10. Wave Optics

Big idea: phase determines how light waves reinforce or cancel, while polarization describes their transverse field direction.

Use wavefront geometry to connect reflection and refraction with interference and diffraction.

Christiaan Huygens developed a wavefront construction in the seventeenth century. Thomas Young’s interference work later supported a wave description, and Augustin-Jean Fresnel developed a fuller wave theory explaining diffraction. These ideas complemented, rather than removed, the usefulness of ray diagrams.

Rays locate an image, but waves explain why light can make bright and dark bands.

A wavefront joins points at the same phase of a wave. In an isotropic medium, a ray is locally perpendicular to the wavefront. A point source produces spherical wavefronts; sufficiently far away, a small part of such a front can be treated as plane. A long line source gives approximately cylindrical fronts.

Phase tells where an oscillation lies in its cycle. Equal phase does not require equal amplitude. The wave description accounts for interference, diffraction and polarization, which cannot all be explained by tracing independent geometrical rays.

A point source is placed at the principal focus of an ideal converging lens. Describe the outgoing wavefront in the paraxial region.

The lens makes the outgoing rays parallel. A wavefront perpendicular to parallel rays is planar.

Result: An approximately plane wavefront, neglecting aperture diffraction and aberrations.

1. What geometric relation connects rays and wavefronts in an isotropic medium?
2. What is the wavefront shape near an ideal point source in a uniform medium?

Answers

1. They are locally perpendicular.
2. Spherical.

Huygens’ construction predicts a new wavefront from an old one.

Treat each point of an existing wavefront as a source of secondary wavelets. After a short interval, construct the forward envelope tangent to these wavelets. This envelope gives the next wavefront. In a uniform isotropic medium each wavelet expands the same distance vΔt.

For a plane front, equal-radius wavelets have a new plane forward envelope, so rays remain parallel. For a spherical front, the envelope is another sphere with a larger radius. The construction gives propagation directions, while quantitative intensity needs the fuller interference of wavelets and appropriate amplitudes.

In a uniform medium, wave speed is $2.0\times10^8\,\mathrm{m\,s^{-1}}$. How far does a secondary wavelet travel in $2.0\,\mathrm{ns}$?

Multiply speed by the time interval in seconds.

Result: $0.4\,\mathrm m$.

1. What determines wavelet radius after a chosen interval?
2. Is elementary Huygens geometry alone a complete formula for diffraction intensity?

Answers

1. The medium’s speed multiplied by the same elapsed time.
2. No. Quantitative intensity also requires phase and amplitude superposition.

At a boundary, one side of a wavefront enters the next medium before the other side.

Let one end reach the boundary first. During a time interval Δt, the remaining incident part travels $v_1\Delta t$ while the transmitted wavelet grows to $v_2\Delta t$. The two right triangles give $\sin i/\sin r=v_1/v_2=n_2/n_1$, hence Snell’s law. Angles refer to ray directions measured from the normal.

The boundary is stationary, so successive crests cross it at the source frequency. A lower speed therefore means a shorter wavelength. When entering higher index, the wavefront turns so its normal ray bends towards the interface normal. When entering lower index, it bends away if a propagating transmitted ray exists.

For higher-to-lower index, $\sin r=(n_1/n_2)\sin i$. Increasing i eventually makes r reach a right angle. Above that critical condition no ordinary propagating transmitted ray satisfies Snell’s law, producing total internal reflection in the ideal ray description.

At a reflecting plane, construct wavelets in the incident medium. Equal propagation speed for incident and reflected fronts gives equality of incidence and reflection angles. Incident ray, reflected ray and normal share one plane. This result agrees with geometrical reflection.

Vacuum wavelength is $600\,\mathrm{nm}$ and a transparent medium has $n=1.50$ at that frequency. Find its wavelength there.

Frequency stays fixed, speed becomes c/n, and therefore $\lambda_{\rm medium}=\lambda_0/n$.

Result: $\lambda=400\,\mathrm{nm}$.

1. Why does wavelength shorten in higher index at a stationary boundary?
2. What does the Huygens reflection construction predict?

Answers

1. Speed decreases while frequency remains fixed.
2. Equal incidence and reflection angles.

Add wave amplitudes first; intensity follows from the square of the resultant amplitude.

For two waves of the same frequency and fixed relative phase δ, $I=I_1+I_2+2\sqrt{I_1I_2}\cos\delta$, assuming compatible polarization and proportional amplitude-to-intensity constants. Constructive phase gives $I_{\max}=(\sqrt{I_1}+\sqrt{I_2})^2$; destructive phase gives $I_{\min}=(\sqrt{I_1}-\sqrt{I_2})^2$. Complete darkness requires equal individual intensities as well as the destructive phase.

Coherent sources have a stable relative phase over the observation time. Two ordinary independent lamps do not maintain that relation, so the cross term averages away and their mean intensities add. Splitting one source into two paths is a practical way to obtain mutual coherence; equal frequency alone is insufficient.

