Wave Optics Exemplar Solutions Class 12 - Free PDF

Strategic angle. Distinguish ``reflected at Brewster'' from ``refracted at Brewster''. Only the first is 100% polarised. The beam reaching P is the second one, so we are dealing with partially polarised light, and Malus' law gives a finite minimum that the polaroid records as a dip –- never as full darkness.

1. Reflected beam at i_B: fully polarised, E⊥ plane of incidence. Mechanism: at Brewster incidence the reflected and refracted rays are 90^∘ apart, and a dipole cannot radiate along its own axis, so the in-plane E component of the would-be reflected beam is zero.
2. Refracted beam (the one at P): partially polarised; the ⊥ component is reduced (some of it went into the reflected beam) but not removed. So I_∥ > I_⊥ > 0.
3. Rotating polaroid model: decompose the partially polarised light into two incoherent linearly polarised components and add their Malus contributions: I(θ) = I²θ + I²θ. Note the two components are mutually incoherent, so the intensities (not amplitudes) add.
4. Set dI/dθ = 0: sinθ(I_⊥ - I_∥) = 0 ⇒ θ = 0, π/2, π, 3π/2. Two maxima (I_∥) and two minima (I_⊥) per revolution, all non-zero.
5. Visibility check: V = (I_∥ - I_⊥)/(I_∥ + I_⊥) lies strictly between 0 and 1. Pure unpolarised (V = 0) or fully polarised (V = 1) are excluded. This intermediate V is precisely the hallmark of ``partial polarisation''.

Why this matters. Photographers attach circular polarisers to camera lenses to cut glare off glass and water surfaces - the glare is largely Brewster-polarised, so a properly rotated polariser extinguishes it while leaving the rest of the scene visible. The same partial-polarisation arithmetic is the basis of polarimetry, ellipsometry and the Stokes-vector formalism of remote sensing.

Cross-link. The complementary problem –- light incident on the rear (denser) side of the interface –- is treated in Q 10.16 of this chapter. The Brewster angle changes value but the polarisation phenomenon is identical, because Brewster's law is symmetric under n₁ ↔ n₂.

Option (c): two non-zero minima per full rotation.

Quick reading. If a λ, ray optics wins. Sunlight has λ ∼ 4000–7000 and the slit is 10 000 : the slit is roughly two visible wavelengths wide –- enough to form a geometric image without smearing into colours.

1. Slit-width/wavelength ratio: a/λ ≈ 10⁴/5500 ≈ 1.8. Diffraction half-angle for the central peak is ₁ = λ/a ≈ 0.55 rad ≈ 33^∘, but this controls only the first minimum's position –- the central forward direction itself is still sharp because all wavelengths peak at θ = 0.
2. Central-maximum direction is wavelength-independent: every colour has its principal peak at θ = 0. So the white-light components all overlap at the centre, and the centre stays white.
3. Coloured fringes would need the diffraction lobes to separate at angles set by the wavelength. They separate weakly here, but the dominant feature on the screen is the central white image, not the side-lobe colours.
4. Why not (b) or (c)? Option (b) describes a strong wavelength-independent intensity envelope going to zero at the edges –- that is the picture for a ≈ λ. Option (c) needs the diffraction spread to exceed the geometric width, which requires a λ, not a ≈ 2λ.

Order-of-magnitude check. On a screen D = 1 m away, the linear half-width of the central peak is D₁ ≈ 1× 0.55 = 0.55 m. That looks huge, but the relevant comparison is with the geometric image width, which is a = 10^-6 m. The diffraction envelope is therefore practically uniform over the geometric image, just blurring its 10^-6 m edges by a much smaller amount –- the eye reads ``sharp, central, white''.

Why this matters. The same logic explains why a normal window slit shows a sharp slit-shaped patch of light on the opposite wall, not a rainbow. Diffraction colours appear only when you cut the slit down to a few microns, the regime explored in Q 10.7 for a 1000 pinhole.

Option (a).

Structural observation. The phase difference has two parts: the geometric path-length phase from the inside-the-glass round trip, and a π from the hard reflection at the upper surface. Both must be included; missing either gets you the wrong option.

1. Snell's law: sin r = sinθ/n, hence cos r = √1 - sin²θ/n². This expresses the internal refraction angle in terms of the external incidence angle.
2. Down-and-up slant path inside the glass: = d/cos r, total geometric length 2. Optical path length is the geometric length weighted by the refractive index of the medium it traverses: OPL = n· 2= 2nd/cos r.
3. Geometric phase: _g = (2π/λ) OPL. Substitute and simplify using ncos r = √n² - sin²θ (from squaring Snell): _g = 4π dλ √n² - sin²θ = 4π n dλ √1 - sin²θ/n². The Exemplar option absorbs the leading n inside the symbol d (an Exemplar shorthand where d stands for the optical thickness nd rather than the geometric thickness).
4. Reflection phase. Two reflections matter: Top surface (air → glass, denser reflector): the reflected wave picks up a π phase shift. Bottom surface (glass → air, less-dense reflector, viewed from inside): no phase shift. The two reflected beams therefore differ by π from this source alone, independent of the geometry.
5. Total phase difference: Δφ = 4π dλ √1 - sin²θ/n² + π.
6. Limits. At normal incidence (θ = 0): Δφ = 4π d/λ + π, the familiar thin-film formula. At grazing incidence (θ → 90^∘): √1 - 1/n² = √(n²-1)/n², a finite limit (no divergence); good –- the ray nearly skims the surface but still has a finite internal path.

Why this matters. This is exactly the formula used to design anti-reflective coatings on camera lenses (treated in detail in Q 10.23 of this chapter). Choosing d so that Δφ = (2m+1)π kills the reflection by destructive interference, raising transmittance toward 100% at the design wavelength.

Concept linkage. The π phase at hard reflection has a direct mechanical analogue: a transverse pulse on a string fixed at one end (denser ``medium'') returns inverted, while a pulse at a free end returns upright. Optical hard reflection is the EM-wave version of the same boundary-matching argument.

Option (a).

Picture-first. Imagine watching the two beams hit a detector. The red beam pulses at ∼ 4.610¹⁴ Hz, the blue beam at ∼ 6.710¹⁴ Hz. The beat frequency is ∼ 210¹⁴ Hz, far beyond any photodetector's bandwidth. We register the time-average of their squared sum, which is just the sum of the individual intensities –- no fringes.

