Corrected exercises: wave optics (203-SN3-RE)

Bright: $\delta=m\lambda$ (the two waves arrive in phase, crest on crest). Dark: $\delta=(m+\frac{1}{2})\lambda$ (opposite phase, crest on trough). A dark fringe is a place where light PLUS light produces darkness: the energy is not destroyed, it is redistributed toward the bright fringes; this is the signature of a wave, and it cannot be explained with classical particles.

$i=\frac{\lambda L}{d}=\frac{(600\times 10^{-9})(2.0)}{0.20\times 10^{-3}}=6.0\times 10^{-3}$ m, that is 6.0 mm: measurable with a ruler, whereas $\lambda$ is submicron. The setup is a geometric amplifier.

Bright fringes at $y_{m}=mi$: the third is at $y_{3}=18$ mm. Dark fringes at $y=(m+\frac{1}{2})i$: the first ($m=0$) at 3.0 mm, the second ($m=1$) at $1.5\,i=9.0$ mm.

In water, light slows down: $v=\frac{c}{n}$, and since the frequency is set by the source (it NEVER changes when the light changes medium), it is the wavelength that shrinks: $\lambda'=\frac{\lambda}{n}$. The fringe spacing follows: $i'=\frac{i}{n}=\frac{6.0}{1.33}\approx 4.5$ mm. The fringes close up by a quarter, the whole pattern contracts, and the perceived colour, which depends on the frequency, stays the same.

$d=\frac{1}{500}$ mm $=2.0\times 10^{-6}$ m, that is 2.0 μm: barely three wavelengths.

$d\sin\theta_{m}=m\lambda$. Order 1: $\sin\theta_{1}=\frac{600\times 10^{-9}}{2.0\times 10^{-6}}=0.30$, so $\theta_{1}\approx 17.5^{\circ}$. Order 2: $\sin\theta_{2}=0.60$, so $\theta_{2}\approx 36.9^{\circ}$. The angles are NOT proportional to the order: it is the sine that is.

We need $\sin\theta_{m}=\frac{m\lambda}{d}\leq 1$, that is $m\leq\frac{d}{\lambda}=\frac{2.0\times 10^{-6}}{600\times 10^{-9}}\approx 3.3$: orders 1, 2 and 3 exist ($\sin\theta_{3}=0.90$, $\theta_{3}\approx 64^{\circ}$), while order 4 would require $\sin\theta=1.2$: impossible. Three orders on each side, seven spots in all with the central beam.

With thousands of slits, the phase condition must be satisfied by ALL of them at once: at the slightest change of angle, the thousands of waves scramble one another and the intensity collapses, so the maxima become sharp, bright needles. Sharp lines mean the ability to separate two nearby wavelengths: that is spectroscopy. And since $\sin\theta=\frac{m\lambda}{d}$ grows with $\lambda$, RED comes out farther than blue, the exact opposite of a prism: a foolproof way to tell which of the two instruments dispersed the light.

Minima at $a\sin\theta=m\lambda$ ($m=1,\ 2,\ldots$). The formula LOOKS like that of Young's maxima, but here it gives the zeros of intensity: in a single slit, when the path difference between the two edges is $\lambda$, each ray from the upper half finds a partner in the lower half that is in opposite phase, and everything cancels. Same geometry, opposite physics: reciting the formula without its meaning leads straight to the wrong conclusion.

First minimum: $\sin\theta=\frac{\lambda}{a}=\frac{500\times 10^{-9}}{0.10\times 10^{-3}}=5.0\times 10^{-3}$ (small angle). Position: $y=L\tan\theta\approx(3.0)(5.0\times 10^{-3})=15$ mm. The central maximum extends from one minimum to the other: $30$ mm wide.

$a$ is in the denominator: a slit half as wide gives a maximum twice as wide, 60 mm. General rule: the more you squeeze the opening, the more the light spreads out. This is the reverse of the geometric shadow (where shrinking the hole shrinks the spot): below a certain size, trying to confine a wave makes it diverge, so there is a fundamental limit to how fine a beam can be.

Diffraction is pronounced only if the opening is comparable to the wavelength. For a 1 m doorway: light ($\lambda\approx 5\times 10^{-7}$ m) gives $\frac{\lambda}{a}\approx 5\times 10^{-7}$, a spread of only a few ten-millionths of a radian, invisible: light travels straight and casts sharp shadows. Sound ($\lambda\approx 1$ m at 343 Hz) has $\frac{\lambda}{a}\approx 1$: it fans out widely behind the doorway. That is why you can hear a conversation from the hallway without seeing the people talking.