Two coherent equally polarized contributions have intensities $9$ and $4$ in the same units. Find extreme resultant intensities.

Their amplitudes are proportional to 3 and 2. Add or subtract amplitudes, then square.

Result: $I_{\max}=25$ and $I_{\min}=1$ in the original intensity units.

A path difference Δ in a uniform medium contributes phase difference $2\pi\Delta/\lambda$, in addition to any initial source or reflection phase. Constructive and destructive conditions must include all such phase shifts. Bright bands gain energy redistributed from dark bands; interference does not create extra net energy.

1. Can unequal coherent beams produce a perfectly dark minimum?
2. Does matching source frequency alone guarantee a stable fringe pattern?

Answers

1. Not in this ideal two-beam model; amplitudes must be equal.
2. No. Their relative phase must also remain suitably stable.

Two narrow coherent slits turn a small path difference into a measurable sequence of fringes.

Let slit separation be d, screen distance D, and transverse coordinate y. For $D\gg d$ and small angles, path difference is $\Delta\approx d\sin\theta\approx dy/D$. For equal-phase sources, bright positions satisfy $y_m=m\lambda D/d$ and dark positions satisfy $y=(m+1/2)\lambda D/d$. The fringe spacing is $\beta=\lambda D/d$.

The central bright fringe corresponds to equal paths. Fringe spacing grows with wavelength or screen distance and shrinks with slit separation. Keep both slits illuminated coherently, with matching polarization and sufficient overlap on the screen. Finite slit width places a diffraction envelope over the interference pattern.

Use $\lambda=520\,\mathrm{nm}$, $D=1.8\,\mathrm m$ and $d=0.30\,\mathrm{mm}$. Find fringe spacing and the third bright-fringe distance from the centre.

Convert both small lengths to metres. Compute β, then multiply by the order m.

Result: $\beta=0.00312\,\mathrm m$ and $y_3=0.00936\,\mathrm m$.

If one slit is blocked, the two-path interference disappears while the remaining slit still diffracts. With white light the equal-path centre can appear white and neighbouring fringes separate by colour; far from the centre overlapping orders reduce clear contrast.

1. What happens to fringe spacing if d doubles?
2. Why must the small-angle approximation be stated?

Answers

1. It halves, keeping wavelength and screen distance fixed.
2. The simple y-position formula uses sinθ approximately equal to tanθ approximately equal to y/D.

Even one slit contains many coherent contributions, so it can make a spread-out bright-and-dark pattern.

For a slit of width a and a far observation direction θ, the edge-to-edge path difference is $a\sin\theta$. At the first minimum this is one wavelength. Pair each point in one half of the slit with a point in the other half: their path difference is half a wavelength, so equal contributions cancel pairwise. The general minima satisfy $a\sin\theta=m\lambda$ for nonzero integer m.

For small angles the first minima lie near $y=\pm\lambda D/a$, so the central maximum has width $2\lambda D/a$. It is twice the spacing between adjacent minima away from the centre in this approximation. Secondary maxima are weaker; do not place them exactly halfway using a universal rule.

A $0.12\,\mathrm{mm}$ slit is illuminated by $600\,\mathrm{nm}$ light. The far screen is $2.0\,\mathrm m$ away. Find the approximate first-minimum angle and central width.

Use $\theta\approx\lambda/a$ in radians, then central width $2D\lambda/a$.

Result: $\theta\approx0.005\,\mathrm{rad}$ and width $0.02\,\mathrm m$.

Narrowing the slit broadens diffraction. This explains a limit of ray optics: an arbitrarily narrow opening does not make an arbitrarily narrow outgoing beam. Any finite optical aperture spreads an image of a point. Two nearby point images become difficult to distinguish when their diffraction patterns overlap strongly.

Interference and diffraction both arise from superposition. The names usually emphasize separate sources or paths for interference and different parts of one aperture for diffraction. They are not competing physical mechanisms. The far-field formulas here assume a suitable observation distance or a lens that forms a Fourier-plane pattern.

1. Why does the centre require special treatment in the single-slit intensity formula?
2. How does halving slit width affect central diffraction width?

Answers

1. The ratio sin u/u has limiting value one at u=0, not a physical zero or infinity.
2. It doubles in the small-angle far-field approximation.

Polarization reveals that light’s field oscillation is transverse.

Linear polarization means the electric field oscillates along one fixed line perpendicular to propagation. Unpolarized light has no fixed preferred transverse orientation over the observation interval. A longitudinal wave has no analogous choice of transverse oscillation direction, so polarization provides evidence for light’s transverse nature.

An ideal linear polarizer transmits the component along its transmission axis. If linearly polarized input makes angle θ with an ideal analyser axis, the transmitted intensity is $I=I_0\cos^2\theta$ (Malus’s law). For unpolarized input, the first ideal polarizer transmits half the mean intensity because the average of $\cos^2\theta$ is $1/2$.