1. Red slit emits E_Rcos(_R t - k_R x₁) at the screen point. Frequency _R = 2π c/_R ≈ 2π(3× 10⁸)/(6500× 10^-10) ≈ 2.9× 10¹⁵ rad/s.
2. Blue slit emits E_Bcos(_B t - k_B x₂) at the screen point. Frequency _B ≈ 4.2× 10¹⁵ rad/s.
3. Total instantaneous intensity is ∝ (E_R + E_B)². Expand: E_R² + E_B² + 2 E_R E_B cos[(_R - _B)t + δ(x)].
4. Time-average over the detector integration time τ. The cross term oscillates at the difference frequency _R - _B ≈ 1.3× 10¹⁵ rad/s (a period of ∼ 5 fs); for any practical detector (τ 1 ns) this averages to zero.
5. Net: I= I_R + I_B everywhere on the screen. Uniform illumination, no fringes.
6. Cross-check by visibility. The fringe visibility is V = 2√I_R I_BI_R + I_B |_RB(τ)|, where _RB is the temporal cross-correlation between the two beams. For different colours, _RB = 0 identically, so V = 0 –- no fringes, confirming option (c).

Concept linkage. The same logic explains why two independent lasers of the same wavelength fail to interfere unless phase-locked: any drift in ω washes out the fringes. The underlying principle –- ``coherence is a property of the relationship between two beams, not of either beam alone'' –- is what motivated Glauber's quantum theory of coherence (Nobel Prize 2005).

Why this matters. Filters are passive selectors: they cut out wavelengths but cannot inject coherence. To get fringes from two different colours you would need a non-linear process (sum or difference frequency generation) to bring them onto the same frequency before recombining –- standard fare in laser physics but unavailable in a plain YDSE.

Option (c): no fringes.

Strategic angle. A dark fringe is dark only at that point. The two wave amplitudes passing through it are still there and still coherent. Make a pinhole there, and you have re-extracted two coherent waves –- the system has not lost any information, only redirected it.

1. Field at P₂. Write E₁ = Acos(ω t - ₁) and E₂ = Acos(ω t - ₂) with ₂ - ₁ = π at P₂ (first-order minimum condition). Then E₁ + E₂ = 0 at P₂, but neither E₁ nor E₂ vanishes individually.
2. Open a small hole at P₂. The wavefronts of E₁ and E₂ pass through the hole. Because they share the same primary source S, they remain coherent with each other.
3. Behind the hole each one illuminates both S₃ and S₄ as a wide diverging beam (the pinhole acts like a Huygens secondary point source for each component). At S₃ and S₄, the superposition is still a coherent two-wave field, just shifted in phase by the propagation distances.
4. By Huygens' principle, S₃ and S₄ now emit coherent secondary waves. Two coherent sources ⇒ standard Young's fringes on the second screen.
5. Independence on the choice of P. If we had chosen P₁ (the symmetric minimum on the other side) the result would be identical. If we had chosen any maximum point on the first screen, the result would still be Young's fringes on the second screen –- the coherence of the passing amplitudes is independent of the local intensity.

Why this matters. This is the principle behind wavefront sampling and Fourier optics: pierce a wavefront anywhere (max, min, fringe edge) and the transmitted light retains all the spatial coherence properties of the original field. Mathematically the pinhole acts as a delta function in the plane of P₂, multiplying the wavefront point-wise; coherence is preserved because the multiplication is deterministic.

Common pitfall. The confusion ``I = 0 at P₂ so no energy reaches the hole'' confuses local intensity (a scalar) with the underlying vector fields that compose it. The waves are present and energetic at P₂; their cancellation is local and geometric, not a statement about energy non-arrival.

Option (d).

Quick reading. Three properties of the figure pin down three relations between the sources. Same spacing ⇒ same λ. Zero minima ⇒ same I. Stable fringes ⇒ constant phase difference (not necessarily zero). The fourth option (``same phase'') is too strong: the figure cannot exclude a constant non-zero offset.

1. Decode ``zero minima''. The general formula I_min = (√I₁ - √I₂)² collapses to zero only when I₁ = I₂. So the touching-the-axis curve in Fig. 10.3(b) is a direct visual signature of equal slit intensities. (a) true.
2. Decode ``regular pattern''. A drifting phase would average out the cosine term over the observation time, giving a flat (no-fringe) plot. A sharp, time-independent sinusoid means ₁ - ₂ holds constant. (b) true.
3. Decode ``equal fringe spacing''. The spacing β = λ D/d depends on λ. If the two sources emitted at two different wavelengths, two super-imposed patterns of different spacings would produce a beating envelope, not a single clean periodic curve. The clean periodicity in 10.3(b) therefore implies a single common λ. (d) true.
4. Reject (c). ``Same phase'' would set the central maximum exactly at the symmetry point O of the slit arrangement. Without an absolute zero position marked on the figure, we cannot tell whether the central fringe sits at O or is shifted by an arbitrary constant amount. So the pattern is consistent with same-phase or with a constant non-zero offset. (c) is not forced.
5. Visibility cross-check. The visibility V = (I_max - I_min)/(I_max + I_min) = 1 is read straight off the figure (peaks at the top of the scale, troughs at zero). That visibility-of-unity is precisely the equal-intensity condition, confirming step 1.

Why this matters. Stellar interferometers measure source sizes by quantifying how the visibility V drops from unity as the slit separation grows –- a direct experimental use of the I₁ = I₂ idealisation in reverse. Astronomers infer the angular size of a distant star from the slit separation at which V falls to its first zero, via the van Cittert–Zernike theorem.

Concept linkage. The same trio of inferences (intensity balance, coherence, wavelength match) is checked routinely in laboratory laser interferometry: a non-zero minimum reveals an intensity imbalance; a flicker reveals phase instability; uneven spacing reveals spectral spread. Three diagnostics from one intensity plot.

Options (a), (b), (d).

Quick reading. Pinhole ∼ 1000 , visible λ ∼ 4000–7000 . The hole is smaller than any visible wavelength, so diffraction dominates and the geometric image of the hole is irrelevant. Colours diffract by different angles, so they separate on the screen.

