Complete Optics

The Nature of Light

Core ideas

Optics studies light as rays, waves, photons, and electromagnetic fields. Which model is useful depends on scale: rays work when wavelengths are tiny, waves explain interference and diffraction, and photons explain detection, emission, and quantum noise.

For review, be able to choose the right model for a problem, relate frequency, wavelength, and energy, and connect intensity to field amplitude. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$c=\lambda\nu,\qquad E_\gamma=h\nu,\qquad I=\frac{1}{2}c\epsilon_0E_0^2$$

Worked example

A He—Ne laser emits light at $\lambda = 632.8\,\text{nm}$ with a measured power of $P = 5.0\,\text{mW}$ spread uniformly over a beam of radius $w = 1.0\,\text{mm}$. Find (a) the frequency, (b) the photon energy, and (c) the peak electric field $E_0$.

(a) Frequency: Using $c = \lambda\nu$,

$$\nu = \frac{c}{\lambda} = \frac{3.00\times10^8\,\text{m/s}}{632.8\times10^{-9}\,\text{m}} \approx 4.74\times10^{14}\,\text{Hz}.$$

(b) Photon energy: Using $E_\gamma = h\nu$ with $h = 6.626\times10^{-34}\,\text{J\cdot s}$,

$$E_\gamma = (6.626\times10^{-34})(4.74\times10^{14}) \approx 3.14\times10^{-19}\,\text{J} \approx 1.96\,\text{eV}.$$

(c) Peak electric field: The intensity is $I = P/(\pi w^2) = (5.0\times10^{-3})/(\pi(1.0\times10^{-3})^2) \approx 1.59\times10^{3}\,\text{W/m}^2$. Solving $I = \frac{1}{2}c\epsilon_0 E_0^2$ for $E_0$ with $\epsilon_0 = 8.85\times10^{-12}\,\text{F/m}$,

$$E_0 = \sqrt{\frac{2I}{c\epsilon_0}} = \sqrt{\frac{2(1.59\times10^3)}{(3.00\times10^8)(8.85\times10^{-12})}} \approx 1.09\times10^{3}\,\text{V/m}.$$

Problems with Solutions

1. Wavelength and frequency of blue light. Blue light has a wavelength of $\lambda = 450\,\text{nm}$. What is its frequency and photon energy in eV?

Solution.

$$\nu = \frac{3.00\times10^8}{450\times10^{-9}} = 6.67\times10^{14}\,\text{Hz},$$ $$E = h\nu = (6.626\times10^{-34})(6.67\times10^{14}) = 4.42\times10^{-19}\,\text{J} = \frac{4.42\times10^{-19}}{1.602\times10^{-19}} \approx 2.76\,\text{eV}.$$

2. Solar intensity. At Earth’s orbit the solar intensity is $I = 1.36\,\text{kW/m}^2$. What is the peak electric field?

Solution.

$$E_0 = \sqrt{\frac{2(1.36\times10^3)}{(3.00\times10^8)(8.85\times10^{-12})}} \approx 1.01\times10^{3}\,\text{V/m}.$$

3. Photon flux. How many photons per second from the He—Ne laser in the main example?

Solution.

$$N = \frac{P}{E_\gamma} = \frac{5.0\times10^{-3}}{3.14\times10^{-19}} \approx 1.6\times10^{16}\,\text{photons/s}.$$

Section summary. Light is one phenomenon with ray, wave, field, and photon descriptions.

Geometrical Optics

Core ideas

Geometrical optics follows rays through reflection, refraction, mirrors, and lenses. Fermat’s principle says rays make optical path length stationary. Paraxial approximations give simple imaging formulas, magnification, apertures, aberrations, and optical instruments.

For review, be able to use Snell’s law, draw ray diagrams, apply lens and mirror equations, and know when the paraxial approximation fails. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$n_1\sin\theta_1=n_2\sin\theta_2,\qquad \frac{1}{f}=\frac{1}{s}+\frac{1}{s'}$$

Worked example

A biconvex lens in air has focal length $f = +10.0\,\text{cm}$. An object of height $h_o = 2.0\,\text{cm}$ is placed at $s = 15.0\,\text{cm}$ from the lens. Find the image position, magnification, and image height, and describe the image.

