NCERT Plus Two (Class 12) Physics • Chapter 10

Wave Optics

Q: Using Huygens' wave theory, draw a neat wavefront diagram and derive Snell's law of refraction for a plane wave entering a denser medium.

Let plane wavefront AB hit interface XY at angle i with speed v₁ . In time τ , edge B advances BC = v₁τ to hit C , while edge A emits secondary wavelet of radius AE = v₂τ inside medium 2; tangent EC is refracted wavefront at angle r . From right triangles △ABC and △AEC : sin i = BC/AC = v₁τ / AC , sin r = AE/AC = v₂τ / AC . Dividing: sin i / sin r = v₁ / v₂ = n₂ / n₁ (Snell's Law!).

Q: Does interference of light violate the law of conservation of energy? Explain briefly.

No. At dark fringes, light energy is NOT destroyed — it is merely redistributed from positions of destructive interference ( I = 0 ) to positions of constructive interference ( I = 4I₀ ). Average intensity across the screen equals (4I₀ + 0)/2 = 2I₀ = I₁ + I₂ , strictly conserving energy.

Q: Derive an expression for the fringe width β in Young's Double Slit Experiment (YDSE).

Slits S₁, S₂ separated by d , screen at distance D ≫ d , point P at distance y from centre. S₂P² − S₁P² = [D² + (y + d/2)²] − [D² + (y − d/2)²] = 2yd . Using S₂P + S₁P ≈ 2D : Path difference Δx = S₂P − S₁P = yd / D . • Bright fringes ( Δx = nλ ): y n = nλD/d . • Dark fringes ( Δx = (n − ½)λ ): y' n = (n − ½)λD/d . Fringe width β = y n+1 − y n = λD / d (identical for bright and dark).

Q: In YDSE, two slits 0.28 mm apart are placed 1.4 m from a screen. Distance between the central bright fringe and the fourth bright fringe is 1.2 cm. Find the wavelength of light used.

y₄ = 4β = 4λD / d = 1.2 × 10⁻² m . λ = (1.2 × 10⁻² × 0.28 × 10⁻³) / (4 × 1.4) = (3.36 × 10⁻⁶) / 5.6 = 6.0 × 10⁻⁷ m = 600 nm .

Q: Explain the formation of the first dark minimum in single-slit diffraction and write the formula for angular and linear width of the central maximum.

Divide slit width a into two equal halves of width a/2 . At angle θ₁ where path diff between extremes is a sinθ₁ = λ , top edge and midpoint differ by λ/2 — every pair separated by a/2 cancels destructively, yielding the first dark minimum ( a sinθ = nλ ). • Angular width of central maximum: 2θ₁ = 2λ / a . • Linear width on screen at distance D : W₀ = 2λD / a .

Q: Write any three differences between an interference pattern and a single-slit diffraction pattern.

1. Fringe Width: Equal width for all fringes in interference; central maximum is twice as wide as secondary fringes in diffraction. 2. Intensity of Maxima: All bright fringes have equal intensity in interference; intensity drops rapidly away from central peak in diffraction. 3. Minima Contrast: Minima are completely dark (zero intensity for equal sources) in interference; minima are generally not completely dark in diffraction.

Q: State Malus's law. Sketch the variation of transmitted intensity I with the angle θ between the pass axes of polariser and analyser (from 0° to 180°).

Malus's Law: When completely plane-polarised light of intensity I₀ falls on an analyser, transmitted intensity varies as the square of the cosine of angle θ between their pass axes: I = I₀ cos²θ . Graph: Bell-like curve starting at I₀ ( 0° ), dropping to 0 at 90° , and rising symmetrically back to I₀ at 180° .

Q: Two polaroids P₁ and P₂ are placed with their pass axes perpendicular (90°) to each other. Unpolarised light of intensity I₀ is incident on P₁. A third polaroid P₃ is inserted between P₁ and P₂ at 45° to P₁. Find the final transmitted intensity.

• After P₁ (unpolarised → polarised): I₁ = I₀ / 2 . • After P₃ (at 45° to P₁ ): I₃ = I₁ cos²45° = (I₀/2)(1/2) = I₀ / 4 . • Angle between P₃ and P₂ is 90° − 45° = 45° . • After P₂ : I₂ = I₃ cos²45° = (I₀/4)(1/2) = I₀ / 8 !

