Ray Optics NCERT Exemplar Solutions - Class 12 Physics Free PDF

Quick reading. The ray leaves the second face normally, so inside the prism it travels parallel to the normal of that face. The geometry of the prism then forces the inside ray to make exactly the angle A with the normal of the first face.

1. Read off r₁ = A = 5^∘ directly from the geometry (the inside ray is perpendicular to face 2, hence makes angle A with the normal to face 1).
2. Apply Snell at face 1 in the small-angle regime (sin x ≈ x in radians): θ = μ r₁ = 1.5 × 5^∘ = 7.5^∘.
3. Cross-check via deviation. The total deviation is D = (i₁ - r₁) + (i₂ - r₂) = (7.5 - 5) + (0 - 0) = 2.5^∘, which agrees with the small-angle prism formula D = (μ - 1)A = 0.5 × 5 = 2.5^∘. Two independent routes give the same answer: the result is robust.
4. Alternative geometric method. Drop a perpendicular from the exit point to face 1: the inside ray, normal to face 2, and normal to face 1 form a right triangle with the prism's apex angle A as the relevant angle, so the inside ray sees face 1's normal at exactly A = 5^∘. No algebra needed.

Why this matters. Recognising that one face is normal-exit collapses the two-face problem to a single Snell's-law step. This trick reappears in minimum-deviation prism problems and in the design of spectrometer Littrow prisms.

Unit check. μ is dimensionless, r₁ is in degrees, θ comes out in degrees. The small-angle linearisation sin x ≈ x requires x in radians, but the equation θ ≈ μ r₁ remains dimensionally consistent because both sides carry the same angular unit.

θ = 7.5^∘ (a).

Structural observation. The slab acts as a group-delay medium for the pulse. The slower a colour, the longer its delay; the faster, the earlier it pops out the other side.

1. Rank refractive indices by wavelength (Cauchy): _red < _yellow < _green < _blue < _violet.
2. The transit time scales like μ; red has the smallest μ and so the smallest transit time.
3. Numerical feel. For crown glass at visible wavelengths, _red ≈ 1.513 and _violet ≈ 1.532. The transit-time difference over a slab of thickness t = 1 cm is Δτ = (_v - _r) t / c ≈ 0.019 × 10^-2/(3× 10⁸) ≈ 6.3 × 10^-13 s = 0.63 ps. Tiny but real, and now routinely measured with ultrafast lasers.
4. No-bending caveat. At normal incidence, none of the colours separates laterally. Dispersion happens in time, not space. Lateral separation requires oblique incidence (or a prism geometry with non-parallel faces).

Why this matters. The same principle explains why prisms spread white light into a spectrum with red bent the least and violet the most. It also lies behind chromatic dispersion in optical fibres, the dominant pulse-broadening mechanism that limits long-haul data rates.

Red (d).

Picture-first. As the object slides from far away towards the focal plane, the image position swings from near F' on the right side out to infinity. So v goes from ∼ f to ∞ while u goes from large to f. That alone tells us motion is not uniform.

1. Map limits: u → ∞ ⇒ v → f; u → f⁺ ⇒ v → ∞.
2. The chain rule on v(u(t)) gives dvdt = f²(u+f)² dudt. Even if du/dt is constant, dv/dt grows without bound as u → f.
3. Magnification gives the same verdict. Lateral magnification m = v/u also diverges as u → f: at u = 2f, |m| = 1; at u = 1.1 f, |m| = 11; at u = 1.01 f, |m| = 101. The non-linear blow-up of m is just the integrated form of the non-linear blow-up of dv/du.
4. Numerical sanity. Take f = 10 cm and v_O = 5 m/s. At u = 20 cm, |dv/du| = (10/30)² = 1/9, so v_I = 5/9 ≈ 0.56 m/s. At u = 11 cm, |dv/du| = (10/21)² ≈ 0.227, so v_I ≈ 1.13 m/s – the image has more than doubled its speed even though the object speed is fixed. Confirms non-uniform acceleration.

Why this matters. Non-linear maps turn uniform inputs into accelerated outputs. The same trick underlies the magnification blowing up at F, autofocus tracking in cameras at close range, and the difficulty of imaging at a microscope's working distance.

(c) non-uniform acceleration, image moves away.

Strategic angle. The geometry of a rainbow is a cone with half-angle 42^∘ (primary) about the anti-solar line. Whether you see a full circle or just an arc depends only on how much of that cone the ground cuts off.

1. Ground observer: half the cone is below the horizon, so an arc results.
2. Aerial observer: rain drops both above and below the observer's altitude exist, so the full cone is visible and a complete circle is seen.
3. Why the two cones share an axis. The anti-solar direction is defined by the Sun (a single direction), so every rainbow seen by a given observer is centred on that same line. The primary forms at 42^∘ from this line (one internal reflection inside each drop) and the secondary at 51^∘ (two internal reflections); both are coaxial cones, hence concentric circles when seen in full.
4. Colour order check. In the primary the inner edge is violet and the outer edge red. In the secondary the ordering reverses (inner red, outer violet) because the extra internal reflection flips the colour-versus-deviation ranking. From a plane both rings should be clearly visible with their distinctive colour orderings.

Why this matters. Pilots and astronauts routinely photograph full-circle rainbows; this is the only natural setting where you can. The same circle-vs-arc reasoning explains the Brocken spectre and the glory seen around an aircraft's shadow on cloud tops.

(b) concentric circles.

Quick reading. The condition ``refraction at 90^∘'' means the incidence angle equals the critical angle for that colour. Whether other colours still refract depends on whether their own critical angles are larger or smaller.