A polarized beam of intensity $80\,\mathrm{W\,m^{-2}}$ meets an ideal analyser at $60^\circ$ to its axis. Find transmitted intensity.

Use Malus’s law with the angle between the two transverse directions.

Result: $I=20\,\mathrm{W\,m^{-2}}$.

At Brewster incidence on an ideal dielectric interface, reflected light is fully linearly polarized perpendicular to the plane of incidence for unpolarized illumination. Reflected and refracted rays are perpendicular then. Combining Snell’s law with $i_B+r=90^\circ$ gives $\tan i_B=n_2/n_1$. This ideal result assumes suitable transparent, nonmagnetic media; real surfaces can depart from it.

Find Brewster angle from air into index $1.50$.

$i_B=\tan^{-1}(1.50)$.

Result: $i_B=56.3099^\circ$.

The Brewster relation follows explicitly from Snell’s law: $n_1\sin i_B=n_2\sin r$. Since $r=90^\circ-i_B$, $\sin r=\cos i_B$. Dividing by $n_1\cos i_B$ gives $\tan i_B=n_2/n_1$. The plane of incidence is defined by the incoming ray and interface normal; the reflected electric field at this angle is perpendicular to that plane.

Two crossed ideal polarizers block a linearly polarized input after the second. Inserting another polarizer between them at an intermediate angle can allow some light through because each element selects a new component. This does not create energy: each transmitted intensity is no greater than the intensity incident on that element.

1. What is transmitted by the first ideal polarizer from unpolarized light?
2. At Brewster angle, how are reflected and refracted rays related?
3. Does the polarization direction refer to E or the propagation direction?

Answers

1. Half the mean incident intensity.
2. They are perpendicular.
3. To the transverse electric-field oscillation direction.

• A wavefront joins equal-phase points; rays are locally normal to it in an isotropic medium.
• Coherent addition: $I=I_1+I_2+2\sqrt{I_1I_2}\cos\delta$.
• Young fringes: $\beta=\lambda D/d$ in the far-screen small-angle limit.
• Single-slit minima: $a\sin\theta=m\lambda$, $m\ne0$; central width approximately $2\lambda D/a$.
• Malus: $I=I_0\cos^2\theta$. Brewster: $\tan i_B=n_2/n_1$.

• A dark interference band redistributes light; it does not destroy energy.
• The labels interference and diffraction emphasize different source arrangements of the same superposition principle.
• A figure’s equal apparent wave heights do not imply equal physical quantities or units.

Try these newly written questions before reading the answers. Show a diagram, the governing equation, substitutions and units where appropriate.

10.1 Describe spherical, cylindrical and plane wavefront sources.

10.2 State Huygens’ construction.

10.3 Vacuum light of $540\,\mathrm{nm}$ enters index $1.35$. Find wavelength there.

10.4 What stays fixed when light crosses a stationary boundary?

10.5 Why do two independent lamps not give a stable ordinary two-beam fringe pattern?

10.6 For coherent intensities $16$ and $1$, find maximum and minimum resultant intensity.

10.7 Find fringe spacing for $\lambda=640\,\mathrm{nm}$, $D=1.5\,\mathrm m$, $d=0.40\,\mathrm{mm}$.

10.8 What happens to two-slit fringes if one slit closes?

10.9 A slit width is $0.20\,\mathrm{mm}$ and wavelength $500\,\mathrm{nm}$. Find the approximate first-minimum angle.

10.10 Explain why narrowing a slit increases beam spread.

10.11 Find transmitted intensity for a polarized $120\,\mathrm{W\,m^{-2}}$ beam at $30^\circ$ to an ideal analyser axis.

10.12 Unpolarized intensity is $100\,\mathrm{W\,m^{-2}}$. It passes an ideal polarizer and then an analyser at $45^\circ$. Find final intensity.

10.13 Find Brewster angle from air into index $1.40$.

10.14 Why does polarization distinguish a transverse from a longitudinal wave?

Answers

10.1 An ideal point source gives spherical fronts; a long line source approximately cylindrical fronts; a collimated beam approximately plane fronts.

10.2 Every front point contributes a secondary wavelet; their common forward envelope is the next wavefront.

10.3 $400\,\mathrm{nm}$.

10.4 Frequency; speed and wavelength can change.

10.5 Their relative phase fluctuates, averaging the cross term away.

10.6 $25$ and $9$ in the same units.

10.7 $0.0024\,\mathrm m$.

10.8 Two-path interference vanishes; the remaining slit retains its diffraction pattern.

10.9 $0.0025\,\mathrm{rad}$.

10.10 The first-minimum condition a sinθ=λ requires larger angle when a decreases.

10.11 $90\,\mathrm{W\,m^{-2}}$.

10.12 $25\,\mathrm{W\,m^{-2}}$.

10.13 $54.4623^\circ$.

10.14 Transverse oscillation has independent directions perpendicular to travel that an analyser can select; a purely longitudinal oscillation does not.