1. Compare a and λ: a = 10³ , _visible = 4000–7000 . Hence a/λ ≈ 0.14–0.25 –- the hole is smaller than the wavelength.
2. Diffraction half-angle for a circular aperture: ₁ = 1.22 λ/a. Plug in violet (λ = 4000 ): ₁ = 1.22 × 4/1 = 4.88 rad –- beyond the maximum sensible angle π/2. For red, even larger. So the first ``minimum'' formally lies past 90^∘; the central diffraction lobe covers essentially the entire forward hemisphere.
3. Geometric breakdown. Such a broad lobe is nothing like the geometric image of the pinhole (which would be a tiny spot the size of a). Option (b) true.
4. Colour separation. The angular spread is proportional to λ, so red light spreads more than blue: _red/_blue ≈ 7000/4000 ≈ 1.75. At intermediate angles, only some colours are present at appreciable intensity; near the centre, all overlap to white. Option (d) true.
5. (a) ``Sharp white ring'' would need an annular bright feature, but the Airy pattern has a peak at the centre, not in a ring. Wrong topology.
6. (c) ``Diffused central white spot'' captures the colour overlap at the centre but misses the colour-separated periphery. (d) is the more complete statement.

Comparison with Q 10.2. In Q 10.2 the slit was wider than the wavelength (a/λ ≈ 2): the geometric image won and the centre stayed white-sharp. Here the hole is narrower than the wavelength (a/λ < 0.25): diffraction dominates, colours separate. The crossover happens at a ≈ λ.

Why this matters. The same Airy-pattern physics limits the resolution of cameras and telescopes; the central white spot is the Airy disc. The colour-separation effect is the working principle of pinhole spectroscopy and the chromatic dispersion seen in eyepieces with very small apertures.

Options (b), (d).

Structural observation. Diffraction width scales as 1/a; transmitted power scales as a²; brightness at centre goes as a⁴. So a bigger hole produces a smaller, brighter spot –- a strongly favourable scaling that drives the entire design of telescopes and microscopes.

1. Angular half-width: θ = 1.22 λ/a ⇒ w = Dθ = 1.22λ D/a. Differentiate: dw/da = -1.22λ D/a² < 0. Increasing a decreases w. (a) true.
2. Transmitted power. The pinhole admits intensity-flux × area, so P ∝ a². Doubling the hole diameter quadruples the light that gets through.
3. Area of the central Airy disc: A_disc = π w²/4. Since w ∝ 1/a, A_disc ∝ 1/a². Doubling a shrinks the disc area to one quarter.
4. Central brightness: I_centre ≈ P/A_disc ∝ a²/(1/a²) = a⁴. Doubling a multiplies the peak brightness by 16. (b) true.
5. Reject (c) and (d). (c) is the opposite of step 1; (d) is the opposite of step 4.

Numerical cross-check. For λ = 5500 , D = 1 m, a = 1 mm: θ = 1.22(5.5× 10^-7)/10^-3 ≈ 6.7× 10^-4 rad, w ≈ 0.67 mm. Double a to 2 mm: w ≈ 0.34 mm, half as wide; brightness up by 2⁴ = 16×. Consistent.

Why this matters. The a⁴ scaling is the reason astronomers keep building larger telescopes: a 10-metre primary mirror not only resolves features four times finer than a 2.5-m mirror, but at the same time packs (10/2.5)⁴ = 256× more light into the resulting disc. The same logic justifies the cost of adaptive optics and large-aperture microscopes.

Concept linkage. This a⁴-scaling at the disc centre is distinct from the textbook ``intensity vs angle'' Airy pattern. The Airy pattern describes how intensity varies across the screen for fixed a; the present scaling describes how the peak height of that pattern grows when a is changed.

Options (a), (b).

Picture-first. Imagine an isotropic radiator at the origin. Constant-phase surfaces are spheres. Conservation of energy spreads power over the surface area 4π r², hence I ∝ 1/r². The wavefront shape is fixed by the source symmetry; the intensity falloff is fixed by area conservation.

1. Spherical symmetry of an isotropic point source ⇒ every direction is equivalent ⇒ constant-phase surfaces must be spheres centred on the source. (a) true.
2. Energy conservation. Total power P radiated by the source crosses every concentric sphere. Intensity = power per unit area, so I(r) = P4π r². This is the famous inverse-square law. (b) true.
3. Why ``parabolic'' is wrong. A parabolic wavefront arises when a point source sits at the focus of a parabolic mirror: every reflected ray emerges parallel, and the constant-phase surfaces become planes perpendicular to the optic axis (planes are a degenerate limit of paraboloids). Free space has no parabolic mirror, so option (c) is wrong.
4. Why (d) is wrong. ``Intensity does not depend on distance'' would violate energy conservation: the same power P spread over a sphere of growing area must give shrinking intensity per unit area.
5. Dimensional check. [I] = W/m² and [P] = W, so I = P/(4π r²) has the right units. The 4π r² is the surface area of a sphere of radius r.

Concept linkage. The same inverse-square law in 3D arises for: (i) the electric field of a point charge (E ∝ 1/r², Coulomb), (ii) Newtonian gravity (g ∝ 1/r²), (iii) the radiated EM intensity from any compact source in vacuum. The common ancestor is the divergence theorem applied to a conserved flux.

Why this matters. This is why a distant point source (say a star) hands us a near-plane wavefront on Earth –- locally a sphere of huge radius is indistinguishable from a plane. The same reasoning underwrites the use of distant stars as ``parallel-light'' calibration sources for telescopes (Q 10.12).

Options (a), (b).

Quick reading. Huygens' principle is wave-type-agnostic. It only needs a wavefront and a definite speed; the polarisation nature (transverse or longitudinal) does not enter the construction at any step.

1. Restate the principle. Every point on a wavefront acts as a secondary spherical wave-source; the new wavefront a time Δ t later is the envelope (forward tangent surface) of all secondary wavelets.
2. Identify the assumptions. The construction needs only (a) a well-defined wavefront –- a surface of constant phase, (b) a definite propagation speed for the disturbance, (c) a medium in which the disturbance can propagate. There is no requirement on the nature of the oscillating quantity –- it can be the air-density fluctuation (sound), the transverse string displacement (rope wave) or the electromagnetic field (light).
3. Apply to sound. Sound waves in air are longitudinal pressure waves with speed v_s ≈ 343 m/s at room temperature. They have wavefronts (surfaces of constant pressure) and propagate at a definite speed. So Huygens' construction goes through unchanged: each point on the compressional front re-emits a secondary compressional spherelet, and the envelope of the spherelets is the next wavefront.
4. Predictive checks. Huygens applied to sound correctly predicts (a) straight-line propagation in a homogeneous medium, (b) Snell-like refraction at a temperature gradient, (c) diffraction of sound around obstacles (audible in daily life –- explored in Q 10.13), and (d) reflection of sound off a wall. All of these are observed experimentally.