1. Larger μ ⇒ smaller i_c (since sin i_c = 1/μ).
2. Blue has the largest μ among red/yellow/green/blue, hence the smallest i_c. The given i already equals i_c(yellow) > i_c(blue), so blue must TIR.
3. Inequality table. Putting the colours in order: arrayc|cccc colour & red & yellow & green & blue
   μ & lowest & low & high & highest
   i_c & largest & i_c = i & smaller & smallest
   at this i & refracts & grazing & TIR & TIR array So both blue and green TIR; option (c) names the dominant TIR-er.
4. Concept linkage. This is the same physics that keeps short-wavelength light better confined in step-index optical fibres: the numerical aperture NA = √_core² - _clad² is slightly larger for blue than for red, so the acceptance cone narrows but TIR is more robust.

Why this matters. Optical fibres exploit exactly this: shorter-wavelength light has a lower critical angle, easier to confine. The reverse logic also explains why red light leaks out of total-reflection prisms first when they are slightly damaged.

(c) blue undergoes TIR.

Quick reading. A thin lens is reversible: reflect it about its principal plane and the focal length stays the same. So if the lens is converging when light enters one side, it stays converging when entering the other.

1. Apply lens-maker's formula with curved side facing object: 1/f = (1.5-1)(1/20) = 1/40 ⇒ f = 40 cm.
2. By the reversibility theorem the result is unchanged when the lens is flipped.
3. Algebraic verification. Flipping interchanges the roles of the two surfaces and reverses the sign convention for each radius. Define new radii R₁' = -R₂ and R₂' = -R₁. Then 1f' = (μ - 1)(1R₁' - 1R₂') = (μ - 1)(1-R₂ - 1-R₁) = (μ - 1)(1R₁ - 1R₂) = 1f. Two minus signs cancel; f' = f.
4. Power in dioptres. P = 1/f = 1/(0.40 m) = 2.5 D, which is a typical reading-glass strength.

Why this matters. The reversibility principle is the geometric-optics ancestor of Helmholtz reciprocity: source and detector can be swapped without changing the path of light. This symmetry is the basis of antenna reciprocity in radio physics and the symmetry of Green's functions in wave equations.

(c) converging both ways, f = +40 cm.

Strategic angle. ``Ionospheric reflection'' is a misnomer: nothing reflects sharply. The wave gradually turns through a graded index and is sent back when the local critical angle is exceeded. That gradual-graded-TIR picture is exactly the mirage.

1. Identify graded index μ(z) in both cases.
2. Recognise the wave bending until total internal reflection occurs in the rarer layer above.
3. Compare and contrast the four options.
   • Plane-mirror reflection (a): sharp boundary, single interface, _i = _r at one point. No gradient. Different mechanism.
   • Mirage TIR (b): continuous μ(z) gradient, wave turns through many layers until TIR. Same mechanism as ionosphere.
   • Rainbow dispersion (c): wavelength-dependent refraction inside water drops. Different; radiowaves are essentially monochromatic at any given station.
   • Atmospheric scattering (d): Rayleigh scattering off molecules, randomises direction. Different; ionospheric reflection is directional.
4. Plasma frequency check. For a frequency ω below _p the index μ(ω) = √1 - _p²/ω² becomes imaginary – the wave cannot propagate and is evanescently reflected. This is why the ionosphere reflects AM (kHz, low ω) but is transparent to FM and TV (MHz, high ω): the latter punch through to space.

Why this matters. The skywave mode of HF radio communication depends on this; it is how AM stations reach intercontinental distances at night, and why FM/TV is line-of-sight only. Earth's ionosphere is essentially a giant plasma mirror for the right frequency band.

(b) mirage-type TIR.

Picture-first. Use the two standard rules: (i) ray through F ⇒ reflects parallel to axis; (ii) ray parallel to axis ⇒ reflects through F. Here the incoming ray passes through F, so rule (i) applies.

1. Confirm PQ passes through F in the figure.
2. Therefore the reflected ray is horizontal: ray 2.
3. Why each wrong option is wrong.
   • Ray 1 – corresponds to a ray reflecting through C (centre of curvature), which is only correct for a ray heading along the normal radius. The incident ray here passes through F, not C.
   • Ray 3 – corresponds to a ray retracing its incoming path, possible only for a ray that hits the mirror normally, i.e. along the radius through C.
   • Ray 4 – bisects the angle between the normal and the incoming ray, but reflection law says equal angles on opposite sides of the normal, not the bisector itself.
4. Reciprocity cross-check. If a ray comes in parallel to the axis, it reflects through F (the dual rule). Reversing this gives our case: a ray through F reflects parallel to the axis. The two rules are time-reverses of each other – a useful consistency check.

Why this matters. These two rules suffice for almost every ray-diagram problem on concave/convex mirrors and lenses. They form the basis of how telescopes, headlight reflectors, and satellite dishes are designed: any source at F produces a parallel (collimated) beam after reflection.

(b).

Quick reading. Two interfaces, two bends. At interface 1 (air→turpentine) bend towards normal; at interface 2 (turpentine→water) bend away from normal.

1. Scan rays for the correct sign of bend at each interface.
2. Only ray 2 satisfies both: towards-then-away.
3. Numerical illustration. Take _air = 1, _turp = 1.47, _water = 1.33. For an incidence of 30^∘ in air, sin r₁ = sin 30^∘ / 1.47 = 0.340, so r₁ ≈ 19.9^∘ (bent towards normal as expected). At the lower interface, the angle of incidence is r₁ = 19.9^∘, so sin r₂ = 1.47 × sin 19.9^∘ / 1.33 = 0.376, giving r₂ ≈ 22.1^∘ – bent away from the normal. Two distinct bends, both in agreement with ray 2's depiction.
4. Why option (d) is special. Straight-through is only correct at normal incidence (i = 0) because Snell's law is then automatically satisfied at any μ. At any nonzero i, refraction must occur whenever μ changes.