Image position: Using the thin-lens equation,

$$\frac{1}{s'} = \frac{1}{f} - \frac{1}{s} = \frac{1}{10.0} - \frac{1}{15.0} = \frac{3-2}{30.0} = \frac{1}{30.0}\,\text{cm}^{-1},$$

so $s' = +30.0\,\text{cm}$. The positive sign means the image is real and on the opposite side of the lens from the object.

Magnification:

$$M = -\frac{s'}{s} = -\frac{30.0}{15.0} = -2.0.$$

Image height: $h_i = M h_o = (-2.0)(2.0) = -4.0\,\text{cm}$. The image is real, inverted, and magnified by a factor of 2.

Problems with Solutions

1. Snell’s law at a water—air interface. Light in water ($n = 1.33$) hits the surface at $\theta_1 = 30^\circ$ to the normal. What is the angle of transmission into air ($n = 1.00$)?

Solution.

$$n_1\sin\theta_1 = n_2\sin\theta_2 \implies \sin\theta_2 = \frac{1.33\sin 30^\circ}{1.00} = 0.665,$$ $$\theta_2 = \sin^{-1}(0.665) \approx 41.7^\circ.$$

2. Concave mirror. A concave mirror has $f = +20.0\,\text{cm}$. Where should an object be placed to obtain a real image twice as large?

Solution. For a real image, $M = -2 = -s'/s \implies s' = 2s$. Then

$$\frac{1}{20.0} = \frac{1}{s} + \frac{1}{2s} = \frac{3}{2s} \implies s = 30.0\,\text{cm}.$$

3. Critical angle. What is the critical angle for light going from glass ($n = 1.50$) to air ($n = 1.00$)?

Solution.

$$\sin\theta_c = \frac{n_2}{n_1} = \frac{1.00}{1.50} = 0.667 \implies \theta_c \approx 41.8^\circ.$$

Section summary. Ray optics is the short-wavelength limit of wave optics.

Wave Propagation

Core ideas

Wave optics treats light as fields satisfying wave equations. Plane waves, spherical waves, phase velocity, group velocity, impedance, and dispersion describe propagation. Boundary conditions at interfaces determine reflection, transmission, and evanescent waves.

For review, be able to derive plane-wave relations, compute phase accumulation, distinguish phase and group velocity, and apply boundary matching. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$\nabla^2\bm E-\frac{n^2}{c^2}\partial_t^2\bm E=0,\qquad \bm E=\bm E_0e^{i(\bm k\cdot\bm r-\omega t)}$$

Worked example

A plane wave of vacuum wavelength $\lambda_0 = 500\,\text{nm}$ travels through a glass plate of thickness $d = 2.00\,\text{mm}$ with refractive index $n = 1.50$. Find (a) the wavelength in the glass, (b) the wave number in the glass, and (c) the phase accumulated across the plate.

(a) Wavelength in glass:

$$\lambda = \frac{\lambda_0}{n} = \frac{500\,\text{nm}}{1.50} \approx 333\,\text{nm}.$$

(b) Wave number:

$$k = \frac{2\pi}{\lambda} = \frac{2\pi}{333\times10^{-9}\,\text{m}} \approx 1.88\times10^{7}\,\text{rad/m}.$$

(c) Phase accumulation:

$$\Delta\phi = k d = (1.88\times10^{7})(2.00\times10^{-3}) \approx 3.77\times10^{4}\,\text{rad}.$$

Equivalently, the optical path length is $n d = 1.50(2.00\,\text{mm}) = 3.00\,\text{mm}$, and since each vacuum wavelength contributes $2\pi$, the total phase is $2\pi (3.00\times10^{-3})/(500\times10^{-9}) = 1.20\times10^{4}\times 2\pi \approx 3.77\times10^{4}\,\text{rad}$.

Problems with Solutions

1. Phase velocity in glass. What is the phase velocity of the wave in the example above?

Solution.

$$v_p = \frac{c}{n} = \frac{3.00\times10^8}{1.50} = 2.00\times10^8\,\text{m/s}.$$

2. Group velocity. If the dispersion relation gives $n(\lambda_0) = 1.50 + 0.01(\lambda_0/\mu\text{m})$, estimate the group velocity at $\lambda_0 = 500\,\text{nm}$.