Huygens wavefronts & reflection/refraction proofs, coherent vs incoherent superposition, Young's double-slit experiment, single-slit diffraction & Malus's law polarisation.

📄 Official NCERT Textbook PDF

Topic 10.2 – 10.3: Huygens Principle & Plane Wave Reflection/Refraction Proofs

Drop a tiny pebble into a calm Alappuzha canal and watch circular ripples race outward! Dutch physicist Christiaan Huygens asked a brilliant question: how do ripples know where to travel next? Every point on a ripple acts like a fresh baby pebble, launching secondary wavelets whose outer shared envelope forms the next wavefront!

WAVEFRONT SHAPESPoint source → Spherical • Slit → Cylindrical • Far Sun/Star → Plane (Rays ALWAYS ⊥ wavefront!)

HUYGENS RULEEvery point emits secondary wavelets of radius r = v τ; forward tangent envelope = new wavefront!

SNELL VIA WAVESsin isin r = v₁v₂ = n₂n₁ (v₂ < v₁ in denser medium!)

🧠 Memory Trick: Newton's corpuscle model failed by guessing light speeds UP in water; Huygens Wave Theory proved light SLOWS DOWN in denser media (v = c / n) while frequency stays fixed!

💡 Simple Kid-Check: A Wavefront is simply a snapshot connecting all water particles bobbing up together at the exact same beat (same phase)! Light rays are just arrows drawn perpendicular (90°) to those wavefronts showing where energy marches.

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Three Wavefront Shapes: A tiny glowing bulb radiates spherical wavefronts; a long glowing tube or narrow slit radiates cylindrical wavefronts; far from any star or sun, small patches flatten into plane wavefronts!
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Huygens' Construction & Reflection Proof: Let plane wavefront AB hit a mirror at angle i. While edge B travels BC = cτ to the mirror, edge A sends a reflected wavelet of radius AE = cτ. Congruent right triangles △ABC ≅ △AEC prove i = r!
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Refraction Proof (Snell's Law): Crossing from medium 1 (speed v₁) into denser medium 2 (speed v₂): in time τ, BC = v₁τ while refracted wavelet radius AE = v₂τ. Taking ratios: sin i / sin r = (v₁τ/AC) / (v₂τ/AC) = v₁ / v₂ = n₂ / n₁!

🧒 Super-Easy Example: Yellow sodium light (λ = 589 nm, speed 3 × 10⁸ m/s) enters water (n = 4/3):
• Speed drops to v = (3 × 10⁸) / (4/3) = 2.25 × 10⁸ m/s
• Wavelength shrinks to λ' = 589 / (4/3) = 441.8 nm, while frequency ν = c/λ = 5.09 × 10¹⁴ Hz stays IDENTICAL!

Watch three classic DHSE traps: (1) Always draw rays perpendicular (90°) to wavefronts! (2) When proving Snell's law with Huygens, make sure the wavelet radius inside the denser medium is SMALLER (v₂τ < v₁τ) — light slows down in glass/water! (3) Don't forget: frequency never changes on refraction (v₁/λ₁ = v₂/λ₂).

Topic 10.4: Coherent vs Incoherent Addition of Light Waves

Switch on two ordinary study lamps next to each other — the wall simply gets evenly brighter with zero dark stripes! Why? Because ordinary lamps flash random uncoordinated bursts every few nanoseconds. Only waves locked in steady rhythm — Coherent Sources — can cancel and boost each other to paint steady bright-and-dark interference fringes!

COHERENT SOURCESSame frequency + zero or constant phase difference Δφ with time!

RESULTANT INTENSITYI = I₁ + I₂ + 2√(I₁I₂) cos φ
Equal waves (I₀): I = 4 I₀ cos²(φ/2)

MAX & MIN INTENSITYI_max = (√I₁ + √I₂)² (4I₀)
I_min = (√I₁ − √I₂)² (0)

🧠 Memory Trick: Two independent bulbs are NEVER coherent! Always split light from ONE parent source! Phase difference ↔ Path difference: Δφ = (2π / λ) Δx!

💡 Simple Kid-Check: Push a playground swing in sync with your friend's timing — the swing rockets super high (Constructive interference!). Push exact opposite to your friend's timing — the swing stops dead still (Destructive interference!).