Why this matters. Optical density (refractive index) controls ray bending, not mass density. Mercury has very high mass density but μ ≈ 1.0 for visible light; olive oil is less dense than water but has μ ≈ 1.47. The two notions of ``denser'' are independent.

(b).

Structural observation. Two mirror types ⇒ two behaviours. Plane mirror is linear; convex mirror is non-linear.

1. Plane (rear): v = u, v = u. Image speed is constant in time at the object's true speed.
2. Convex (side): v = uf/(u-f). The derivative |dv/du| = f²/(u-f)² blows up as u → f⁺, so the rear car appears to surge ahead in the side mirror as it closes in. Statement (d) is correct.
3. Numerical sanity. Take the side mirror's |f| = 1 m (a reasonable value for a car side mirror). At u = 100 m, |dv/du| = 1²/(101)² ≈ 10^-4 – the image barely moves. At u = 10 m, |dv/du| = 1/121 ≈ 0.008 – still tiny. At u = 2 m, |dv/du| = 1/9 ≈ 0.11 – already significant. As u drops further the apparent speed climbs steeply.
4. Statement (a) revisited. The rear car's actual ground speed is 60 + 5 = 65 km h^-1. But the question asks about appearance in mirrors. The cleaner reading: (a) talks about a ground-frame number and is a distractor; (d) is the optics statement. NCERT's intended choice is (d).

Why this matters. ``Objects in mirror are closer than they appear'' on every car side mirror: the convex mirror diminishes size but accelerates apparent motion at close range. The same non-linear |dv/du| scaling explains the dizzying acceleration of the parallax of a foreground tree when viewed from a car window.

(d).

Quick reading. Negative refractive index flips the sign of the refraction angle. Geometrically: the refracted ray bends to the same side of the normal as the incident ray.

1. Recall Veselago's 1968 prediction: μ < 0 keeps Snell's law in form but reverses the direction of the refracted ray's transverse component.
2. Pick the panel where the refracted ray is on the same side of the normal as the incident ray: panel (a).
3. Sign-of-Snell argument. Write n₁ sin₁ = n₂ sin₂ with n₁ = +1, n₂ < 0. If ₁ > 0 (incident ray on the right of normal), then sin₂ < 0, so ₂ < 0: the refracted ray is on the left of the normal, i.e. on the same side as the reflection of the incident ray about the normal. That is exactly the picture of ``same side, opposite transverse direction.''
4. Energy still flows forward. The negative index flips the phase velocity direction, not the energy (Poynting) flow. Energy still travels from medium 1 into medium 2 normally; only the wave fronts run backwards. This is why μ < 0 materials don't violate energy conservation.

Why this matters. Negative-index materials (metamaterials), demonstrated experimentally around 2000 by Pendry, Smith and others, enable ``perfect lenses'' that beat the classical diffraction limit and form the basis of invisibility-cloak research. They are made of sub-wavelength resonant structures (split-ring resonators, fish-scale arrays), not natural materials.

(a).

Strategic angle. Each option is a distinct optical effect. Score each separately.

1. (a) Apparent depth varies with the line-of-sight angle, so edge-distance variations distort the shape: tick.
2. (b) Image is closer; for a fixed eye-to-object base the angular size shrinks: tick.
3. (c) Critical-angle cutoff hides distant submerged points: tick.
4. (d) Flat surfaces don't magnify; no lens action: cross.
5. Critical-angle number for water. sin i_c = 1/_water = 1/1.33 = 0.752, so i_c ≈ 48.8^∘. Any submerged point whose line-of-sight to the eye exceeds 48.8^∘ from the normal at the surface stays invisible: this defines Snell's window.
6. Angle of view shrinks: a quick derivation. An object at depth h subtends an angle 2arctan(R/h_air) in air, but its image is at apparent depth h/μ = 0.75 h, subtending 2arctan(R/0.75 h) – larger angle? Actually no, because the geometry near the edge also contracts the lateral extent. The clean statement is that for fixed lateral size, the image is closer, and for fixed eye position it subtends a smaller angle. (b) holds.

Why this matters. These three effects combine to produce the familiar ``broken stick in water'' illusion. The same physics governs why a fish sees a circular ``Snell's window'' of the world above water, surrounded by a TIR-mirrored view of the lake bottom.

(a), (b), (c).

Picture-first. Imagine standing on the left of the block looking through face AD. You see the pin only through a cone of rays whose angle inside the glass is less than i_c. Rays bound for the lower part of AD already exceed i_c before reaching it and are TIR-trapped.

1. Compute i_c = sin^-1(1/1.6) ≈ 38.7^∘.
2. Find the cone of escaping rays: the image of the pin sits near the top of AD (near A); the rest is dark (not seen at all).
3. Cone half-angle inside the glass. From the pin at the midpoint of AB, a ray heading towards face AD makes an angle θ with the horizontal. For the ray to emerge into air the angle with the AD-normal must be less than i_c = 38.7^∘. The geometry maps this to a maximum vertical position y_max = (L/2) tan i_c ≈ 0.4 L below A.
4. Coexistence of (a) and (d). Statements (a) and (d) describe complementary outcomes for different viewing angles: a viewer in the right place sees the pin near A; a viewer at an angle outside the escape cone sees nothing. Both are valid claims about some observer position.
5. Why not (b) or (c). For (b) ``near D'' the image would have to emerge near the far corner of AD – but rays reaching that part of the face from the midpoint of AB already make angles exceeding i<sub>c</sub> and TIR. For (c) ``centre of AD'' the same TIR argument rules out the lower half.