Each of the two reflections occurs on a medium of HIGHER index (air 1.00 to coating 1.38, then coating 1.38 to glass 1.50): each undergoes a phase shift of half a wavelength. The two inversions cancel: only the path difference $2nt$ in the coating remains. This is the special case opticians are looking for, precisely because it makes the condition simple: the choice of $n_{coating}$ between $n_{air}$ and $n_{glass}$ is no accident.

With no net phase shift, destruction requires $2nt=(m+\frac{1}{2})\lambda$. Minimum thickness ($m=0$): $t=\frac{\lambda}{4n}=\frac{550}{4(1.38)}\approx 100$ nm: the famous “quarter-wave” coating, a hundred millionths of a millimetre deposited on every lens of the glasses.

The condition is exact ONLY for 550 nm (the yellow-green, at the peak of the eye's sensitivity). At the two ends of the spectrum, red and blue-violet, the destruction is only partial: a little of those colours is still reflected. Their mixture gives the purplish reflection characteristic of coated optics: the tint is not a defect, it is the signature of the coating.

The bubble (a film of soapy water in air) has only ONE inverted reflection: that of the outer face (air to water); the inner face (water to air) reflects with no phase shift. As the film thins and $t$ tends to zero, the path difference $2nt$ becomes negligible: only the half wavelength of the inversion remains, and the two reflections cancel FOR ALL wavelengths at once: no reflection at all, and the film looks black against a dark background. The black bubble is a film much thinner than the light itself ($t\ll\lambda$), a few tens of nanometres from rupture.

The first reflection occurs on the lower face of the upper plate, so going from glass to AIR, a medium of lower index: no phase shift. The second occurs at the bottom of the wedge, going from air to GLASS, a medium of higher index: a phase shift of half a wavelength. The two rays therefore end up with an extra half wavelength of difference on top of the geometric path difference $2t$. Darkness, which normally requires a half-integer difference, therefore occurs when $2t$ equals a WHOLE number of wavelengths: $2t=m\lambda$. Since the thickness grows linearly along the plates, $t=\frac{D}{L}x$, the fringes are straight lines parallel to the edge, spaced by $\Delta x=\frac{\lambda L}{2D}=\frac{(600\times 10^{-9})(0.120)}{2(5.0\times 10^{-5})}\approx 7.2\times 10^{-4}$ m, that is 0.72 mm. Total number: $\frac{2D}{\lambda}=\frac{1.0\times 10^{-4}}{6.0\times 10^{-7}}\approx 167$, so 166 complete dark fringes.

At the edge, $t=0$ and so $2t=0=0\times\lambda$: this is the dark fringe of order zero, and the line of contact is BLACK. The result may be surprising, since the two surfaces touch there and no extra path is travelled; that is exactly what makes the observation decisive. If the two reflections behaved in the same way, the edge would be bright. It is black, so one of the two reflections, and only one, inverts the wave: the rule of the phase shift on a higher-index medium is confirmed experimentally, with no calculation at all. It is exactly the same blackness as the top of a soap bubble just before it bursts.

In a medium of index $n$, the wavelength becomes $\frac{\lambda}{n}$ and the condition reads $2nt=m\lambda$. The spacing is therefore divided by $n$: $\Delta x'=\frac{0.72}{1.33}\approx 0.54$ mm, and the number of fringes is multiplied by $n$: about 222. The two reflections keep their nature, incidentally (glass to water: lower index, no phase shift; water to glass: higher index, phase shift), so the edge stays black.

At a distance $r$ from the axis, the lens is separated from the plate by $t$, and the Pythagorean theorem in the triangle of radius $R$ gives $R^{2}=r^{2}+(R-t)^{2}$, that is $r^{2}=2Rt-t^{2}\approx 2Rt$ since $t\ll R$. Substituting into $2t=m\lambda$: $t=\frac{m\lambda}{2}$ and $r_{m}=\sqrt{m\lambda R}$. First ring: $r_{1}=\sqrt{(600\times 10^{-9})(2.00)}\approx 1.10$ mm. Tenth: $r_{10}=\sqrt{10}\,r_{1}\approx 3.46$ mm. The rings are therefore NOT equally spaced: their radii grow as $\sqrt{m}$, so they crowd together away from the centre, whereas the fringes of the air wedge are regularly spaced. The difference comes from the geometry: the thickness grows linearly in the wedge, quadratically under the lens.