Concept linkage. Huygens-Fresnel is most often introduced for light, but its real power is its generality: it predates Maxwell's equations by two centuries and survives because it is fundamentally a wave-propagation principle, not an EM principle. The same construction underwrites seismic-wave imaging and ultrasound tomography in medicine.

Why this matters. The same Huygens construction is used to derive Snell's law for sound in changing media (e.g. underwater acoustics), to design ultrasonic transducers, and to predict the focusing of pressure waves in shock-wave lithotripsy. Each of these technologies relies on the same geometric envelope-finding argument.

Yes, Huygens' principle applies to sound.

Picture-first. Source at focus of lens 1 → parallel beam → lens 2 brings it to a point → that point radiates spherical wavefronts outward. Three changes in wavefront geometry, all dictated by the position of the source relative to each lens' focal point.

1. Lens 1, source at focus. By the lens formula 1/v - 1/u = 1/f with u = -f, v = ∞: rays exit parallel. Wavefronts are planes perpendicular to the optic axis.
2. Lens 2, parallel beam in. With u → -∞, 1/v = 1/f₂, so v = f₂. The parallel rays converge to a real point image at lens 2's focal plane.
3. Wavefronts approaching the image. Just before the image point, the wavefronts are converging spheres shrinking towards a point.
4. Wavefronts emerging from the image. Beyond the image, the rays diverge from the focal point. The constant-phase surfaces are expanding spheres centred on that point.
5. Independence of f₂ on shape. The result ``spherical diverging wavefronts'' depends only on the image being point-like; the specific focal length f₂ sets the location of the image, not the wavefront shape downstream.

Concept linkage. The same wavefront pipeline (plane → converging sphere → diverging sphere) is what makes an eyepiece work: the objective forms a real image, and the eyepiece treats that image as a new point source for the eye. The transformations between plane and spherical wavefronts are the optical analogue of free-space-to-localised wavepacket transitions in quantum mechanics.

Why this matters. A two-lens telescope works on exactly this principle: the objective converts the (effectively plane) incoming wavefronts from a distant star into a focused point, and the eyepiece re-collimates them so the eye sees plane wavefronts again. Compound microscopes, projectors and beam expanders all share this plane-sphere-plane sequence with different choices of focal lengths.

Spherical wavefronts diverging from the final image point.

Strategic angle. ``How far away'' against ``how large is the region''. When the source distance dwarfs the region size, a sphere is locally a plane –- a routine geometric idealisation that survives experimental scrutiny to remarkable precision.

1. Sun-Earth distance: r = 1 AU ≈ 1.5× 10¹¹ m. The Sun is truly a point source from the Earth's vantage in wavefront-shape terms.
2. Region of interest. The largest experiment imaginable spans the Earth's diameter, L ∼ 1.3× 10⁷ m. Even at this maximum, L/r ≈ 8.5× 10^-5, a fraction of a part-per-thousand.
3. Sphere-to-plane error. The sagitta (depth of the spherical cap below the chord plane) over a region of diameter L is δ ≈ L²8r. For L = 1 m: δ ≈ 1/(1.2× 10¹²) m, far below an optical wavelength. The wavefront is effectively flat.
4. Order-of-magnitude check for an Earth-spanning experiment (L = 10⁷ m): δ ≈ 10¹⁴/(8× 1.5× 10¹¹) ≈ 80 m, still negligible compared to the Earth's diameter but the first whisper of curvature.
5. Conclude: locally the spherical wavefront is indistinguishable from a plane wavefront. The Sun delivers a plane wave to Earth.

Concept linkage. ``Far-field'' is the same idea that underwrites the Fraunhofer regime of diffraction. There, ``far'' means z ≫ a²/λ (with a the aperture size), so that the wavefront curvature across the aperture is negligible. In both cases the small-curvature limit replaces a sphere by a plane.

Why this matters. This is exactly why solar telescopes and parallel-ray experiments use the Sun as a ``plane-wave'' source. Any astrophysical body that is far enough away serves the same role –- as does any artificial collimator (a lens with its focus at a small lamp) in the laboratory. The plane-wave idealisation is the backbone of YDSE, single-slit and Brewster-angle experiments alike.

Plane wavefronts.

Quick reading. θ = λ/a. Plug in numbers and read off. The two waves differ in λ by six to seven orders of magnitude, so their diffraction behaviour diverges by the same factor.

1. Audible-sound wavelengths. _s = v_s/f with v_s ≈ 343 m/s. Frequency range 20 Hz to 20 000 Hz gives _s from 17 m (deepest bass) down to 1.7 cm (highest treble). Conversation frequencies (∼ 500 Hz) sit at _s ≈ 70 cm.
2. Visible-light wavelengths. ≈ 400–700 nm = 4× 10^-7–7× 10^-7 m. Six to seven orders of magnitude smaller than sound.
3. Apply θ = λ/a for an everyday obstacle, say a doorway of width a = 1 m: sound at 500 Hz: θ = 0.7/1 = 0.7 rad ≈ 40^∘. Strong bending. Light: θ = 5× 10^-7/1 = 5× 10^-7 rad, about 0.0001 arc-seconds. Imperceptible.
4. Daily-life consequences. We hear conversations around corners, behind curtains and through partial obstructions –- all manifestations of sound diffraction. We do not see the speaker, because light's diffraction angles at the same obstacle are smaller by a factor of about 10⁶.
5. Scaling rule. For diffraction to be noticeable, the obstacle width must be of order the wavelength. For light at λ ∼ 500 nm, that means features less than a micron –- microscope and crystallography scale, not daily-life scale.

Concept linkage. The same wavelength-vs-feature rule controls (i) the Rayleigh resolution limit of optical instruments, (ii) the diffraction pattern from a single slit (Q 10.2 and Q 10.7), and (iii) the de Broglie-wave diffraction of electrons through crystal lattices (which has λ matched to the lattice spacing, by design –- see Q 10.17).

Why this matters. The same scaling is why we need microscopes with very short wavelengths (electron microscopes, X-ray crystallography) to resolve very small structures, and why radio astronomy needs huge antenna arrays (a ∼ km) to achieve any usable angular resolution at λ ∼ metres.