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Superposition & Intensity: Two coherent waves y₁ = a cos(ωt) and y₂ = a cos(ωt + φ) add to amplitude A = 2a cos(φ/2). Since Intensity I ∝ A², I = 4 I₀ cos²(φ/2) (or generally I = I₁ + I₂ + 2√(I₁I₂) cosφ).
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Constructive vs Destructive Conditions:
• Bright (Max): φ = 0, ±2π, ±4π... ⇒ Path diff Δx = n λ (n = 0, 1, 2...) ⇒ I = 4I₀.
• Dark (Min): φ = ±π, ±3π... ⇒ Path diff Δx = (n + ½) λ ⇒ I = 0.
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Incoherent Waves & Energy Conservation: Two separate lamps suffer random phase jumps every ~10⁻⁸ s, averaging ⟨cos φ⟩ = 0 ⇒ I = I₁ + I₂ = 2I₀ (uniform glow!). Does interference destroy energy at dark fringes? NO! Energy missing from dark spots (0) shifts to bright peaks (4I₀), keeping average intensity (4I₀ + 0)/2 = 2I₀!

🧒 Super-Easy Example: Two coherent sources have intensity ratio I₁ : I₂ = 9 : 1 (amplitudes 3 : 1).
I_max : I_min = (√9 + √1)² : (√9 − √1)² = 4² : 2² = 16 : 4 = 4 : 1!

Top 3 DHSE traps: (1) Why can't two identical independent sodium lamps produce sustained interference? Because atoms emit independent random flashes, so phase difference φ varies randomly with time! (2) Does destructive interference violate Conservation of Energy? NO! Light energy is merely redistributed from dark minima to bright maxima! (3) Don't mix up amplitude ratio (a₁/a₂) with intensity ratio (a₁/a₂)²!

Topic 10.5: Young's Double Slit Experiment & Fringe Width Derivation

In 1801, Thomas Young poked one pinhole S in front of a candle, then split that single beam through two razor-thin slits S₁ and S₂ side by side. Everyone expected just two bright bars on the screen — instead, Young gasped at an entire ladder of equally spaced bright and dark zebra fringes!

PATH DIFFERENCEΔx = S₂P − S₁P ≈ y dD (when d ≪ D)

FRINGE WIDTH (β)β = λ Dd (Same width for ALL bright & dark fringes!)

FRINGE POSITIONSBright: y_n = n β (n=0,1,2...)
Dark: y_n = (n − ½) β (n=1,2...)

🧠 Memory Trick: β = λ D / d — "Large Distance, Long wavelength, tiny d = Wider fringes!" Angular fringe width θ = β / D = λ / d doesn't even depend on screen distance D!

💡 Simple Kid-Check: Why did Young need ONE pinhole S before the two slits S₁ and S₂? Because S₁ and S₂ behave like twin children copying the exact same parent heartbeat — guaranteeing 100% coherence!

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Deriving Path Difference (d ≪ D): For point P at height y from screen centre O: S₂P² − S₁P² = [D² + (y + d/2)²] − [D² + (y − d/2)²] = 2 y d. Since S₂P + S₁P ≈ 2D, dividing gives Δx = S₂P − S₁P = y d / D.
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Bright & Dark Positions + Fringe Width:
• Bright fringes (Δx = nλ): y_n = n (λD / d) (n = 0, ±1, ±2...; central fringe at y = 0 is BRIGHT!).
• Dark fringes (Δx = (n − ½)λ): y_n = (n − ½)(λD / d).
• Fringe Width: β = y_n+1 − y_n = λD / d!
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White Light & Medium Tricks: With white light, central fringe (Δx = 0) is purely WHITE (all colours interfere constructively together!), flanked by violet closest (β_V smallest) and red outermost (β_R largest). Dipping the apparatus in water (n) reduces λ' = λ/n, shrinking β' = β/n!

🧒 Super-Easy Example: In YDSE, slits are d = 0.5 mm apart, screen sits at D = 1.2 m, and light wavelength is λ = 500 nm:
Fringe width β = (500 × 10⁻⁹ × 1.2) / (0.5 × 10⁻³) = 1.2 × 10⁻³ m = 1.2 mm!

Top 4 Kerala DHSE traps: (1) Dip YDSE in water (n = 4/3) → wavelength shrinks to λ/n, so fringe width SHRINKS to 3/4 β! (2) Cover slit S₂ → interference fringes vanish! (3) Replace monochromatic light with white light → central bright fringe stays WHITE, NOT coloured! (4) Move screen farther (↑D) → linear width β increases, yet angular width θ = λ/d stays UNCHANGED!

