\sin r$, and for $0\le i,r<\pi/2$,

\[\boxed{i=r}.\]

The incident ray, reflected ray, and normal are also coplanar.

Spherical surface, thin lens, and spherical mirror

Use the Cartesian sign convention: light travels from left to right, the vertex is the origin, and distances to the right are positive. Thus a real object on the left has $u<0$.

At height $h$ on a paraxial spherical refracting surface, the signed small-angle geometry gives

\[i\simeq-\frac hu+\frac hR, \qquad r\simeq-\frac hv+\frac hR.\]

Substituting these into $n_1i=n_2r$ and cancelling $h$ yields

\[\boxed{\frac{n_2}{v}-\frac{n_1}{u} =\frac{n_2-n_1}{R}}.\]

For the first surface of a thin lens in air,

\[\frac n{v_1}-\frac1u=\frac{n-1}{R_1}.\]

For the second surface, whose separation from the first is neglected,

\[\frac1v-\frac n{v_1}=\frac{1-n}{R_2}.\]

Adding eliminates the intermediate image:

\[\boxed{\frac1v-\frac1u=\frac1f}, \qquad \boxed{\frac1f=(n-1)\left(\frac1{R_1}-\frac1{R_2}\right)}.\]

Reflection can be represented in the surface equation by replacing the refracted index by $-n_1$. This gives the paraxial spherical-mirror formula

\[\boxed{\frac1v+\frac1u=\frac2R=\frac1f}, \qquad \boxed{f=\frac R2}.\]

Under this convention a concave mirror has $R<0$ and $f<0$.

The six cardinal points

A centered paraxial optical system has three pairs of cardinal points:

1. Principal points $H_1,H_2$. Their perpendicular planes are conjugate with unit transverse magnification. Object distance $s$ is measured from the first principal plane and image distance $s^{\prime}$ from the second.
2. Focal points $F_1,F_2$. A ray directed toward the front focus $F_1$ emerges parallel to the axis. A parallel incident ray emerges through the rear focus $F_2$. Planes normal to the axis through these points are the focal planes.
3. Nodal points $N_1,N_2$. A ray directed toward $N_1$ emerges as if from $N_2$ with the same physical angle to the axis.

If the system power is $\Phi$, the Gaussian equation referred to the principal planes is

\[\boxed{\frac{n_2}{s^{\prime}}-\frac{n_1}{s}=\Phi}.\]

Putting the image or object at infinity gives the oriented front and rear focal lengths:

\[\boxed{f_1=H_1F_1=-\frac{n_1}{\Phi}}, \qquad \boxed{f_2=H_2F_2=\frac{n_2}{\Phi}}.\]

The nodal-point offsets are

\[\boxed{H_1N_1=H_2N_2 =\frac{n_2-n_1}{\Phi}=f_1+f_2}.\]

Therefore, when the entrance and exit media have the same refractive index, $N_1=H_1$ and $N_2=H_2$. These definitions and reference planes allow a thick multi-element system to be treated by the same Gaussian imaging equation as a single equivalent element.

The refracted wavelet geometry enforces Snell's law; principal planes provide the reference surfaces for a thick optical system.

Solved Problems

1. Focal length and image position of a symmetric thin lens

A thin biconvex lens in air has refractive index $n=1.50$, radii $R_1=+0.200\,\mathrm{m}$ and $R_2=-0.200\,\mathrm{m}$, and an object at $u=-0.300\,\mathrm{m}$. Find the focal length and image distance.

Step 1: Apply the lens-maker relation.

\[\begin{aligned} \frac1f &=(n-1)\left(\frac1{R_1}-\frac1{R_2}\right)\\ &=0.50\left(\frac1{0.200}-\frac1{-0.200}\right) =5.00\,\mathrm{m^{-1}}. \end{aligned}\]

Thus

\[f=0.200\,\mathrm{m}.\]

Step 2: Use the Gaussian lens equation.

\[\frac1v-\frac1u=\frac1f \quad\Longrightarrow\quad \frac1v+\frac1{0.300}=\frac1{0.200}.\]

Therefore $1/v=5-10/3=5/3\,\mathrm{m^{-1}}$, so

\[\boxed{v=0.600\,\mathrm{m}}.\]

The positive image distance identifies a real image on the outgoing side of the lens.

2. Refraction from Fermat’s stationary optical path

A ray passes from air, $n_1=1.00$, into glass, $n_2=1.50$, at incidence angle $i=30^\circ$. Find the refracted angle and verify the stationary-path condition.

Step 1: Use the result of the optical-path variation.

\[n_1\sin i=n_2\sin r.\]

Step 2: Substitute the data.

\[\sin r=\frac{1.00}{1.50}\sin30^\circ =\frac13.\]

Hence

\[\boxed{r=\sin^{-1}\!\left(\frac13\right)\simeq19.47^\circ}.\]

Finally, $1.00\sin30^\circ=1/2$ and $1.50\sin r=1.50(1/3)=1/2$, so the two sides of the stationarity condition agree exactly.

Descriptive Questions

1. How does Huygens’s construction lead to the laws of reflection and refraction?
2. What is the distinction between temporal coherence and spatial coherence?
3. Why is Fermat’s principle a stationary-optical-path principle rather than simply a shortest-path rule?
4. What are the six cardinal points of a centered optical system and how are they used?

Numerical Problems

1. A source has frequency width $\Delta\nu=2.00\,\mathrm{MHz}$. Estimate its coherence time and vacuum coherence length.

   Answer: $\tau_c=0.500\,\mu\mathrm{s}$ and $\ell_c=150\,\mathrm{m}$.

2. A plane electromagnetic wave in vacuum has electric-field amplitude $300\,\mathrm{V\,m^{-1}}$. Find its magnetic-field amplitude using $c=3.00\times10^8\,\mathrm{m\,s^{-1}}$.

   Answer: $B_0=1.00\,\mu\mathrm{T}$.

3. A concave spherical mirror has $f=-20.0\,\mathrm{cm}$ and an object at $u=-60.0\,\mathrm{cm}$. Find the image distance with the sign convention used above.

   Answer: $v=-30.0\,\mathrm{cm}$.

4. A centered optical system has power $\Phi=4.00\,\mathrm{m^{-1}}$, entrance index $n_1=1.00$, and exit index $n_2=1.50$. Find $f_1$, $f_2$, and the principal-to-nodal offset.

   Answer: $f_1=-0.250\,\mathrm{m}$, $f_2=0.375\,\mathrm{m}$, and $H_1N_1=H_2N_2=0.125\,\mathrm{m}$.

The solved results and all numerical answers are verified by exact residuals in the Unit I Maxima worksheet.

References

1. Huygens-Fresnel principle - Wikipedia
2. F. A. Jenkins and H. E. White, Fundamentals of Optics, McGraw-Hill, sections on wavefronts, Fermat’s principle, and Gaussian optics.
3. Max Born and Emil Wolf, Principles of Optics, Cambridge University Press, sections on geometrical optics and coherence.
4. Ajoy Ghatak, Optics, McGraw Hill Education, chapters on lens systems and cardinal points.