Solution.

$$\frac{dn}{d\lambda_0} = \frac{0.01}{10^{-6}} = 10^{4}\,\text{m}^{-1} \text{ per } \mu\text{m? No: } n = 1.50 + 0.01(\lambda_0/\mu\text{m}) \text{ so at } \lambda_0 = 0.5\,\mu\text{m},$$ $$\frac{dn}{d\lambda_0} = \frac{0.01}{1\,\mu\text{m}} = 10^{4}\,\text{m}^{-1} \text{ is wrong. Actually } dn/d\lambda_0 = 0.01/(10^{-6}\,\text{m}) = 10^{4}\,\text{m}^{-1}.$$

Better: write $\omega = 2\pi c/(n\lambda_0)$. For small dispersion,

$$v_g = \frac{c}{n}\left(1 + \frac{\lambda_0}{n}\frac{dn}{d\lambda_0}\right)^{-1}.$$

With $dn/d\lambda_0 = 0.01/(10^{-6}) = 10^4$? No, the coefficient is $0.01$ per $\mu$m, so $dn/d\lambda_0 = 0.01\,\mu\text{m}^{-1} = 10^4\,\text{m}^{-1}$.

$$\frac{\lambda_0}{n}\frac{dn}{d\lambda_0} = \frac{500\times10^{-9}}{1.50}(10^4) \approx 3.33\times10^{-3},$$ $$v_g \approx \frac{2.00\times10^8}{1.0033} \approx 1.99\times10^8\,\text{m/s}.$$

3. Reflection coefficient at normal incidence. For $n_1 = 1.00$ and $n_2 = 1.50$, what fraction of intensity is reflected at normal incidence?

Solution.

$$r = \frac{n_1 - n_2}{n_1 + n_2} = \frac{-0.50}{2.50} = -0.20,\qquad R = |r|^2 = 0.04.$$

So $4\%$ of the intensity is reflected.

Section summary. Wave propagation is controlled by phase, boundary conditions, and dispersion.

Superposition of Waves

Core ideas

Linear optics is based on superposition. Adding waves explains beats, standing waves, wave packets, normal modes, and interference. Phase relationships matter as much as amplitudes, so complex notation is the natural bookkeeping system.

For review, be able to add complex amplitudes, compute beats and standing waves, form wave packets, and interpret group velocity. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$E_1+E_2=E_{01}e^{i\phi_1}+E_{02}e^{i\phi_2},\qquad I\propto |E_1+E_2|^2$$

Worked example

Two coherent plane waves of the same frequency and equal amplitude $E_0$ overlap at a point. Their phases differ by $\Delta\phi = \phi_2 - \phi_1 = \pi/3$. Write the total field and find the resulting intensity relative to $I_0 \propto E_0^2$.

Total field:

$$E_{\text{tot}} = E_0 e^{i\phi_1} + E_0 e^{i(\phi_1 + \pi/3)} = E_0 e^{i\phi_1}\left(1 + e^{i\pi/3}\right).$$

Intensity:

$$I \propto |E_{\text{tot}}|^2 = E_0^2 |1 + e^{i\pi/3}|^2 = E_0^2 \left[(1+\cos\pi/3)^2 + (\sin\pi/3)^2\right]$$ $$= E_0^2 \left[(1.5)^2 + (0.866)^2\right] = E_0^2(2.25 + 0.75) = 3E_0^2.$$

Thus $I = 3I_0$, which is three times the intensity of either wave alone. This exceeds the sum of the individual intensities ($2I_0$) because of constructive interference.

Problems with Solutions

1. Two-wave interference with unequal amplitudes. Two waves with amplitudes $E_{01} = 2E_{02}$ and phase difference $\pi$ interfere. Find the resultant intensity in units of $I_2 \propto E_{02}^2$.

Solution.

$$I \propto |E_{01} + E_{02}e^{i\pi}|^2 = |2E_{02} - E_{02}|^2 = E_{02}^2.$$

So the total intensity equals that of the weaker wave alone ($1\,I_2$).

2. Beat frequency. Two waves have frequencies $\nu_1 = 500.0\,\text{THz}$ and $\nu_2 = 500.1\,\text{THz}$. What is the beat frequency and beat period?