One fringe corresponds to a thickness change of $\frac{\lambda}{2}$, that is 300 nm in our case. Placing a surface to be tested on a flat reference and looking at the fringes therefore amounts to reading a CONTOUR MAP of the gap between the two, graduated every 300 nm. A perfectly flat surface gives rigorously straight, parallel fringes; the slightest hollow or bump bends them, and the optician reads the defect directly in fractions of a fringe. The eye easily distinguishes a tenth of a fringe, which gives a precision of the order of 30 nanometres, a twentieth of a wavelength: no mechanical measuring instrument comes close, and this is how all precision mirrors and lenses are checked, including those of telescopes.

The uncertainty of the ruler (about half a millimetre) weighs ten times less when spread over ten intervals: measuring long to know short is the first experimental reflex. Fringe spacing: $i=\frac{45}{10}=4.5$ mm.

$i=\frac{\lambda L}{d}$ gives $\lambda=\frac{i\,d}{L}=\frac{(4.5\times 10^{-3})(0.25\times 10^{-3})}{1.8}=6.25\times 10^{-7}$ m $=625$ nm: an orange-red. A ruler graduated in millimetres has just measured a length smaller than a micrometre: all the power of the setup lies in the lever $\frac{L}{d}$, here 7200.

$i=\frac{(450\times 10^{-9})(1.8)}{0.25\times 10^{-3}}=3.24$ mm. The fringe spacing is proportional to $\lambda$: the shorter blue tightens the fringes. The geometry has not moved, only the light has changed: the interference pattern is a portrait of the wavelength.

The central fringe is WHITE: at $y=0$, the path difference is zero for all colours at once, so all of them are constructive there. As soon as you move away, each colour has its own fringe spacing ($i\propto\lambda$): the blue fringes tighten, the red ones spread out, so the first fringes are iridescent, blue on the inside and red on the outside. Farther out, the patterns of all the colours overlap in disorder and the screen becomes uniformly white again: blurring replaces contrast.

Two light beams can ADD to produce darkness, which corpuscles piling up could never do: only a wave, capable of opposite phase, explains the dark fringes.

Fringe spacing: $i=\frac{\lambda L}{d}=\frac{(600\times 10^{-9})(2.0)}{3.0\times 10^{-4}}=4.0\times 10^{-3}$ m, that is 4.0 mm. First diffraction minimum: $a\sin\theta=\lambda$ gives $y=\frac{\lambda L}{a}=\frac{(600\times 10^{-9})(2.0)}{7.5\times 10^{-5}}=1.6\times 10^{-2}$ m, that is 16 mm. The central maximum therefore extends from $-16$ mm to $+16$ mm.

A bright fringe of order $m$ lies at $y=m\frac{\lambda L}{d}$; a diffraction minimum of order $p$ lies at $y=p\frac{\lambda L}{a}$. They coincide when $\frac{m}{d}=\frac{p}{a}$, that is $m=p\frac{d}{a}$. Here $\frac{d}{a}=\frac{0.30}{0.075}=4$: orders $m=4$, $8$, $12$... fall exactly on zeros of the envelope and are therefore ERASED, whatever intensity interference alone would predict there. General rule: if $\frac{d}{a}$ is an integer, all its multiples are missing orders. This is the only case in the whole chapter where two phenomena contradict each other, and diffraction has the last word: it cannot give light where it sends none.

In the central maximum, $|y|<16$ mm, with fringes every 4.0 mm: orders $m=0$, $\pm 1$, $\pm 2$, $\pm 3$ fit inside, order $\pm 4$ falling precisely on the edge and being erased. We therefore count SEVEN bright fringes in the central maximum, of unequal intensities since they follow the decay of the envelope.

Narrowing the slits widens the envelope, since its width goes as $\frac{1}{a}$: the missing orders move outward, and in the limit of a very narrow slit the envelope covers the whole screen and we recover the ideal Young pattern, with fringes of equal intensity. Conversely, widening the slits narrows the envelope and makes the higher orders disappear. The case $a=d$ is the extreme limit: the two slits touch and form only one slit of width $2a$; the ratio $\frac{d}{a}=1$ also announces that ALL orders except the central one are missing, which is exactly the diffraction pattern of a single slit. The two formulas join up, which is rather reassuring.