Why this matters. Same principle as why a swimming pool looks shorter near you: TIR cuts off the far-end view. The ``shadow region'' below the escape cone is invisible to an outside observer, which is the basis of optical-fibre cladding design.

(a) and (d).

Quick reading. The rainbow caustics are at 42^∘ and 51^∘. Outside these caustics light dims; between them is the dark band.

1. Map angular ranges: <42^∘ has primary's interior contribution; >51^∘ has secondary's exterior; in between, neither.
2. Hence between 42^∘ and 51^∘ no rainbow scattering light reaches the eye: dark.
3. Why ``destructive interference'' is wrong. The primary and secondary rainbows are caustics in intensity, not phase-coherent patterns. Adjacent drops scatter independently and incoherently; there is no well-defined relative phase to destroy. Option (a) is a wave-optics red herring inserted to test conceptual understanding.
4. Why ``absorption'' is wrong. Water absorbs almost equally across the visible spectrum (its main absorption is in the IR). Selective absorption between 42^∘ and 51^∘ would require some angular filter inside the drop, which doesn't exist.
5. Brightness contrast. The interior of the primary is brighter than the dark band; the dark band is darker than the exterior of the secondary. So the dark band is defined by relative darkness compared to its neighbours, not absolute black.

Why this matters. Alexander of Aphrodisias noticed this band in 200 AD, long before geometric-optics models. It's one of the oldest reliably named optical effects, and Descartes's 1637 ray-tracing analysis of rainbows was an early triumph of geometric optics.

(b) and (d).

Strategic angle. A magnifying glass works by enlarging the angular size of the object, not the field of view.

1. Compute angular magnification: M = D/f (image at ∞) or 1 + D/f (image at near point). Both finite.
2. Confirm image character: virtual, erect, magnified for u < f.
3. Field of view vs. angular magnification. The two are inversely related: a stronger magnifier (smaller f) gives larger M but a narrower angular field of view. That's why a 20× jeweller's loupe shows only a tiny patch of a coin even though the patch is highly enlarged. Option (c) confuses these two distinct quantities.
4. Why M_D is finite at the near point. The identity M_D = 1 + D/f shows M_D = D/f + 1, which is bounded above by a number set by the smallest practical f. For a loupe with f = 2.5 cm: M_D = 1 + 25/2.5 = 11. Never infinite. Option (d) confuses ``image at near point'' with ``object at focus'' (the latter gives image at infinity, but observing at ∞ is the relaxed-eye case with M_∞ = D/f, still finite).

Why this matters. The same logic governs the eyepiece in microscopes and telescopes: enlarge angular size, accept reduced field of view. The trade-off is fundamental to optical instrument design.

(a) and (b).

Picture-first. Two lenses, one long focal length, one short. Light from infinity converges at f_o inside the tube; eyepiece picks up that image and re-images it to infinity.

1. Length: just f_o + f_e = 20.02 m. (a) tick.
2. Mag: 20/0.02 = 1000. (b) tick.
3. Two real inversions through the optics; final image inverted from sky. (c) tick.
4. Aperture controls brightness, not chromatic aberration. (d) wrong.
5. Brightness scales with D². If the aperture diameter is D, the light-gathering area is π D²/4. Doubling D quadruples the brightness. This is why professional telescopes have enormous primary mirrors – 8 m class instruments collect 10⁶ times more light than the unaided eye.
6. Why chromatic aberration is aperture-independent. Chromatic aberration arises from df/dλ, which is set by the material's dispersion (Abbe number), not the physical size of the lens. To reduce chromatic aberration you must combine glasses of different dispersions (an achromat), not enlarge the aperture.
7. Caveat on (d). Larger aperture actually worsens two aberrations: spherical aberration (∝ D⁴) and coma. Achromats fix chromatic aberration; asphere or aplanat fixes spherical/coma.

Why this matters. The Yerkes refractor (f_o = 19.4 m, f_e ≈ 1 cm) is the world's largest refracting telescope; modern professional instruments (e.g. Keck, VLT, JWST) use mirrors not lenses to avoid chromatic aberration entirely. Big refractors are physically impossible above ∼ 1 m aperture because their lenses sag under gravity.

(a), (b), (c).

Quick reading. Red bends less than blue in a prism; the same lens has weaker focusing power for red, hence longer focal length.

1. _red < _blue ⇒ f_red > f_blue.
2. Numerical illustration. For crown glass: take _red = 1.513, _blue = 1.528. A double-convex lens with R₁ = -R₂ = 20 cm has f_red = 1(0.513)(2/20) = 10.0513 ≈ 19.49 cm, f_blue = 1(0.528)(2/20) = 10.0528 ≈ 18.94 cm. Difference ≈ 0.55 cm, the longitudinal chromatic aberration. Small but enough to blur a sharp image.
3. Direct consequence: chromatic aberration. A white-light point source forms two slightly displaced images along the axis – red farther from the lens, blue closer. The visible result is coloured fringes (blue fringes inside, red fringes outside) on edges of bright objects.

Why this matters. This is the cause of chromatic aberration: red and blue images focus at slightly different positions, blurring colours at the edge of the image. Camera lens designers spend significant effort building achromatic doublets (crown + flint) to cancel this effect, and apochromats to push it to higher orders.

f_red > f_blue.

Quick reading. For M_D = 10 with D = 0.25 m, solve 1 + D/f = 10.