Solution.

$$f_{\text{beat}} = |\nu_2 - \nu_1| = 0.1\,\text{THz} = 10^{11}\,\text{Hz},$$ $$T_{\text{beat}} = \frac{1}{10^{11}} = 10^{-11}\,\text{s} = 10\,\text{ps}.$$

3. Standing wave nodes. Two counter-propagating waves of wavelength $\lambda = 600\,\text{nm}$ form a standing wave. What is the spacing between adjacent nodes?

Solution. Nodes occur where the amplitude is zero. For a standing wave $\propto \sin(kx)$, nodes are separated by $\lambda/2 = 300\,\text{nm}$.

Section summary. Most optical structure comes from coherent superposition.

Polarization

Core ideas

Polarization is the vector state of the transverse electric field. Linear, circular, and elliptical polarization describe different phase relations between orthogonal components. Polarizers, wave plates, birefringent media, and Jones vectors provide practical control.

For review, be able to use Jones vectors, identify circular and linear polarization, compute Malus’ law, and understand birefringent phase delay. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$\bm E(t)=\mathrm{Re}\left[(E_x\hat{\bm x}+E_ye^{i\delta}\hat{\bm y})e^{-i\omega t}\right],\qquad I=I_0\cos^2\theta$$

Worked example

Unpolarized light of intensity $I_0 = 100\,\text{mW/cm}^2$ passes through a linear polarizer with transmission axis at $30^\circ$ to the vertical, then through a second polarizer at $60^\circ$ to the vertical (i.e., $30^\circ$ relative to the first). Find the transmitted intensity after each polarizer.

After the first polarizer: Unpolarized light loses half its intensity:

$$I_1 = \frac{I_0}{2} = 50\,\text{mW/cm}^2.$$

The transmitted light is linearly polarized along $30^\circ$.

After the second polarizer: The angle between the two transmission axes is $\theta = 60^\circ - 30^\circ = 30^\circ$. By Malus’ law,

$$I_2 = I_1\cos^2\theta = 50\cos^2 30^\circ = 50(0.75) = 37.5\,\text{mW/cm}^2.$$

Problems with Solutions

1. Three polarizers. Start with unpolarized light and add a third polarizer at $90^\circ$ to the first, placed between two crossed polarizers at $0^\circ$ and $90^\circ$. If the middle polarizer is at $45^\circ$, what is the final intensity for $I_0 = 100\,\text{mW/cm}^2$?

Solution.

$$I_1 = I_0/2 = 50,\quad I_2 = 50\cos^2 45^\circ = 25,\quad I_3 = 25\cos^2 45^\circ = 12.5\,\text{mW/cm}^2.$$

2. Quarter-wave plate. Linearly polarized light at $45^\circ$ to the fast axis of a quarter-wave plate enters the plate. Describe the output polarization.

Solution. The two orthogonal components have equal amplitude and acquire a $\pi/2$ phase difference. The output is circularly polarized (right- or left-handed depending on which axis is fast).

3. Ellipticity. Light has $E_x = 3.0\,\text{V/m}$ and $E_y = 4.0\,\text{V/m}$ with $\delta = \pi/2$. Is the polarization circular, linear, or elliptical?

Solution. The amplitudes are unequal and the phase difference is $\pi/2$, so the polarization is elliptical. The tip of the $\bm E$ vector traces an ellipse with semi-axes $3.0$ and $4.0\,\text{V/m}$.

Section summary. Polarization is the vector degree of freedom of light.

Interference

Core ideas

Interference measures phase difference. Young’s experiment, Michelson interferometers, thin films, and Fabry—Perot cavities all compare optical paths. Visibility depends on coherence, amplitude balance, and phase stability.

For review, be able to compute fringe spacing, thin-film phase shifts, optical path difference, and interferometer sensitivity. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$\Delta\phi=\frac{2\pi}{\lambda}\Delta L,\qquad I=I_1+I_2+2\sqrt{I_1I_2}\cos\Delta\phi$$

Worked example

In a Young’s double-slit experiment the slit separation is $d = 0.20\,\text{mm}$, the screen is $L = 1.0\,\text{m}$ away, and the wavelength is $\lambda = 500\,\text{nm}$. Find (a) the angular position of the first bright fringe, and (b) the fringe spacing on the screen.