Number of illuminated lines: $N=(500\text{ lines/mm})(30\text{ mm})=1.5\times 10^{4}$. At order 2: $R=mN=3.0\times 10^{4}$. The smallest wavelength difference that can be told apart is therefore $\Delta\lambda=\frac{\lambda}{R}=\frac{589.0}{3.0\times 10^{4}}\approx 0.020$ nm. The sodium doublet is separated by 0.6 nm, thirty times more: it is very comfortably resolved, and the two lines of the doublet would be seen well apart. Remember what the formula $R=mN$ says: the fineness of a spectroscope depends not on the quality of its grating but on the NUMBER of lines actually illuminated, and so on the width of the beam. Widening the beam, or going up in order, are the only two levers.

Moving the mirror by $\Delta d$ changes the path difference by $2\Delta d$, and each time it changes by one wavelength, the pattern returns to its initial appearance: one fringe has gone by. So $2\Delta d=N\lambda$, that is $\Delta d=\frac{N\lambda}{2}$. Here: $\Delta d=\frac{(792)(546.1\times 10^{-9})}{2}\approx 2.16\times 10^{-4}$ m, that is 0.216 mm. Notice what the instrument does: it converts a length into a NUMBER, which you just have to count.

One fringe corresponds to $\frac{\lambda}{2}=273$ nm, a little more than a quarter of a micrometre. At a tenth of a fringe, the precision reaches about 27 nm. A laboratory micrometer caliper, for its part, tops out at the micrometre: the interferometer is therefore about forty times finer, and above all it does not touch the object measured. This is what allowed Michelson to tie the standard platinum-iridium metre to a wavelength of cadmium, and it is the principle that still governs precision machine tools today.

In the cell of length $L$, the light makes the round trip, so it travels $2L$ through the gas. Replacing vacuum by a gas of index $n$ changes the OPTICAL path by $2L(n-1)$, so $2L(n-1)=N\lambda$ and $n-1=\frac{N\lambda}{2L}=\frac{(46)(589\times 10^{-9})}{2(0.0500)}\approx 2.71\times 10^{-4}$, that is $n\approx 1.00027$. The accepted value for air is 1.000293: the agreement is excellent for a measurement that, once again, requires nothing more than counting fringes. No other method gives the index of a gas so easily, precisely because the difference from 1 is so tiny.

$\Delta L=(10^{-21})(4.0\times 10^{3})=4.0\times 10^{-18}$ m, about 0.4 % of the radius of a proton. A whole fringe would correspond to a few hundred nanometres: LIGO must therefore detect something of the order of $10^{-11}$ fringe. This is possible only by stacking up advantages: resonant cavities that make the beam travel hundreds of round trips and lengthen the effective arm to hundreds of kilometres, enormous laser power that reduces the statistical noise of the photons, a multistage seismic suspension, and two detectors 3000 km apart of which only coincident signals are accepted. The first detection, in 2015, was of the merger of two black holes 1.3 billion light-years away.

If light propagated in a motionless ether, the Earth, racing at 30 km/s along its orbit, would have had to pass through an ether wind, and the two perpendicular arms of the interferometer would have been affected differently: rotating the apparatus by 90° would then have shifted the pattern. The result was rigorously null, to within a twentieth of a fringe, and it has remained so in every season and at every repetition, each more precise. The conclusion is unavoidable: there is no detectable motion relative to an ether; in other words, the speed of light is the SAME in all directions and for all observers, whatever their motion. This is precisely the second postulate that Einstein raised to a principle in 1905, and from which he derived special relativity, with time dilation and length contraction. The most famous experiment in the history of physics is an experiment that found nothing.

$\frac{I_{0}}{2}$. Natural light contains all polarization directions in equal shares: averaging Malus's law over all angles, $\langle\cos^{2}\theta\rangle=\frac{1}{2}$. The orientation of the axis does not change this average: that is the nature of perfect disorder.

After the first polarizer, the light is polarized along its axis, with $\frac{I_{0}}{2}$. Malus at the second: $I=\frac{I_{0}}{2}\cos^{2}30^{\circ}=\frac{I_{0}}{2}\times\frac{3}{4}=0.375\,I_{0}$.