Topic 10.6: Single-Slit Diffraction & Central Maximum Width vs Slit Width

Pinch your thumb and index finger almost touching and peer through the hairline gap at a distant bulb — you see soft fuzzy bright and dark streaks fanning sideways! Even through ONE single slit, light refuses to stay in a razor-sharp beam and flares around the edges — that spreading is Diffraction!

DARK MINIMA (SINGLE SLIT)a sin θ = n λ (n = ±1, ±2..., NOT 0!)

SECONDARY MAXIMAa sin θ ≈ (n + ½) λ (Rapidly dimmer!)

CENTRAL MAX WIDTHLinear: W₀ = 2 λ Da | Angular: 2θ₁ = 2 λa

🧠 Memory Trick: Single-slit diffraction FLIPS the YDSE condition! a sin θ = n λ is DARK (opposite to YDSE bright!), and narrower slit ↓a ⇒ WIDER central peak ↑W₀!

💡 Simple Kid-Check: Sound waves (λ ~ 1 m) bend right around classroom doorways because door width matches sound's wavelength! Light has tiny wavelength (λ ~ 0.5 μm), so light only flares noticeably through hairline slits!

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Why a sin θ = λ Makes First DARK Band: Imagine dividing slit of width a into two equal halves (top a/2 and bottom a/2). At angle sin θ = λ/a, the path difference between top edge and midpoint is (a/2) sinθ = λ/2! Every wavelet in the top half cancels its exact partner a/2 below it — producing complete DARKNESS!
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Higher Minima & Secondary Peaks: Divide the slit into 2n parts to get the n-th minimum at a sinθ = nλ. Between minima sit weak secondary maxima at (n + ½)λ (where only 1/3, 1/5... of the slit remains uncancelled!).
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Central Maximum Width = 2× Side Bands: All wavelets at θ = 0° arrive in phase (giant peak I₀), spanning from −θ₁ = −λ/a to +θ₁ = +λ/a! So angular width = 2λ/a and linear width = 2λD/a (twice every secondary fringe λD/a!).

🧒 Super-Easy Example: Red light (λ = 650 nm) passes through a single slit of width a = 0.2 mm onto a screen D = 2 m away.
First dark fringe sits at y₁ = λD/a = (650×10⁻⁹ × 2)/(0.2×10⁻³) = 6.5 mm → Full central bright width W₀ = 2y₁ = 13 mm!

Watch the #1 Kerala HSE exam mistake: Students write a sinθ = nλ as a bright maximum just because YDSE used d sinθ = nλ! Remember: in single-slit diffraction, a sinθ = nλ gives DARK MINIMA (and never plug n = 0, because θ = 0 is the brightest Central Maximum!). Also, squeezing slit width a smaller makes the central bright band WIDER!

Topic 10.6 (Compare): Comparison: Interference Fringes vs Diffraction Pattern

Hold two striped ribbons side by side: one has identical crisp zebra stripes from top to bottom; the other has one huge glowing king stripe in the middle with timid, fading baby stripes on each side! In Kerala Plus Two Physics exams, comparing Interference vs Diffraction is one of the highest-yield 3-mark questions!

FRINGE WIDTHInterference: Equal (β = λD/d)
Diffraction: Unequal (Central 2× wider!)

BRIGHTNESSInterference: All maxima equally bright
Diffraction: Drops sharply (1 : 0.045 : 0.016)

MINIMA CONTRASTInterference: Perfectly dark (I = 0)
Diffraction: Never strictly zero / lower contrast

🧠 Memory Trick: I = Identical, D = Decreasing & Double-wide! Real double-slit fringes are cos² Interference bars tucked INSIDE a (sin α / α)² Diffraction umbrella!

💡 Simple Kid-Check: Interference is a duet between two slits (S₁ and S₂); Diffraction is a full choir across every tiny point inside ONE slit! Because a real YDSE slit isn't infinitely thin, real lab patterns show BOTH working together!