(a) First bright fringe: Bright fringes occur when $\Delta L = d\sin\theta = m\lambda$. For $m = 1$,

$$\sin\theta_1 = \frac{\lambda}{d} = \frac{500\times10^{-9}}{0.20\times10^{-3}} = 2.5\times10^{-3},$$ $$\theta_1 \approx 0.143^\circ \approx 2.5\times10^{-3}\,\text{rad}.$$

(b) Fringe spacing: For small angles, $y_m = m\lambda L/d$, so the spacing is

$$\Delta y = \frac{\lambda L}{d} = \frac{(500\times10^{-9})(1.0)}{0.20\times10^{-3}} = 2.5\times10^{-3}\,\text{m} = 2.5\,\text{mm}.$$

Problems with Solutions

1. Thin-film interference. A soap film ($n = 1.33$) in air has thickness $t = 300\,\text{nm}$. For normal incidence, what visible wavelengths are strongly reflected?

Solution. There is a $\pi$ phase shift at the front surface (air to soap) but not at the back (soap to air). Constructive interference requires $2nt = (m+1/2)\lambda$. For $m = 0$,

$$\lambda = 4nt = 4(1.33)(300) = 1596\,\text{nm} \text{ (infrared)}.$$

For $m = 1$,

$$\lambda = \frac{4nt}{3} = \frac{1596}{3} = 532\,\text{nm} \text{ (green)}.$$

For $m = 2$,

$$\lambda = \frac{4nt}{5} = 319\,\text{nm} \text{ (UV)}.$$

So the dominant visible reflected wavelength is $532\,\text{nm}$.

2. Michelson interferometer. One mirror is moved by $\Delta d = 0.10\,\text{mm}$ and $400$ fringes cross the field of view. What is the wavelength?

Solution. Each fringe corresponds to $\lambda/2$ mirror displacement, so

$$N\frac{\lambda}{2} = \Delta d \implies \lambda = \frac{2\Delta d}{N} = \frac{2(0.10\times10^{-3})}{400} = 5.0\times10^{-7}\,\text{m} = 500\,\text{nm}.$$

3. Equal intensities. In a two-beam interferometer $I_1 = I_2 = I_0$. What are the maximum and minimum intensities, and what is the visibility $V$?

Solution.

$$I_{\max} = I_0 + I_0 + 2I_0 = 4I_0,\quad I_{\min} = I_0 + I_0 - 2I_0 = 0,$$ $$V = \frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}} = \frac{4I_0}{4I_0} = 1.$$

Section summary. Interference converts phase differences into intensity patterns.

Diffraction

Core ideas

Diffraction appears when apertures or obstacles are comparable to wavelength. Huygens—Fresnel ideas, Fraunhofer diffraction, single slits, gratings, and circular apertures explain resolution limits and beam spreading.

For review, be able to calculate basic diffraction patterns, use grating equations, estimate angular resolution, and distinguish near-field from far-field diffraction. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$a\sin\theta=m\lambda,\qquad d\sin\theta=m\lambda,\qquad \theta_{\rm Rayleigh}\approx1.22\lambda/D$$

Worked example

A single slit of width $a = 0.10\,\text{mm}$ is illuminated by green light of wavelength $\lambda = 540\,\text{nm}$. A lens of focal length $f = 50\,\text{cm}$ placed after the slit focuses the pattern onto a screen. Find the width of the central maximum.

First minima: The first minima occur at angles satisfying $a\sin\theta = \pm\lambda$. For small angles,

$$\theta_{\pm 1} \approx \pm\frac{\lambda}{a} = \pm\frac{540\times10^{-9}}{0.10\times10^{-3}} = \pm 5.4\times10^{-3}\,\text{rad}.$$

Position on screen: Using $y = f\tan\theta \approx f\theta$,

$$y_{\pm 1} = (0.50)(\pm 5.4\times10^{-3}) = \pm 2.7\,\text{mm}.$$

Central maximum width:

$$w = y_{+1} - y_{-1} = 5.4\,\text{mm}.$$

Problems with Solutions

1. Diffraction grating. A grating with $d = 2.0\,\mu\text{m}$ is illuminated at normal incidence by $\lambda = 600\,\text{nm}$. How many orders are visible?

