Class 9 Physics Chapter 10 Wave Optics MCQ with Answers – 30 Questions (Easy, Medium, Hard)

The 2024-25 CBSE Physics syllabus has shifted emphasis toward conceptual MCQs over calculation-heavy long-answer questions. Wave Optics, Chapter 10, is no exception—examiners use MCQs to test whether you *understand* why light behaves in specific ways, not just memorise formulas. A typical CBSE Class 9 Physics paper allocates 25–30% weightage to Chapter 10, with at least 40% of that in MCQ format (roughly 5–6 questions out of 12–15 total). MCQs also appear in board pre-boards, school unit tests, and competitive entrance exams like JEE. The advantage: if you master MCQs here, you've built a strong conceptual foundation that transfers to long-answer and problem-solving questions. Wave Optics MCQs often test your ability to distinguish between reflection/refraction (ray optics) and diffraction/interference (wave optics), or to correctly apply Huygens' principle to predict wavefront direction. The new pattern rewards precision—one wrong tick loses marks without partial credit, so strategic reading and elimination are crucial.

**Question 1:** Which scientist proposed the principle that every point on a wavefront acts as a source of secondary wavelets? (A) Snell (B) Huygens (C) Young (D) Fresnel **Answer:** (B) Huygens — Huygens' principle is the foundation of wave optics and explains diffraction and refraction. **Question 2:** What is the SI unit of wavelength? (A) Hertz (B) Metre (C) Joule (D) Newton **Answer:** (B) Metre — Wavelength (λ) is a distance; 1 metre = 10⁹ nanometres (visible light: 400–700 nm). **Question 3:** When two coherent light waves overlap and cancel each other completely, it is called: (A) Constructive interference (B) Destructive interference (C) Diffraction (D) Refraction **Answer:** (B) Destructive interference — Occurs when path difference = (2n+1)λ/2, where n = 0, 1, 2… **Question 4:** Which phenomenon best explains why we can hear a sound even if the speaker is hidden behind a wall? (A) Reflection (B) Refraction (C) Diffraction (D) Polarisation **Answer:** (C) Diffraction — Sound waves bend around obstacles; light waves do too, but wavelength differences affect the degree. **Question 5:** Polarised light is one in which: (A) Light vibrates in all directions (B) Light vibrates in only one direction perpendicular to propagation (C) Light vibrates parallel to propagation (D) Light has infinite wavelengths **Answer:** (B) Light vibrates in only one direction perpendicular to propagation — Unpolarised light vibrates in all perpendicular directions; polarisers transmit only one. **Question 6:** In Young's double-slit experiment, what is the condition for constructive interference? (A) Path difference = λ (B) Path difference = λ/2 (C) Path difference = nλ (n = 0, 1, 2…) (D) Path difference = (2n+1)λ/2 **Answer:** (C) Path difference = nλ (n = 0, 1, 2…) — Bright fringes form when waves arrive in phase. **Question 7:** The phenomenon of bending of light at the edges of an obstacle is: (A) Refraction (B) Reflection (C) Diffraction (D) Dispersion **Answer:** (C) Diffraction — Light's wave nature causes it to bend around sharp edges. **Question 8:** Which colour of visible light has the longest wavelength? (A) Violet (B) Green (C) Red (D) Blue **Answer:** (C) Red — Red ≈ 650–700 nm; violet ≈ 400–420 nm; frequency and wavelength are inversely related. **Question 9:** Huygens' principle can be used to derive: (A) Snell's law only (B) Laws of reflection only (C) Both Snell's law and laws of reflection (D) Neither law **Answer:** (C) Both Snell's law and laws of reflection — The principle treats wavefront propagation consistently for all phenomena. **Question 10:** In interference, the intensity at a point is maximum when: (A) Crests of both waves overlap (B) Crest of one overlaps trough of the other (C) Waves are incoherent (D) Waves have different frequencies **Answer:** (A) Crests of both waves overlap — Constructive interference; amplitude sums, intensity ∝ (A₁ + A₂)².