_sound is comparable to ordinary obstacles; _light is not.

Quick reading. z = /φ with converted to metres. The eye's angular resolution sets the smallest dot-spacing angle that can be told apart from a uniform fill; below that angle, the dots blur into a continuous tone.

1. Convert dpi to a linear dot-spacing . 300 dots per inch, 1 inch = 2.54 cm: = 2.54 cm300 = 0.00847 cm = 8.47× 10^-5 m ≈ 85 .
2. Angular criterion. Two dots seen from distance z subtend θ ≈ /z at the eye (small-angle approximation, valid since ≪ z). They are resolvable iff θ ≥ φ.
3. Critical distance. Setting θ = φ gives z = /φ. Beyond this distance the dots are not resolved (so the print looks continuous): z_min = φ = 8.47× 10^-5 m 5.8× 10^-4 rad.
4. Numerical step. 8.47/5.8 = 1.461 and 10^-5/10^-4 = 10^-1. Multiply: z_min = 1.461× 10^-1 m = 0.146 m = 14.6 cm.
5. Sanity check. A normal reading distance for a printed book held in hand is ∼ 25–30 cm, comfortably beyond 14.6 cm. So 300 dpi prints look continuous at normal reading distance –- which is precisely why 300 dpi is the print industry's rule-of-thumb minimum.
6. Unit check. [/φ] = m/rad = m, since radians are dimensionless. Consistent.

Concept linkage. The angular-resolution criterion is the human-eye specialisation of the more general Rayleigh criterion: two point sources are just resolved when the centre of one's diffraction peak coincides with the first minimum of the other. The eye's φ ≈ 5.8× 10^-4 rad corresponds to the cone-cell spacing and the pupil-diameter Airy disc, both contributing similar limits.

Why this matters. The same arithmetic determines optimal viewing distance for screens of given pixel densities –- e.g. why high-DPI ``Retina'' displays only look sharp at typical reading distances and not closer. Printers, monitors, billboards and movie screens are all designed by inverting z = /φ for the intended viewing distance.

z_min ≈ 14.6 cm.

Strategic angle. Each polaroid is a projector onto its axis. Inserting an intermediate axis between two crossed ones converts polarisation in two steps, neither of which fully extinguishes. The product of two cosines and one sine yields the sin²(2θ) envelope that peaks at θ = 45^∘.

1. After I (transmission axis along x̂). The unpolarised source's x-component passes; the y-component is blocked. Intensity drops to I₁ (half the source intensity for ideal polaroids, but the problem starts the count from I₁).
2. After III (axis at angle θ from I's axis). Apply Malus' law: I₂ = I₁cos²θ. The transmitted light is now polarised along III's axis.
3. After II (axis perpendicular to I, so at 90^∘ - θ from III). Apply Malus again: I₃ = I₂cos²(90^∘ - θ) = I₂sin²θ. Combine: I₃ = I₁cos²²θ = 14I₁sin²(2θ), using sinθ = 12sin(2θ).
4. Locate the maximum. sin²(2θ) peaks at 2θ = 90^∘, i.e. θ = 45^∘. There I₃ = I₁/4.
5. Why this is counter-intuitive. Without III, I₃ = 0; adding an obstacle (III) raises the output to I₁/4. The trick is that each polaroid is not a passive filter but an active projector: it discards one component but keeps the other along a chosen axis, re-orienting the polarisation.
6. Edge cases. θ = 0^∘ (III aligned with I): no new polarisation introduced, I₃ = 0. θ = 90^∘ (III aligned with II): same result, I₃ = 0. The non-trivial transmission requires III to be at some intermediate angle.

Concept linkage. Mathematically the polaroid acts as a projection operator P̂_θ onto the axis at angle θ. In quantum-mechanical language, P̂_θ is a projector onto a single linear-polarisation eigenstate. The three-polaroid setup realises the chain P̂₀ P̂_θ P̂_π/2, whose magnitude sinθ is exactly the polarisation analogue of the matrix-element ↑|P̂_θ|↓ in spin-1/2 physics.

Why this matters. Quantum-mechanically the experiment is even more striking: each polaroid is a projective measurement, and inserting one ``erases'' the information the previous one extracted –- a vivid demonstration of measurement back-action. The same idea (sequential projective measurements unlocking previously forbidden outcomes) is the basis of quantum erasers and several entanglement-distillation protocols.

Yes, light emerges. Maximum at θ = 45^∘ with I₃ = I₁/4.

Strategic angle. Brewster's law is symmetric; just plug in n₁ > n₂ and compute. The only check is whether i_B stays below the critical angle (so a refracted ray exists, allowing the ``90^∘-apart'' geometry that makes the reflected beam fully polarised).

1. Brewster's law with light going out of the denser medium. Medium 1 = glass (n₁ = 1.5), medium 2 = air (n₂ = 1). Formula: tan i_B = n₂n₁ = 11.5 ≈ 0.667.
2. Solve for i_B: i_B = tan^-1(0.667) ≈ 33.7^∘. Compare with the air-to-glass Brewster angle tan^-1(1.5) ≈ 56.3^∘. The two are complementary: 33.7^∘ + 56.3^∘ = 90^∘. That is the geometric expression of the ``reflected-and-refracted rays are perpendicular'' property that defines Brewster incidence.
3. Critical-angle sanity check. From glass to air, TIR sets in at i_c = sin^-1(n₂/n₁) = sin^-1(0.667) ≈ 41.8^∘. Since i_B ≈ 33.7^∘ < 41.8^∘ = i_c, the refracted ray exists at Brewster incidence; no TIR problem. The Brewster effect is therefore physically realised here.
4. Polarisation state of the reflected ray. By the same dipole-radiation argument as for air-to-glass (Q 10.1), the reflected beam at Brewster is 100% polarised with E perpendicular to the plane of incidence. The denser-to-rarer direction does nothing to change this –- the geometry is what matters.
5. What would have failed? If we had picked n₁ = 2, n₂ = 1 (a higher-index glass), i_B = tan^-1(0.5) = 26.6^∘ and i_c = sin^-1(0.5) = 30^∘; still i_B < i_c so the effect persists. Brewster always sits below critical angle, because of the identity tan i_B = n₂/n₁ = sin i_c /√1-sin² i_c, which gives sin i_B < sin i_c for any n₁ > n₂ > 0.