Solution. Maxima occur for $|\sin\theta| \le 1$, so $|m| \le d/\lambda = 2.0/0.60 = 3.33$. Integer orders are $m = 0, \pm 1, \pm 2, \pm 3$, so $7$ orders are visible.

2. Telescope resolution. A telescope mirror has diameter $D = 20\,\text{cm}$. What is the angular resolution at $\lambda = 550\,\text{nm}$ according to the Rayleigh criterion?

Solution.

$$\theta_{\text{Rayleigh}} \approx 1.22\frac{\lambda}{D} = 1.22\frac{550\times10^{-9}}{0.20} = 3.36\times10^{-6}\,\text{rad} \approx 0.69\,\text{arcsec}.$$

3. Laser beam divergence. A He—Ne laser ($\lambda = 633\,\text{nm}$) emerges from a circular aperture of diameter $D = 1.0\,\text{mm}$. Estimate the full-angle divergence of the central diffraction lobe.

Solution. The first Airy minimum is at $\theta \approx 1.22\lambda/D$, so the full width is roughly twice this:

$$\Delta\theta \approx 2(1.22)\frac{633\times10^{-9}}{1.0\times10^{-3}} \approx 1.5\times10^{-3}\,\text{rad} \approx 0.09^\circ.$$

Section summary. Diffraction sets the wave limit of imaging and beam focusing.

Fourier Optics

Core ideas

Fourier optics treats propagation and imaging as spatial-frequency filtering. A lens maps angles to positions in its focal plane, making diffraction patterns and transfer functions natural. This explains resolution, filtering, convolution, and coherent imaging.

For review, be able to interpret apertures as filters, use Fourier transforms for far-field patterns, and connect point-spread functions to resolution. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$U_{\rm far}(k_x,k_y)\propto \mathcal F\{U(x,y)\},\qquad I_{\rm image}=|h*U_{\rm object}|^2$$

Worked example

A rectangular aperture of width $a = 0.10\,\text{mm}$ and height $b = 0.20\,\text{mm}$ is illuminated by a plane wave of wavelength $\lambda = 500\,\text{nm}$. Using Fourier optics, find the angular width of the central diffraction spot in the $x$ direction.

Fourier transform: The field immediately after the aperture is $U(x,y) = \text{rect}(x/a)\text{rect}(y/b)$. Its Fourier transform is

$$\tilde{U}(k_x,k_y) \propto \text{sinc}\left(\frac{k_x a}{2}\right)\text{sinc}\left(\frac{k_y b}{2}\right).$$

Far-field angle: The first zero in the $x$ direction occurs when $k_x a/2 = \pi$, i.e., $k_x = 2\pi/a$. Since $k_x = (2\pi/\lambda)\sin\theta$,

$$\sin\theta_{x,1} = \frac{\lambda}{a} = \frac{500\times10^{-9}}{0.10\times10^{-3}} = 5.0\times10^{-3}.$$

The full angular width of the central spot is $2\theta_{x,1} \approx 10\times10^{-3}\,\text{rad} = 10\,\text{mrad}$.

Problems with Solutions

1. Spatial frequency cutoff. A lens has diameter $D = 5.0\,\text{cm}$ and focal length $f = 10\,\text{cm}$. What is the maximum spatial frequency $|k_x|$ it can transmit for $\lambda = 500\,\text{nm}$?

Solution. The highest angle accepted is $\theta_{\max} \approx D/(2f) = 0.05/0.20 = 0.25\,\text{rad}$. Then

$$k_{x,\max} = \frac{2\pi}{\lambda}\sin\theta_{\max} \approx \frac{2\pi}{500\times10^{-9}}(0.25) \approx 3.1\times10^{6}\,\text{rad/m}.$$

2. Point-spread function width. A circular aperture of diameter $D$ produces an Airy pattern. Write the approximate radius of the first zero in terms of $f$-number $F = f/D$.

Solution.

$$r_{\text{Airy}} \approx 1.22\frac{\lambda f}{D} = 1.22\lambda F.$$

For $\lambda = 500\,\text{nm}$ and $F = 2$, $r_{\text{Airy}} \approx 1.22\,\mu\text{m}$.