1. f = D/9 = 0.25/9 ≈ 0.0278 m.
2. P = 1/f = 36 D.
3. Unit consistency check. D is in metres (0.25 m), so f comes out in metres and P = 1/f is in dioptres (m^-1). If you instead use D = 25 cm you'd get f = 25/9 cm = 2.78 cm and P = 1/0.0278 = 36 D, same answer. The trick is to convert to SI before computing P.
4. Sanity check using the M_∞ formula. If we had used M_∞ = D/f = 10 (image at infinity, relaxed eye), we'd get f = D/10 = 2.5 cm and P = 40 D. Slightly different. The problem specifies ``angular magnification of 10'' and the standard convention for ``simple microscope'' is near-point image, so we use M_D = 1 + D/f.

Why this matters. A magnifying glass of 36 D is a strong hand lens (focal length under 3 cm), the kind used by jewellers (``loupe''). Typical eyeglasses are ± 2 to ± 4 D in comparison; reading glasses are about +2.5 D.

P = 36 D.

Structural observation. Thin-lens focal length is invariant under reversal because lens-maker's formula is symmetric.

1. 1/f depends only on (1/R₁ - 1/R₂), which flips sign twice under reversal, returning to itself.
2. Same f ⇒ same image at same u.
3. Concrete example. Take R₁ = +10 cm, R₂ = -20 cm (a thicker-on-the-left bi-convex lens), μ = 1.5. Then 1/f = 0.5(1/10 - 1/(-20)) = 0.5(1/10 + 1/20) = 0.5 × 0.15 = 0.075, so f ≈ 13.33 cm. After reversal: new R₁' = +20, new R₂' = -10, and 1/f' = 0.5(1/20 - 1/(-10)) = 0.5 × 0.15 = 0.075. Identical.
4. When does reversal matter? Only for a thick lens (where the optical centre is offset from the geometric centre) does reversal change the image position. For thin lenses (the standard NCERT setting) reversal is a pure symmetry.

Why this matters. This is a thin-lens property only. For a thick lens, the principal planes shift on reversal even though the focal length itself can stay the same. Real camera lenses are deliberately asymmetric to control aberrations – but they are far from ``thin'' in the geometric-optics sense.

No change in image position.

Quick reading. Each layer shrinks by 1/μ for near-normal viewing; total apparent depth is the sum.

1. Apply the layer-additive apparent-depth formula.
2. Each thickness is h/3, indices ₁, ₂, ₃.
3. Where the formula comes from. For a single layer of thickness t and index μ viewed near-normally, the first-surface refraction relation ₁/v - ₂/u = (₁ - ₂)/R with R = ∞ (flat) gives |v|/|u| = ₁/₂. Treating the eye as being in air (₁ = 1) and the object inside the liquid (₂ = μ), the image shifts to depth t/μ. Stacked layers are independent because each refraction happens at a flat interface where the previous image is the object.
4. Numerical example. Take h = 30 cm with ₁ = 1.6 (densest), ₂ = 1.4, ₃ = 1.2. Apparent depth = 10 × (1/1.6 + 1/1.4 + 1/1.2) = 10 × (0.625 + 0.714 + 0.833) = 21.7 cm. The real depth 30 cm has shrunk by about a third.
5. Why near-normal viewing matters. The formula t/μ is the small-angle (i → 0) limit; for larger viewing angles a more complex formula applies (and the image acquires lateral distortion). The problem statement ``near normal vision'' guarantees the simple sum.

Why this matters. The same logic applies when peering through a stack of plates of different glasses or to atmospheric layers above us: stars appear higher in the sky than they really are because of cumulative apparent-depth shifts.

h3(1₁ + 1₂ + 1₃).

Quick reading. Use the D_m = A shortcut: it reduces the prism formula to μ = 2 cos(A/2).

1. Solve 2 cos(A/2) = √3: A/2 = 30^∘, A = 60^∘.
2. Verification via angles. At minimum deviation r₁ = r₂ = A/2 = 30^∘ and i₁ = i₂ = (A + D_m)/2 = 60^∘. Check Snell at face 1: sin 60^∘ = μ sin 30^∘, i.e. √3/2 = √3 × 1/2. Equality holds.
3. Geometric interpretation. The condition D_m = A with an equilateral prism means the entry and exit angles (60^∘ each) coincide with the apex angle. The ray passes symmetrically through the prism, making the same angle with both faces.

Why this matters. Equilateral prisms (A = 60^∘) at μ = √3 have the elegant property D_m = A, often used in spectroscope calibration. The same prism geometry with μ = 1.5 (crown glass) gives D_m ≈ 38^∘, and optical prism designs aim to operate near the minimum-deviation point where dD/di = 0 – meaning slight alignment errors don't shift the spectrum.

A = 60^∘.

Strategic angle. Treat the short axial object as two end points separated by L. Their image separation comes from linearising the mirror map v(u).

1. Lateral magnification is m = -v/u; the longitudinal magnification is -dv/du = -m² (in magnitude m²).
2. Using m = f/(f-u) for mirrors, |m|² = f²/(f-u)² = f²/(u-f)², so L' = f² L/(u-f)².
3. Alternative direct method. Apply the mirror formula to the near end (u) and the far end (u + L) of the object, find both images v(u) and v(u+L), and take the difference. To first order in L ≪ |u - f|, v(u+L) - v(u) ≈ (dv/du) L = -f² L/(u-f)². The minus sign means the image is inverted longitudinally relative to the object.
4. Numerical check. Take f = -20 cm (concave mirror, f negative in Cartesian convention), u = -60 cm (object beyond focus). Then u - f = -60 - (-20) = -40, (u-f)² = 1600, f² = 400. L' = (400/1600) L = L/4. So an object of length L = 4 cm images to length 1 cm – a 4-fold longitudinal shrinkage at u = 3f.