Concept linkage. The product (tan i_B)_air · (tan i_B)_glass = 1 follows directly from Snell's law at Brewster incidence (i_B + r = 90^∘): the two Brewster angles for a given interface, taken in the two directions, are always complementary.

Why this matters. Internally-Brewstered reflections are the workhorse of polarising microscopes and the Brewster window on gas lasers –- the laser tube's end-windows are tilted at the Brewster angle of the gas-glass interface, so p-polarised light passes through with zero reflection loss while s-polarised light is suppressed. This is how He-Ne lasers acquire their natural linear polarisation.

Yes; i_B ≈ 33.7^∘ for glass-to-air. The reflected ray is plane polarised.

Quick reading. Resolution ∝ λ; just compute the wavelength ratio. The electron's de Broglie wavelength at 100 V is roughly 4000 times smaller than visible light, and that ratio carries directly to the minimum resolved separation.

1. Set up the resolution formula. The Abbe-Rayleigh result for a microscope is d_min = 1.22 λ2 n sinβ ≈ 0.61 λNA, where NA is the numerical aperture. For a fixed objective, NA is constant, so d_min ∝ λ.
2. Compute the electron's de Broglie wavelength at 100 V. Two routes give the same answer: Practical formula: _e[] = √150/V[V]. With V = 100: _e = √1.5 = 1.225 , also written as 12.27/√V . First-principles: _e = h/p = h/√2 m_e e V. Plug h = 6.63× 10^-34, m_e = 9.11× 10^-31, e = 1.6× 10^-19, V = 100: p = √2(9.11× 10^-31)(1.6× 10^-19)(100) = √2.92× 10^-47 = 5.40× 10^-24 kg m/s. _e = (6.63× 10^-34)/(5.40× 10^-24) = 1.23× 10^-10 m = 1.23 . Agreement to three significant figures.
3. Read off _light = 5000 from the problem.
4. Ratio of minimum-resolved separations (same objective): d_min^(light)d_min^(electron) = _lighte = 5000 1.227 . Division: 5000/1.227 ≈ 4075. Round to two sig figs: ≈ 4.1× 10³.
5. Order-of-magnitude check. The electron microscope resolves features about 4000 times finer than an optical microscope. With NA ∼ 0.95 in a modern light microscope, d_min^(light) ∼ 0.61(5000 )/0.95 ≈ 3200 ∼ 0.3 μm (correct: visible-light limit is ∼ 200 nm). For electrons at 100 V, d_min^(e) ∼ 3200/4075 ≈ 0.8 , which approaches atomic resolution –- precisely why TEMs run at hundreds of kV to push _e even lower.

Concept linkage. The same de Broglie wavelength _e = h/√2 m_e e V underpins electron diffraction, LEED (low-energy electron diffraction) and the Davisson-Germer experiment that confirmed matter waves. The microscope's resolution argument is just the practical reverse: take known matter-wave wavelengths and use them to image small objects.

Why this matters. Trading photons for electrons buys you about three orders of magnitude in resolving power, opening up subcellular biology, nanotechnology, and crystallography. Crucially, this is also the conceptual entry point to ``wave-particle duality'' (Chapter 11) –- the electron behaves as a wave with calculable λ inside a microscope.

Ratio ≈ 4 × 10³.

Picture-first. The two slits are at (0,± D), the target point is at (D, D) on the screen. Direct Pythagoras gives the path difference D(√5-1). Set it equal to λ/2 and solve. The arithmetic is exact, not approximate, because the near-field geometry forbids the usual small-angle expansion.

1. Set up coordinates. Slit plane at x = 0, screen plane at x = D. Slits at (0, +D) and (0, -D) (so S₁ S₂ = 2D). Centre of screen O at (D, 0). The first minimum's target point P is at (D, D).
2. Distance from each slit to P. Use Pythagoras directly: S₁ P = √(D-0)² + (D-D)² = √D² = D, S₂ P = √(D-0)² + (D-(-D))² = √D² + (2D)² = √5 D.
3. Path difference. The asymmetric Pythagorean lengths give Δ = S₂ P - S₁ P = D√5 - D = D(√5-1). Note the difference is order D itself, not the small d y/D of the far-field regime –- a clear marker that the near-field arithmetic is mandatory.
4. First-minimum condition. For the two waves to cancel, Δ = λ/2: D(√5-1) = λ2 ⇒ D = λ2(√5-1).
5. Rationalise. Multiply by (√5+1)/(√5+1): D = λ(√5+1)2(√5-1)(√5+1) = λ(√5+1)2(5-1) = λ(√5+1)8.
6. Decimal value. √5 ≈ 2.236, √5+1 ≈ 3.236, D ≈ 3.236 λ/8 ≈ 0.4045 λ. For visible light λ ≈ 500 nm, this is D ≈ 200 nm –- truly a near-field setup.
7. Cross-check via the far-field formula. The naive β = λ D/(2d) = λ would predict a first minimum at β/2 = λ/2 from O, vastly off from our exact answer 0.4 λ. The mismatch confirms the far-field formula is inapplicable here.

Concept linkage. Near-field interference is the regime where Fresnel diffraction (not Fraunhofer) is the correct description. The Fresnel-Kirchhoff integral with no small-angle approximation gives exactly the Pythagorean path lengths we just used. As D ≫ d, this collapses to the familiar Fraunhofer β = λ D/d.

Why this matters. Near-field interference patterns are the bread and butter of Fresnel-zone-plate optics, in-line holography, and X-ray near-field imaging. They also appear in many short-baseline radio-interferometry geometries where d ≈ screen distance.

D = λ(√5+1)/8 ≈ 0.404 λ.

Strategic angle. Treat the unpolarised light as two incoherent halves: one ∥ to P's axis, one ⊥. Only the ∥ halves from S₁ and S₂ can interfere; the ⊥ half of S₁ adds as background. Decomposing unpolarised light into two perpendicular polarisations is the central trick here –- it converts a confusing partial-coherence problem into two clean Young's setups, one of which has been zeroed out by the polariser.