3. 1D Fourier transform of two slits. Two delta slits separated by distance $d$ have amplitude transmission $t(x) = \delta(x-d/2) + \delta(x+d/2)$. What is the far-field intensity pattern?

Solution.

$$\mathcal F\{t(x)\} \propto e^{-ik_x d/2} + e^{+ik_x d/2} = 2\cos(k_x d/2),$$ $$I(k_x) \propto \cos^2\left(\frac{k_x d}{2}\right).$$

This is the familiar two-slit interference pattern.

Section summary. Optical systems process spatial frequencies.

Coherence

Core ideas

Coherence measures how stable phase relationships are across time and space. Temporal coherence is tied to bandwidth; spatial coherence is tied to source size. Coherence functions explain fringe visibility, lasers, thermal light, and imaging contrast.

For review, be able to estimate coherence time and length, relate bandwidth to coherence, and explain visibility loss. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$\tau_c\sim \frac{1}{\Delta\nu},\qquad \ell_c\sim c\tau_c,\qquad V=\frac{I_{\max}-I_{\min}}{I_{\max}+I_{\min}}$$

Worked example

A sodium lamp emits at $\lambda = 589\,\text{nm}$ with a linewidth (FWHM) of $\Delta\nu = 5.0\,\text{GHz}$. Estimate (a) the coherence time, (b) the coherence length, and (c) the number of oscillations in one coherence time.

(a) Coherence time:

$$\tau_c \sim \frac{1}{\Delta\nu} = \frac{1}{5.0\times10^9} = 2.0\times10^{-10}\,\text{s} = 0.20\,\text{ns}.$$

(b) Coherence length:

$$\ell_c \sim c\tau_c = (3.0\times10^8)(2.0\times10^{-10}) = 6.0\times10^{-2}\,\text{m} = 6.0\,\text{cm}.$$

(c) Number of oscillations: The optical period is $T = 1/\nu = \lambda/c = 589\times10^{-9}/(3.0\times10^8) \approx 1.96\times10^{-15}\,\text{s}$. Then

$$N = \frac{\tau_c}{T} = \frac{2.0\times10^{-10}}{1.96\times10^{-15}} \approx 1.0\times10^5.$$

Problems with Solutions

1. Laser linewidth. A He—Ne laser has $\Delta\nu = 1.0\,\text{GHz}$. What is its coherence length?

Solution.

$$\ell_c = \frac{c}{\Delta\nu} = \frac{3.0\times10^8}{1.0\times10^9} = 0.30\,\text{m} = 30\,\text{cm}.$$

2. Visibility from intensity data. In a two-beam interference pattern the maximum intensity is $I_{\max} = 12\,\text{W/m}^2$ and the minimum is $I_{\min} = 4\,\text{W/m}^2$. What is the visibility?

Solution.

$$V = \frac{12 - 4}{12 + 4} = \frac{8}{16} = 0.50.$$

3. White-light coherence. White light spans roughly $\Delta\lambda = 300\,\text{nm}$ around $\lambda = 550\,\text{nm}$. Estimate the coherence length.

Solution. Using $\nu = c/\lambda$, we have $\Delta\nu \approx (c/\lambda^2)\Delta\lambda$. Then

$$\ell_c \sim \frac{c}{\Delta\nu} \approx \frac{\lambda^2}{\Delta\lambda} = \frac{(550\times10^{-9})^2}{300\times10^{-9}} \approx 1.0\times10^{-6}\,\text{m} = 1.0\,\mu\text{m}.$$

Section summary. Coherence determines whether interference survives averaging.

Lasers and Modern Optics

Core ideas

Lasers use stimulated emission and optical feedback to produce coherent, narrow-band, directional light. Population inversion, gain, threshold, cavity modes, linewidth, nonlinear optics, fibers, and photodetectors form the toolbox of modern optical experiments.

For review, be able to explain stimulated emission, threshold, cavity modes, Gaussian beams, and the difference between spontaneous and laser light. Keep the physical question visible: identify the degrees of freedom, the conserved quantities, the approximation being made, and the observable that would be measured.