Feature	Interference (YDSE, 2 Slits)	Diffraction (Single Slit)
Origin	Superposition of waves from two coherent sources	Superposition of secondary wavelets across one slit width
Fringe Width	All fringes equal width (`β = λD/d`)	Unequal! Central max (`2λD/a`) is 2× wider than side bands
Maxima Intensity	All bright fringes equally bright (`4I₀`)	Central max brightest; secondary peaks fade rapidly
Minima Darkness	Completely dark (`I = 0`) → super crisp contrast	Never completely zero / weaker contrast
`nλ` Condition	`d sinθ = nλ` gives BRIGHT maxima	`a sinθ = nλ` gives DARK minima!

🧒 Super-Easy Combined YDSE Check: In real double slits of width a separated by d, intensity follows I(θ) = I₀ [sin α / α]² cos² β. How many interference bright fringes fit inside the central diffraction peak? Divide widths: (2λ/a) / (λ/d) = 2d / a! (If d = 4a, ~7–8 bright bands fit inside!)

Watch two classic exam misconceptions: (1) Saying "Interference and Diffraction are totally different phenomena" — actually BOTH arise from wave superposition! (2) Forgetting that in a real double-slit experiment, the bright fringes gradually fade toward the edges because they sit underneath the single-slit diffraction envelope!

Topic 10.7: Polarisation by Polaroids & Malus's Law (I = I₀ cos² θ)

Put on Polaroid sunglasses on a sunny afternoon ride along a wet Kochi street and the blinding glare off car hoods and puddles magically vanishes! How? Light waves vibrate sideways to their travel direction (transverse), and a Polaroid film acts like a narrow picket fence letting only one vibration direction pass!

TRANSVERSE PROOFPolarisation proves light is a TRANSVERSE wave! (Longitudinal sound CANNOT be polarised!)

FIRST POLARISERUnpolarised light (I_un) passing ideal Polariser P₁ ALWAYS drops to I₀ = ½ I_un!

MALUS'S LAWI = I₀ cos² θ (θ = 0° ⇒ I₀; θ = 90° crossed ⇒ 0!)

🧠 Memory Trick: Gate 1 halves unpolarised light (÷ 2), Gate 2 uses Malus (× cos² θ)! Put a 45° middle Polaroid between two crossed (90°) ones and light magically returns (I_un / 8)!

💡 Simple Kid-Check: Shake a skipping rope through a vertical wooden garden fence — up-down waves slip straight through, while side-to-side waves slam into the slats and get blocked! A Polaroid sheet has aligned molecular chains acting just like that fence for light's electric field!

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Unpolarised vs Linearly Polarised Light: Ordinary bulbs emit electric field vibrations in all perpendicular planes symmetrically. Passing Polaroid P₁ (Polariser) filters vibrations down to one single plane along the pass-axis, halving intensity: I₀ = ½ I_un.
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Malus's Law (I = I₀ cos²θ): When plane-polarised light of amplitude A₀ (intensity I₀ ∝ A₀²) hits Analyser P₂ tilted at angle θ to P₁, only the parallel component A = A₀ cosθ gets through ⇒ transmitted intensity I = I₀ cos²θ!
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Applications & Why Sound Fails: Sound waves vibrate along propagation (longitudinal), so no fence slit can pick a perpendicular plane — sound CANNOT be polarised! Polaroids power glare-free sunglasses, camera sky filters, LCD screens, and 3D movie glasses.

🧒 Super-Easy Example: Unpolarised sunlight of intensity 64 W/m² passes Polariser P₁, then Analyser P₂ rotated 30° relative to P₁:
1. After P₁: I₀ = 64 / 2 = 32 W/m² (NOT 64 cos²30°!).
2. After P₂: I = 32 cos²(30°) = 32 × (√3/2)² = 32 × 3/4 = 24 W/m²!

Watch the #1 Kerala DHSE Polaroid numerical trap: (1) Students forget the ÷ 2 on the FIRST polariser when unpolarised light enters! Unpolarised light has average ⟨cos²φ⟩ = ½, so intensity ALWAYS halves at P₁ regardless of P₁'s orientation! (2) Classic 3-Polaroid puzzle: Two crossed polaroids (90°) block 100% light (I = 0). Insert a 3rd polaroid at 45° in between → light passes with I = (I_un/2) cos²45° cos²45° = I_un / 8!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 3 Marks): Using Huygens' wave theory, draw a neat wavefront diagram and derive Snell's law of refraction for a plane wave entering a denser medium.