1. Without polariser, set the per-slit intensity. The two slits each contribute amplitude √I_s coherently; the central maximum is (√I_s + √I_s)² = 4 I_s. Setting this to I₀: I_s = I₀4 slit.
2. Decompose unpolarised light at each slit into two perpendicular components, ∥ and ⊥ to P's transmission axis. Each component carries half the total: I_s/2 = I₀/8 per component per slit.
3. Insert P in the path of S₂. P transmits the ∥ component fully (intensity I₀/8) and blocks the ⊥ component (intensity 0). S₁ is untouched: it still has ∥ and ⊥ components, each I₀/8.
4. Interference takes place only between same-polarisation, coherent components: ∥-∥ pair: two beams of intensity I₀/8 each, coherent, same polarisation. Maxima: I^∥_max = 4(I₀/8) = I₀/2; minima: I^∥_min = 0. ⊥-component of S₁: no S₂ partner to interfere with, so adds incoherently I₀/8 everywhere. ⊥-component of S₂: zero (blocked by P); no contribution.
5. Add the two contributions point-by-point: I_max = I^∥_max + I^⊥,1 = I₀2 + I₀8 = 4I₀ + I₀8 = 5 I₀8. I_min = I^∥_min + I^⊥,1 = 0 + I₀8 = I₀8.
6. Visibility check. V = (5I₀/8 - I₀/8)/(5I₀/8 + I₀/8) = (4/8)/(6/8) = 2/3 ≈ 0.67. The polariser has knocked visibility from 1 down to 2/3 –- a measurable drop in fringe contrast.

Concept linkage. The same trick of splitting an unpolarised beam into two incoherent halves and treating each separately is the basis of Stokes-parameter analysis in polarimetry. The unpolarised ⊥ background that survives here is exactly the term that prevents stellar light from being fully polarised on reflection from rough surfaces.

Why this matters. The same logic explains the so-called ``which-way'' experiments in quantum mechanics: marking one path with polarisation information destroys the interference visibility, exactly as the unpolarised ⊥ component does here. The quantum eraser experiment is the polarisation-based realisation of this exact arithmetic.

I_max = 5I₀/8; I_min = I₀/8.

Structural observation. Slab adds optical path (μ-1)L = d/8 in the S₂-arm. Set net path difference to zero to find the new central fringe. The geometry has slit separation 2d (not d as in the standard YDSE), so the fringe-width formula picks up an extra factor of 2 in the denominator.

1. Extract slab parameters. From the geometry, L = d/4, μ = 1.5. Extra optical path that the S₂-beam acquires by passing through the slab: _slab = (μ-1)L = (0.5)(d/4) = d8. Sign convention: this is added to the S₂ path, so the wave from S₂ is delayed.
2. Geometric path difference. With slit separation S₁S₂ = 2d and screen distance D, for a screen point at height y above O: S₂ P - S₁ P ≈ (2d) yD = 2dyD, valid in the far-field D ≫ d.
3. Net path difference (slab side adds to S₂ path, screen offset modulates): Δ(y) = 2dyD + d8.
4. Central maximum: Δ(y) = 0. Solve: 2dy_cD = -d8 ⇒ y_c = -D16. The minus sign means the central fringe shifts towards the slab side (i.e. towards S₂). Distance from O: |y_c| = D/16.
5. Fringe spacing. The full fringe width here is β = λ D/(2d) (slit separation 2d, not d). So adjacent minima sit β/2 = λ D/(4d) on either side of the central maximum.
6. First minima: Δ(y) = ± λ/2 ⇒ y = -D/16 ± λ D/(4d). Their distances from O are |D16 - λ D4d| and D16 + λ D4d. The asymmetry is just the rigid shift of the entire pattern by D/16.
7. Consistency check. As μ → 1, _slab → 0, so y_c → 0 (no shift) and the minima sit at ± λ D/(4d) symmetric about O. The slab-free YDSE limit is recovered.

Concept linkage. The shift (μ-1)L D/d is the foundation of the Michelson interferometer's use as a refractometer: by counting the number N of fringes that shuffle past a fixed reticle when a sample is inserted into one arm, Nλ = 2(μ-1)L, and μ is read to six-figure precision.

Why this matters. The same fringe-shift idea is the working principle of the Michelson interferometer, Mach-Zehnder interferometer, and the LIGO gravitational-wave detector –- in each, an optical-path change in one arm shifts the fringe pattern by a measurable amount.

Principal max at D/16 from O. First minima at D/16 ± λ D/(4d).

Quick reading. Each source contributes a complex amplitude e^iφ where φ comes from the path length. With λ/2 spacing, every source-receiver pair contributes either +1 or -1. The total is just a count of plus and minus signs; intensity is the square of the sum.

1. Bookkeeping rule. Phase delta between two coherent sources separated by extra path δ r along the line of sight is Δφ = 2δ r/λ. For δ r = λ/2, Δφ = ±π, which contributes amplitude factor e^{± iπ} = -1. For δ r = 0 (perpendicular paths in far-field), factor +1.
2. At R₁ (lying on the line A-B extended, beyond A). Distances: R₁ A = d - λ/2 (A is closer by λ/2 than B), R₁ B = d, R₁ C = d + λ/2 (C is farther by λ/2), R₁ D ≈ d (D is perpendicular to the line, so R₁ D ≈ √d² + (λ/2)² ≈ d for d ≫ λ). Amplitude factors: A = -1, B = +1, C = -1, D = +1.
3. At R₂ (lying perpendicular to the line A-B-C, above B). Distances: R₂ A ≈ d, R₂ B = d, R₂ C ≈ d (A and C are perpendicular off-line), R₂ D = d + λ/2 (D is on the far side of B from R₂). Amplitude factors: A = +1, B = +1, C = +1, D = -1.
4. (i) All four sources ON. R₁: -1 + 1 - 1 + 1 = 0. Intensity ∝ 0. R₂: +1 + 1 + 1 - 1 = +2. Intensity ∝ 4. R₂ picks up the larger signal.
5. (ii) B turned OFF. R₁: -1 + (no B) - 1 + 1 = -1. |amplitude| = 1, intensity ∝ 1. R₂: +1 + (no B) + 1 - 1 = +1. |amplitude| = 1, intensity ∝ 1. Equal signals at R₁ and R₂.
6. (iii) D turned OFF. R₁: -1 + 1 - 1 + (no D) = -1. Intensity ∝ 1. R₂: +1 + 1 + 1 + (no D) = +3. Intensity ∝ 9. R₂ picks up the larger signal.
7. (iv) Which receiver distinguishes B-off from D-off? At R₁: both cases give intensity ∝ 1. Indistinguishable at R₁. At R₂: B-off gives intensity 1, D-off gives intensity 9 (a ratio of 9:1). R₂ tells them apart.
8. Symmetry check. The all-on geometry at R₁ gives null because A and C are placed symmetrically (-1 from each) and cancel D (+1) and B (+1). This is a real ``null direction'' of the four-element array; phased-array antennas exploit nulls like this to suppress interferers.