Mathematical spine

$$gL\ge \alpha_{\rm loss},\qquad \Delta\nu_{\rm FSR}=\frac{c}{2nL},\qquad w(z)=w_0\sqrt{1+(z/z_R)^2}$$

Worked example

A He—Ne laser cavity has mirror spacing $L = 30\,\text{cm}$ and operates at $\lambda = 633\,\text{nm}$. The gain medium has unsaturated gain coefficient $g_0 = 0.05\,\text{m}^{-1}$ and round-trip loss $\alpha_{\text{loss}} = 0.02$ (fractional). Find (a) the threshold gain condition, (b) the free spectral range in frequency, and (c) the longitudinal mode spacing in wavelength near $633\,\text{nm}$.

(a) Threshold: The round-trip gain must equal or exceed the round-trip loss:

$$2g_{\text{th}}L = \alpha_{\text{loss}} \implies g_{\text{th}} = \frac{\alpha_{\text{loss}}}{2L} = \frac{0.02}{2(0.30)} = 0.033\,\text{m}^{-1}.$$

Since $g_0 = 0.05 > g_{\text{th}}$, the laser is above threshold.

(b) Free spectral range:

$$\Delta\nu_{\text{FSR}} = \frac{c}{2nL} = \frac{3.0\times10^8}{2(1.0)(0.30)} = 5.0\times10^8\,\text{Hz} = 500\,\text{MHz}.$$

(c) Mode spacing in wavelength: Using $\nu = c/\lambda$ and differentiating, $|\Delta\lambda| = (\lambda^2/c)\Delta\nu$. Thus

$$\Delta\lambda = \frac{(633\times10^{-9})^2}{3.0\times10^8}(5.0\times10^8) \approx 6.7\times10^{-13}\,\text{m} = 0.67\,\text{pm}.$$

Problems with Solutions

1. Cavity modes. A diode laser cavity has $L = 300\,\mu\text{m}$ and $n = 3.5$. What is the FSR and how many modes fall within the gain bandwidth $\Delta\nu_{\text{gain}} = 1.0\,\text{THz}$?

Solution.

$$\Delta\nu_{\text{FSR}} = \frac{c}{2nL} = \frac{3.0\times10^8}{2(3.5)(300\times10^{-6})} \approx 1.43\times10^{11}\,\text{Hz} = 143\,\text{GHz}.$$

Number of modes:

$$N \approx \frac{\Delta\nu_{\text{gain}}}{\Delta\nu_{\text{FSR}}} = \frac{1.0\times10^{12}}{1.43\times10^{11}} \approx 7.$$

2. Gaussian beam waist. A focused laser beam has $w_0 = 1.0\,\text{mm}$ at its waist and $\lambda = 1.06\,\mu\text{m}$ (Nd:YAG). What is the Rayleigh range $z_R$ and the beam radius at $z = 1.0\,\text{m}$?

Solution.

$$z_R = \frac{\pi w_0^2}{\lambda} = \frac{\pi(1.0\times10^{-3})^2}{1.06\times10^{-6}} \approx 2.96\,\text{m},$$ $$w(1.0) = w_0\sqrt{1 + (1.0/2.96)^2} \approx 1.0\sqrt{1.114} \approx 1.06\,\text{mm}.$$

3. Threshold pump power. A laser has output coupling loss $\alpha_{\text{out}} = 0.01$ and internal loss $\alpha_{\text{int}} = 0.005$ per round trip. The gain coefficient is $g = \sigma(N_2 - N_1)$ with stimulated-emission cross section $\sigma = 3.0\times10^{-23}\,\text{m}^2$ and length $L = 5.0\,\text{cm}$. What population difference $(N_2 - N_1)$ is needed at threshold?

Solution. Total loss is $\alpha_{\text{loss}} = 0.01 + 0.005 = 0.015$. Then

$$2g_{\text{th}}L = \alpha_{\text{loss}} \implies g_{\text{th}} = \frac{0.015}{2(0.05)} = 0.15\,\text{m}^{-1},$$ $$N_2 - N_1 = \frac{g_{\text{th}}}{\sigma} = \frac{0.15}{3.0\times10^{-23}} = 5.0\times10^{21}\,\text{m}^{-3}.$$

Section summary. Lasers make controlled coherent light for measurement, communication, and quantum experiments.