Why this matters. Longitudinal magnification equals the square of the lateral magnification: an important rule for axial extent. This is why a sphere imaged by a microscope looks flatter than its actual depth (longitudinal m = m² vs. lateral m: for |m| < 1, m < m).

L' = f² L(u - f)².

Picture-first. Before filling, the rim-to-far-edge sight line just grazes the bowl. After filling, refraction at the rim lets the rim-to-near-edge sight line bend to that same angle inside the liquid.

1. Geometry: inside the liquid, the ray goes from the near edge to the rim; the angle with the vertical normal satisfies tan i = (a-R)/d. In air, the line of sight from the eye to the (now hidden) far edge keeps the same angle as in the empty bowl: tanα = (a+R)/d.
2. Snell (liquid→air): μ sin i = sinα, with sinθ = tanθ/√1+tan²θ.
3. Algebra: d = (a²-R²)√μ²-1/√(a+R)² - μ²(a-R)².
4. Two-stage geometric reading. The problem combines (i) a pure-geometry step (where the line of sight is in the empty bowl) with (ii) a Snell's-law step (after filling). Recognising that the same angle α governs both pre-fill geometry and post-fill in-liquid path is the key trick. The angle α is fixed by the bowl-disc geometry; only the air-side angle i changes when liquid is added.
5. Limiting checks.
   • μ → 1 (no liquid): numerator √μ² - 1 → 0, so d → 0. Makes sense – no liquid means no refraction, so no near-edge visibility shift.
   • R → a (disc fills the bowl rim): numerator (a² - R²) → 0, so d → 0. Again sensible.
   • R → 0: d → a²√μ² - 1/√μ² a² - a² = a√μ² - 1/√μ² - 1 · stuff; careful algebra gives a finite small-disc limit.

Why this matters. The same idea makes a coin in a cup ``appear'' when water is added: refraction lifts the line of sight. The classic ``coin in the bowl'' demonstration (a coin invisible in an empty cup becomes visible once water is poured in) is exactly this geometry.

d = (a²-R²)√μ²-1√(a+R)² - μ²(a-R)².

Structural observation. A horizontally-cut top half of a thin lens behaves like a normal lens whose principal axis is along the cut line, shifted downward by 0.5 cm from the top edge.

1. New principal axis passes through (0, -0.5) cm.
2. Object at (-50, 0) has effective coordinates (-50, +0.5) relative to the new axis.
3. Use 1/v = 1/f + 1/u gives v = +50 cm and m = -1; image is at (50, -0.5) in new-axis frame, i.e. (50, -1) in the original frame.
4. Cut lens ≡ shifted full lens. Cutting a thin lens parallel to its axis does not change its focal length – the cut piece behaves like the full lens but with the optical centre on the cut line. This is a consequence of paraxial-ray theory: the focal-length formula depends on R₁, R₂ and μ, all of which are unaltered by the cut.
5. Magnification interpretation. m = -1 means the image is the same size as the object but inverted. The ``inversion'' across the new axis (at y = -0.5) takes a height +0.5 above the new axis to a height -0.5 below the new axis, putting the image at y_abs = -1.
6. Why object at (-50, 0) gives u = 50 cm. The new lens plane is the y-axis (the cut is at the origin in x); the new principal axis is offset in y only. The x-distance from the lens to the object is 50 cm, unaffected by the y-offset.

Why this matters. Cut lenses are used to make ``D-shaped'' optics for prism couplers; this thinking generalises directly to Fresnel half-lenses, axicon prisms, and the classic Billet's split lens interference experiment.

(50, -1) cm.

Strategic angle. The quadratic u² - Du + fD = 0 encodes everything: two solutions u₁, u₂ with sum D and product fD.

1. Discriminant gives d = u₂ - u₁ = √D(D - 4f).
2. Magnification ratio uses m ∝ v/u and the swap u₁ ↔ v₂ implied by the symmetric roots.
3. Object/image swap symmetry. At one lens position the lens is closer to the source (u₁ small, v₁ large); at the other position it is closer to the screen (u₂ large, v₂ small). In fact u₁ = v₂ and v₁ = u₂. The two images are mirror swaps of each other in terms of size: one is magnified, the other diminished.
4. Focal length from d and D. Inverting, f = D² - d²4D. This is the practical Bessel formula: measure D and d, compute f without ever locating the optical centre.
5. Existence condition D > 4f. If D ≤ 4f the discriminant is non-positive: no real lens position gives an image on the screen. Physically, the lens isn't strong enough to converge in such a short throw.
6. Numerical example. For f = 10 cm and D = 60 cm: d = √60(60-40) = √1200 ≈ 34.6 cm. The two lens positions are u₁ = (60-34.6)/2 = 12.7 cm and u₂ = (60+34.6)/2 = 47.3 cm. Sizes ratio ((D+d)/(D-d))² = (94.6/25.4)² ≈ 13.9.

Why this matters. The displacement method is the standard lab technique to measure unknown focal lengths to high precision, avoiding the need to locate the optical centre. It also illustrates the optical reciprocity principle: u₁ ↔ v₂ under reversal of the light path.

d = √D(D-4f), ratio = (D+dD-d)².

Picture-first. The dot sends out rays in a cone. Rays within the critical-angle cone escape; rays outside it TIR. To hide the dot completely, cover the escape circle on the top surface.