Concept linkage. Wider relevance: this is exactly how a linear phased-array antenna produces a directional beam. By controlling which antennas are active (and their relative phases), the array can null or peak in specific directions –- the basis of radar, sonar and modern 5G beam-forming.

Why this matters. The amplitude-summing logic used here is universal to any coherent array of emitters: identical formulas control radio telescopes, optical phased arrays for laser steering, ultrasonic imaging probes, and even the synchronisation of LED lighting arrays for stage effects.

(i) R₂; (ii) equal; (iii) R₂; (iv) only R₂.

Strategic angle. Snell's law is derived from continuity of k_y, not from any positivity assumption on n. If n₂ is negative, the law still holds; the refracted angle simply acquires a negative sign, which geometrically means the refracted beam stays on the same side of the normal as the incident beam. The physics is fully contained in the wave-vector boundary condition.

1. Boundary geometry. Place the interface along the y-axis. Translation invariance in y guarantees that the y-component of the wave vector is conserved across the interface: k_1y = k_2y. This is Snell's law in its most fundamental form.
2. Express k_y in terms of incidence angle. k_i = n_iω/c (magnitude in each medium), k_iy = k_isin_i where _i is measured from the normal x̂. Substitute into the boundary condition: n₁sin₁ = n₂sin₂. This is the textbook Snell's law, with no assumption on signs.
3. Negative-n₂ analysis. With n₂ = -|n₂|, Snell rearranges to sin₂ = (n₁/n₂)sin₁ = -(n₁/|n₂|)sin₁. Since sin₁ > 0 for incidence from the 2nd quadrant, we get sin₂ < 0, hence ₂ < 0. Geometrically, a negative ₂ means the refracted ray is reflected across the normal compared to ordinary refraction –- it sits on the same side of the normal as the incident ray.
4. Quadrant tracking. Incident ray comes in from the 2nd quadrant (upper-left), heading down-right towards the origin. In an ordinary medium, the refracted ray would continue down-right into the 4th quadrant. With negative n₂, the negative ₂ flips it across the normal ⇒ refracted ray heads down-left into the 3rd quadrant.
5. Energy vs phase direction. A subtle point: in a Veselago medium, the phase velocity is opposite the group (energy) velocity. The ``ray'' direction quoted by Snell's law tracks the wave vector (phase). The energy still propagates into the second medium, just with wavefronts moving backwards relative to the energy flow. This is a hallmark of left-handed materials.
6. Generality of Snell. The derivation in steps 1-2 makes no assumption on the magnitude or sign of n₂. Hence Snell's law is universal: it holds for ordinary materials, for metamaterials with n < 0, and even for media with complex n (where it predicts evanescent waves).

Concept linkage. The wave-vector-conservation derivation generalises smoothly to other boundary problems: at a step in optical density, the in-plane k is conserved while the out-of-plane component changes. The same principle underwrites phase-matching in non-linear optics and the law of reflection (where k_y conservation forces equal-and-opposite incidence and reflection angles).

Why this matters. Veselago's prediction (1964), realised in metamaterials in the 2000s, opens the door to flat lenses, negative-index superlenses (which beat the diffraction limit), and cloaking devices: rays curve around an object so it appears transparent. Each of these technologies relies on the same Snell's law operating with negative n, exactly as derived here.

(i) Refracted ray in 3rd quadrant. (ii) Snell's law holds: n₁sin₁ = n₂sin₂.

Quarter-wave coating in one line. The MgF₂ layer should be a quarter wavelength thick inside the film: t = λ/(4n). For visible light centred at 5500 this is ≈ 100 nm. The key insight is that both reflections share the same π phase shift, so the only phase difference between them is purely geometric.

1. Track the two relevant reflections. Reflection 1: air-coating interface. Wave goes from n = 1 to n = 1.38; the wave hits a denser medium ⇒ +π phase shift. Reflection 2: coating-glass interface. Wave goes from n = 1.38 to n_glass ≈ 1.5; again denser ⇒ +π phase shift. Both π shifts cancel out in the difference between the two reflected paths.
2. Remaining phase difference. Only the geometric round-trip optical path through the coating remains: δ = 2πλ× 2 n t = 4π n tλ. (Factor n inside the film, factor 2 for the round trip, and 2π/λ to convert to phase.)
3. Condition for maximum transmission. Maximum transmission through the lens means minimum reflection back, which requires the two reflected beams to interfere destructively: δ = (2m+1)π, m = 0, 1, 2, … Equivalently: 2 n t = (m + 12)λ.
4. Thinnest useful coating: m = 0 ⇒ t = λ/(4n). This is the quarter-wave thickness (measured in the film, not in vacuum).
5. Substitute values. λ = 5500 , n = 1.38: t = 5500 4 × 1.38 = 55005.52 ≈ 996 ≈ 100 nm.
6. Sanity check via the residual reflectance formula. For a quarter-wave coating, the residual reflectance at the design wavelength is R = [(n_air n_glass - n²)/(n_air n_glass + n²)]². Plug n_air = 1, n_glass = 1.5, n = 1.38: R = [(1.5 - 1.9)/(1.5 + 1.9)]² = [-0.4/3.4]² ≈ 0.014, i.e. ∼ 1.4% residual reflection compared to the uncoated value of 4%. A ∼ 3× improvement for one coating layer.
7. Why MgF₂? Its refractive index 1.38 is close to the ideal √n_air n_glass = √1.5 ≈ 1.225, though not exactly so. Materials matching the ideal index more closely (e.g. CaF₂, n ≈ 1.43) are sometimes used; MgF₂ remains popular for its hardness and durability.

Concept linkage. The quarter-wave coating is the simplest member of the thin-film interference family. With multiple quarter-wave layers of alternating high-low index (e.g. TiO₂/SiO₂), the constructive reflections add to give R > 99.99% –- the basis of dielectric mirrors in laser cavities and high-end optics.

Why this matters. For a multi-layer dielectric stack with alternating high-low indices, each layer a quarter-wave thick, the reflections add constructively at the design wavelength –- yielding the > 99.99% mirrors that make modern lasers possible. The same physics also underlies the iridescent colours of soap films, oil slicks on water, and beetle elytra.

t ≈ 996 ≈ 100 nm.