Let plane wavefront AB hit interface XY at angle i with speed v₁.
In time τ, edge B advances BC = v₁τ to hit C, while edge A emits secondary wavelet of radius AE = v₂τ inside medium 2; tangent EC is refracted wavefront at angle r.
From right triangles △ABC and △AEC:
sin i = BC/AC = v₁τ / AC, sin r = AE/AC = v₂τ / AC.
Dividing: sin i / sin r = v₁ / v₂ = n₂ / n₁ (Snell's Law!).

Q2 (Kerala HSE Say • 2 Marks): Does interference of light violate the law of conservation of energy? Explain briefly.

No. At dark fringes, light energy is NOT destroyed — it is merely redistributed from positions of destructive interference (I = 0) to positions of constructive interference (I = 4I₀). Average intensity across the screen equals (4I₀ + 0)/2 = 2I₀ = I₁ + I₂, strictly conserving energy.

Q3 (Kerala HSE March • 4 Marks): Derive an expression for the fringe width β in Young's Double Slit Experiment (YDSE).

Slits S₁, S₂ separated by d, screen at distance D ≫ d, point P at distance y from centre.
S₂P² − S₁P² = [D² + (y + d/2)²] − [D² + (y − d/2)²] = 2yd.
Using S₂P + S₁P ≈ 2D: Path difference Δx = S₂P − S₁P = yd / D.
• Bright fringes (Δx = nλ): y_n = nλD/d.
• Dark fringes (Δx = (n − ½)λ): y'_n = (n − ½)λD/d.
Fringe width β = y_n+1 − y_n = λD / d (identical for bright and dark).

Q4 (Kerala HSE Model • 2 Marks): In YDSE, two slits 0.28 mm apart are placed 1.4 m from a screen. Distance between the central bright fringe and the fourth bright fringe is 1.2 cm. Find the wavelength of light used.

y₄ = 4β = 4λD / d = 1.2 × 10⁻² m.
λ = (1.2 × 10⁻² × 0.28 × 10⁻³) / (4 × 1.4) = (3.36 × 10⁻⁶) / 5.6 = 6.0 × 10⁻⁷ m = 600 nm.

Q5 (Kerala HSE March • 3 Marks): Explain the formation of the first dark minimum in single-slit diffraction and write the formula for angular and linear width of the central maximum.

Divide slit width a into two equal halves of width a/2. At angle θ₁ where path diff between extremes is a sinθ₁ = λ, top edge and midpoint differ by λ/2 — every pair separated by a/2 cancels destructively, yielding the first dark minimum (a sinθ = nλ).
• Angular width of central maximum: 2θ₁ = 2λ / a.
• Linear width on screen at distance D: W₀ = 2λD / a.

Q6 (Kerala HSE Say • 3 Marks): Write any three differences between an interference pattern and a single-slit diffraction pattern.

1. Fringe Width: Equal width for all fringes in interference; central maximum is twice as wide as secondary fringes in diffraction.
2. Intensity of Maxima: All bright fringes have equal intensity in interference; intensity drops rapidly away from central peak in diffraction.
3. Minima Contrast: Minima are completely dark (zero intensity for equal sources) in interference; minima are generally not completely dark in diffraction.

Q7 (Kerala HSE March • 2 Marks): State Malus's law. Sketch the variation of transmitted intensity I with the angle θ between the pass axes of polariser and analyser (from 0° to 180°).

Malus's Law: When completely plane-polarised light of intensity I₀ falls on an analyser, transmitted intensity varies as the square of the cosine of angle θ between their pass axes: I = I₀ cos²θ.
Graph: Bell-like curve starting at I₀ (0°), dropping to 0 at 90°, and rising symmetrically back to I₀ at 180°.

Q8 (NCERT / DHSE Tricky • 3 Marks): Two polaroids P₁ and P₂ are placed with their pass axes perpendicular (90°) to each other. Unpolarised light of intensity I₀ is incident on P₁. A third polaroid P₃ is inserted between P₁ and P₂ at 45° to P₁. Find the final transmitted intensity.

• After P₁ (unpolarised → polarised): I₁ = I₀ / 2.
• After P₃ (at 45° to P₁): I₃ = I₁ cos²45° = (I₀/2)(1/2) = I₀ / 4.
• Angle between P₃ and P₂ is 90° − 45° = 45°.
• After P₂: I₂ = I₃ cos²45° = (I₀/4)(1/2) = I₀ / 8!