1. Critical-angle radius on the top: r = h tan i_c = h/√μ² - 1.
2. Disc diameter = 2r = 2h/√μ² - 1.
3. Why TIR alone isn't enough. Outside the escape cone, rays already TIR – no disc needed. Inside the cone, rays would escape – the disc must block them. So the disc only has to cover the escape circle, no more, no less.
4. Numerical case: water. For μ = 4/3 and h = 10 cm: d_min = 20/√16/9 - 1 = 20/√7/9 = 20 × 3/√7 ≈ 22.7 cm. Substantial disc – more than twice the depth.
5. Snell's window in disguise. The same circle of radius h/√μ² - 1 on the water surface is the boundary of Snell's window: looking up from under water, the entire sky is compressed into a cone of this diameter. Here we cover that window from above to hide the dot below.
6. Limiting case μ → 1. The disc diameter → ∞, meaning that with no refractive-index contrast, no disc can hide the dot from a sufficiently oblique observer.

Why this matters. Same geometry sets the maximum viewing cone of an underwater fish (Snell's window) and the design of optical fibre acceptance cones. The numerical aperture of a fibre is NA = _core sin_cone, exactly the analogous quantity.

d_min = 2 h√μ² - 1.

Strategic angle. Compute the relaxed-eye power, add accommodation, and use the spectacle lens to translate object positions to image positions inside the eye's working range.

1. Relaxed eye: P₀ = 1/v_retina + 1/u_far = 50 + 10 = 60 D.
2. Spectacle: P = -10 D to bring infinity to far point.
3. Near point: with full accommodation P_max = 64 D ⇒ |u_n| = 1/14 m ≈ 7.14 cm (no glasses); 25 cm with glasses.
4. Sign-convention discipline. The Cartesian convention here puts object distances on the left of the lens as negative. Switching to all-positive ``magnitudes'' is convenient for eye-physics problems where every distance has a known direction. Keep consistent: either Cartesian throughout, or magnitudes throughout, never mix.
5. Verifying with glasses. The -10 D lens takes any object at finite distance and produces a virtual image closer to the lens. Specifically: object at ∞ → image at -10 cm (the far point), perfect. Object at 25 cm in air → image where? 1/v - 1/(-0.25) = -10, so 1/v = -10 - 4 = -14, v = -7.14 cm. Image at near point: matches. Confirms that a 25 cm object is just visible with full accommodation.
6. Concept linkage to eye defects. Myopia: spectacles are diverging (P < 0). Hypermetropia: spectacles are converging (P > 0). Presbyopia (age-related loss of accommodation): bifocal/varifocal lenses combining both powers.

Why this matters. The myopic eye trades far-point sharpness for a closer-than-normal near point: a feature, not just a deficit. Some watchmakers and microsurgeons mildly myopic without correction can do fine work without external magnifiers.

(i) -10 D, (ii) 7.14 cm, (iii) 25 cm.

Picture-first. Inside the slab, the worst angle for side-face TIR is when the refracted ray is steepest (entry at 90^∘). Even then, the angle with the side-face normal must still exceed the critical angle.

1. Worst-case refracted angle: sin r = 1/μ.
2. Need cos r ≥ 1/μ for TIR at side face.
3. Combine: √1 - 1/μ² ≥ 1/μ ⇒ μ² ≥ 2.
4. Geometric picture. The worst-case entry is a ray grazing the front face at i = 90^∘. Inside the slab this ray makes angle r = sin^-1(1/μ) with the front-face normal, i.e. 90^∘ - r with the side-face normal. The TIR condition at the side face requires this angle to be at least i_c = sin^-1(1/μ). The clean bound on μ comes from 90^∘ - r ≥ r, i.e. r ≤ 45^∘, i.e. sin r ≤ 1/√2, i.e. 1/μ ≤ 1/√2, i.e. μ ≥ √2.
5. Cross-check with the cosine inequality. cos r ≥ 1/μ with sin r = 1/μ gives cos² r + sin² r = 1, so (1/μ)² + (1/μ)² ≤ 1 ⇒ μ² ≥ 2. Same answer, reached differently.
6. Numerical for common glasses. Crown glass μ = 1.5 < √2 ≈ 1.414? No – √2 ≈ 1.414 and 1.5 > 1.414. So crown glass does satisfy μ ≥ √2 marginally. Water (μ = 1.33) does not, hence cannot guide all incidence angles. Optical fibre cores (μ ≈ 1.46) do.

Why this matters. This is the principle behind optical fibres: with μ ≥ √2, fibres of any shape can guide light without leakage, regardless of input angle. Real fibres use a cladding (lower-μ glass) around the core to relax this condition: the relevant μ is the index contrast _core/_clad, not the absolute _core.

μ ≥ √2 guarantees full guiding.

Strategic angle. Continuous refractive-index gradients bend rays like gravity bends massive particles. Use the small-angle ray equation y'' = μ'/μ.

1. Integrate twice with y(0) = 0, y'(0) = 0: y(d) = 12(μ'/μ) d².
2. Where the ray equation comes from. Fermat's principle says rays minimise optical path length ∫ μ ds. The Euler-Lagrange equation of this functional, in the paraxial limit, reads d² ydx² = 1μ ∂ μ∂ y. For nearly horizontal rays in a vertically-varying index, this reduces to y'' = μ'/μ.
3. Constant-acceleration analogue. The trajectory y(d) = 12 β d² with β = μ'/μ is formally identical to projectile motion under constant ``acceleration'' β. Rays bending in graded-index media are just like balls bending in gravity.
4. Direction of bending. If dμ/dy > 0 (denser higher), the ray bends upward. If dμ/dy < 0 (denser lower, the typical solution-diffusion case), the ray bends downward. Light always bends towards regions of higher μ.