Without the middle polarizer: $\cos^{2}90^{\circ}=0$, complete blackness. With it: the first gives $\frac{I_{0}}{2}$ polarized at 0°; the filter at 45° transmits $\cos^{2}45^{\circ}=\frac{1}{2}$ of it, AND REPOLARIZES the light at 45°; the last one, at 45° from this new direction, again transmits $\frac{1}{2}$. Total: $\frac{I_{0}}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{I_{0}}{8}$. The paradox dissolves as soon as we accept that a polarizer does not merely filter: it PROJECTS. Polarization at 45° has a component along 90° that the initial light, polarized at 0°, did not have. Measuring, here, transforms the state: this idea will come back unchanged in quantum physics.

Light reflected by a horizontal surface (water, pavement) is partially polarized HORIZONTALLY, that is, parallel to the surface (totally, at Brewster's angle). The glasses therefore place their transmission axis VERTICAL: the horizontal glare is almost extinguished by Malus's law, while the rest of the scene, unpolarized, loses only its standard half. This is why the angler suddenly sees BELOW the surface of the water: the reflection of the sky, which hid everything, has been filtered out.

Let $\theta_{B}$ be the angle of incidence and $\theta_{r}$ the angle of refraction. The reflected ray leaves at $\theta_{B}$ on the other side of the normal; if it is perpendicular to the refracted ray, then $\theta_{B}+\theta_{r}=90°$, so $\theta_{r}=90°-\theta_{B}$. Snell's law gives $n_{1}\sin\theta_{B}=n_{2}\sin(90°-\theta_{B})=n_{2}\cos\theta_{B}$, hence $\tan\theta_{B}=\frac{n_{2}}{n_{1}}$. For air-glass: $\theta_{B}=\arctan(1.50)\approx 56.3°$. The physical reason for the total polarization is that the electrons of the surface, set oscillating along the direction of the refracted field, CANNOT radiate along their own axis of oscillation: the component that should have left in the reflected direction is exactly that one, and it is missing.

$\theta_{B}=\arctan(1.33)\approx 53.1°$. The component that survives the reflection is the one perpendicular to the plane of incidence, that is, for a HORIZONTAL surface such as a lake or a road, a HORIZONTAL oscillation of the electric field. The dazzling glare off a body of water is therefore polarized horizontally, and the more completely so the closer you look to 53° from the vertical, which corresponds to the natural glance toward a lake some tens of metres away.

The axis of the polarizer must be VERTICAL, to stop precisely the horizontal component of the glare. In the ideal case of glare totally polarized horizontally and a perfect polarizer, Malus's law gives $I=I_{0}\cos^{2}(90°)=0$: the glare disappears entirely, while the natural light of the landscape, unpolarized, keeps half its intensity. It is this contrast, and not a simple darkening, that makes the scene legible. On a VERTICAL surface such as a shop window, the plane of incidence is horizontal and the glare ends up polarized VERTICALLY, that is, along the axis of the glasses: they then let it through completely and bring no benefit. The vertical orientation of the axis is a deliberate choice, optimized for horizontal surfaces, which are the most troublesome in practice.

Sunlight sets the electrons of air molecules oscillating, and they in turn radiate: this is Rayleigh scattering, more efficient the shorter the wavelength, hence the blue of the sky. Now an electron radiates nothing along its axis of oscillation: looking at 90° from the Sun, we therefore receive only the component perpendicular to the Sun-observer direction, and the sky light there is strongly polarized, up to 70 or 80 %. Bees perceive this polarization and use it as a compass to find the direction of the Sun even through a gap in the clouds, and Viking navigators perhaps used a birefringent crystal for the same purpose. In photography, a polarizing filter turned the right way darkens the sky spectacularly, but only in the band located at 90° from the Sun: a sky photographed facing the Sun or with your back to it does not respond to the filter, which often baffles beginners.

A plastic under stress becomes BIREFRINGENT: its index differs according to the direction of oscillation of the field, the more so the greater the stress. The two components of the polarized light then travel through it at different speeds, come out out of phase, and their recombination gives a polarization rotated by an angle that depends on the wavelength. The second, crossed polarizer therefore transmits some colours and blocks others: coloured fringes appear that draw the map of the stresses. This is photoelasticity, for a long time the only way to visualize stress concentrations in a mechanical part, around a hole or a sharp corner, before numerical simulation; it is still practised on plastic models. The everyday device based on the same principle is the liquid-crystal display: each pixel is a cell placed between two crossed polarizers, and a voltage orients the molecules in it, and so the rotation they impose on the polarization. With no voltage, the light is rotated by 90° and passes; with voltage, it no longer is and the pixel turns black. The whole display rests on Malus's law.