Why this matters. This is exactly the mirage equation; graded-index optical components like GRIN lenses (used in fibre couplers, photocopiers, endoscopes) work on the same principle. The same equation describes star-light bending through the Earth's atmosphere and the slow drift of laser pulses through chirped optical fibres.

Δ y = 12 (1/μ)(dμ/dy) d².

Strategic angle. Grazing deflection by a sphere of mass M and radius R comes from the integrated transverse gradient of n(r) along the ray.

1. Compute ∂ n/∂ y from n(r) = 1 + 2GM/(rc²).
2. Integrate along -∞ to ∞ with y = R: ∫ dx/(x²+R²)^3/2 = 2/R².
3. Multiply: α = 4GM/(c² R).
4. Numerical: light grazing the Sun. M = 2 × 10³⁰ kg, R = 7 × 10⁸ m, G = 6.67 × 10^-11, c = 3 × 10⁸. Then α = 4 × 6.67 × 10^-11 × 2 × 10³⁰ (3 × 10⁸)² × 7 × 10⁸ ≈ 8.5 × 10^-6 rad ≈ 1.75''. This is the famous 1.75-arcsecond deflection that Eddington measured during the 1919 solar eclipse.
5. Newtonian half-result. A purely Newtonian calculation (treating light as a non-relativistic projectile at speed c) gives only 2GM/(c² R) = 0.87'' – half the correct value. General relativity adds an equal contribution from spatial curvature, yielding the observed 1.75''. The factor-of-2 mismatch was the decisive test in 1919.
6. Alternative: integral table check. The integral _-∞^∞ dx/(x² + R²)^3/2 = 2/R² is a standard table integral (let x = R tanθ). Always verify such integrals before plugging into physics formulas.

Why this matters. This is exactly Einstein's 1919 prediction (confirmed by Eddington), which gave us general relativity's first decisive test. The same formula applies in gravitational lensing of distant galaxies, providing the dominant modern probe of dark matter distributions in the universe.

α = 4 G Mc² R.

Picture-first. Negative refraction at the curved interface flips the ray's transverse component. For a μ = -1 cylinder, each refraction adds θ to the running deviation, with θ = sin^-1(|x|/R) set by the entry geometry. Two refractions accumulate 4θ of deviation; the ray is sent back down whenever 4θ lies in the back-half angular range [π/2, 3π/2].

1. Identify the Snell condition sin_i = -sin_r at each curved interface.
2. Total deviation across the two refractions is 4θ at the cylinder centre.
3. Back-scatter condition: π/2 ≤ 4θ ≤ 3π/2, i.e. π/8 ≤ θ ≤ 3π/8.
4. Convert to |x| via |x| = Rsinθ ≈ Rθ (paraxial): forbidden range Rπ8 ≤ |x| ≤ 3Rπ8.
5. Outside the forbidden range.
   • For |x| < Rπ/8: the deviation 4θ < π/2, so the ray still emerges into the upper half-space and reaches the top plate.
   • For |x| > 3Rπ/8 (but |x| ≤ R): the deviation 4θ > 3π/2, again the ray escapes upward.
   • For |x| > R: the ray misses the cylinder entirely and goes straight up.
6. Boundary check at |x| = Rπ/8. Here θ = π/8, 4θ = π/2: the exit direction is exactly horizontal, the borderline between ``reaches the top'' and ``goes back''.

Why this matters. Pendry's perfect lens (2000) uses exactly this μ = -1 slab geometry to achieve sub-wavelength imaging. The cylindrical version explored here is the basis of transformation-optics invisibility-cloak designs.

Rπ8 ≤ |x| ≤ 3Rπ8.

Strategic angle. Fermat reduces the lens problem to an extremisation of OPL over b. For a parabolic lens you recover the thin-lens formula; for a log-profile lens you get the Einstein ring.

1. Write OPL(b) = √u²+b² + (n-1)w(b) + √v²+b².
2. Paraxial expand; set d(OPL)/db = 0.
3. (i) Parabolic w gives 1/f = 2(n-1)/α.
4. (ii) Log w gives b² = (n-1)k₁ uv/(u+v) and β = b/v.
5. Why a parabolic lens is special. The parabolic thickness profile w(b) = w₀ - b²/α is exactly the shape that focuses paraxial rays to a single point on the axis. Any deviation from parabolic (e.g. a true spherical surface) introduces spherical aberration: rays at different b focus at slightly different points. This is why high-quality lenses are aspheric.
6. Einstein-ring geometry. For the log profile, the stationarity condition gives a ring of values of b where d(OPL)/db = 0, not a single point. Every direction around the symmetry axis is equally ``focused'', so the image of a point source is a ring of angular radius β. This is exactly the strong gravitational-lensing geometry: a perfectly-aligned source behind a point mass produces a luminous ring image.
7. Sanity check on part (i). Compare 1/u + 1/v = 2(n-1)/α with the thin-lens equation 1/u + 1/v = 1/f. Identifying gives f = α/[2(n-1)]. Larger α (thicker lens) ⇒ longer f. Larger n (denser glass) ⇒ shorter f. Both physically reasonable.

Why this matters. The log-profile is exactly the gravitational lens around a point mass; the predicted Einstein ring has been observed for distant galaxies behind closer ones, most famously by the Hubble Space Telescope. Modern surveys (e.g. DES, Euclid) map cosmic dark matter via the statistics of gravitational lensing distortions.

(i) f = α/[2(n-1)]. (ii) β = √(n-1)k₁ u/[v(u+v)].