**Question 11:** In Young's double-slit experiment, if the wavelength of light is 600 nm and the distance between two consecutive bright fringes is 3 mm, what is the relationship between slit separation (d), distance to screen (D), and these values? (A) d = λD/Δx (B) Δx = λD/d (C) D = d·Δx/λ (D) λ = d·Δx/D **Answer:** (B) Δx = λD/d — Fringe width Δx = λD/d; here, 3 mm = (600 × 10⁻⁹ × D)/d, confirming this inverse proportionality to d. **Question 12:** A diffraction pattern is observed when light passes through a single slit. The central maximum is widest when: (A) Slit width is largest (B) Slit width is smallest (C) Wavelength is largest (D) Both (B) and (C) **Answer:** (D) Both (B) and (C) — Central maximum width ∝ λ/b; smaller b and larger λ broaden it (e.g., red light through narrow slit). **Question 13:** Unpolarised light of intensity I₀ passes through a polariser and then an analyser whose transmission axis is at 30° to the polariser. What is the intensity after the analyser? (A) I₀/2 (B) I₀/4 (C) 3I₀/8 (D) I₀/8 **Answer:** (C) 3I₀/8 — After polariser: I = I₀/2; after analyser: I = (I₀/2)·cos²(30°) = (I₀/2)·(√3/2)² = 3I₀/8 (Malus' law). **Question 14:** When light travels from a denser medium (glass, n = 1.5) to a rarer medium (air, n = 1), Huygens' principle predicts that the refracted ray: (A) Bends towards the normal (B) Bends away from the normal (C) Travels along the normal (D) Undergoes total internal reflection **Answer:** (B) Bends away from the normal — Secondary wavelets in air travel faster; wavefront tilts away from normal, consistent with Snell's law (n₁sinθ₁ = n₂sinθ₂). **Question 15:** Two sources of light are said to be coherent if: (A) They have the same intensity (B) They have the same frequency and constant phase difference (C) They emit light in the same direction (D) They are at the same distance from the observer **Answer:** (B) They have the same frequency and constant phase difference — Coherence is essential for stable interference patterns; incoherent sources produce no fringes. **Question 16:** In a diffraction grating, if the number of slits increases, the diffraction maxima become: (A) Broader and more intense (B) Narrower and less intense (C) Narrower and more intense (D) Unchanged **Answer:** (C) Narrower and more intense — More slits increase constructive interference at maxima (sharp, bright peaks) and stronger destructive interference between them. **Question 17:** When unpolarised light is reflected from a glass surface at Brewster's angle, the reflected light is: (A) Partially polarised (B) Completely polarised (C) Unpolarised (D) Scattered randomly **Answer:** (B) Completely polarised — At Brewster's angle (θ_B ≈ 56° for glass), reflected light contains only perpendicular vibrations; the refracted ray also becomes completely polarised. **Question 18:** In Young's double-slit experiment, if the distance from the slits to the screen is doubled while keeping slit separation fixed, the fringe width: (A) Halves (B) Doubles (C) Remains the same (D) Becomes zero **Answer:** (B) Doubles — Δx = λD/d; D increases by factor 2 → Δx doubles (fringes spread out). **Question 19:** Diffraction is more prominent for: (A) Shorter wavelengths and larger obstacles (B) Longer wavelengths and smaller obstacles (C) Shorter wavelengths and smaller obstacles (D) All wavelengths and obstacle sizes equally **Answer:** (B) Longer wavelengths and smaller obstacles — Diffraction effect ∝ λ/b; red light ≈ 700 nm diffracts more than violet ≈ 400 nm through the same slit. **Question 20:** A polaroid sheet transmits only 20% of incident unpolarised light. This polaroid is: (A) Ideal (100% transmission) (B) Absorbing due to impurities (C) Damaged or has high absorption (D) Correctly functioning at 50% efficiency **Answer:** (C) Damaged or has high absorption — An ideal polaroid transmits 50% of unpolarised light (first pass). Transmission < 50% indicates material defects or high dichroic absorption.

**Question 21 (Assertion-Reason):** **Assertion (A):** Light diffracts significantly when passing through a narrow slit but not when passing through a large door. **Reason (R):** Diffraction occurs when the wavelength of light is comparable to or larger than the size of the obstacle. (A) Both A and R are true; R is the correct explanation of A (B) Both A and R are true; R is NOT the correct explanation (C) A is true; R is false (D) A is false; R is true **Answer:** (A) Both A and R are true; R is the correct explanation of A — Light (λ ≈ 500 nm) is tiny compared to a door (metres) but comparable to slit width (micrometres), making diffraction prominent in the latter. **Question 22 (Calculation-Based):** In a single-slit diffraction pattern, the first minimum is at an angle θ = 30° from the central axis. If the slit width is 2 μm, what is the approximate wavelength of light? (Use sinθ ≈ θ for small angles, or direct substitution here.) **Condition for first minimum:** b·sinθ = λ (A) 100 nm (B) 1000 nm (C) 500 nm (D) 2000 nm **Answer:** (B) 1000 nm — λ = b·sinθ = 2 × 10⁻⁶ × sin(30°) = 2 × 10⁻⁶ × 0.5 = 1 × 10⁻⁶ m = 1000 nm (infrared, but mathematically valid). **Question 23 (Assertion-Reason):** **Assertion (A):** Huygens' principle explains both refraction and diffraction using the concept of secondary wavelets. **Reason (R):** Refraction and diffraction are fundamentally the same phenomenon occurring at different scales. (A) Both A and R are true; R is the correct explanation of A (B) Both A and R are true; R is NOT the correct explanation (C) A is true; R is false (D) Both A and R are false **Answer:** (C) A is true; R is false — Huygens' principle does unify the mathematical descriptions, but refraction (bending at interface due to speed change) and diffraction (bending around edges due to wave nature) are distinct phenomena; their underlying causes differ. **Question 24 (Application-Based):** A student observes that when light from a laser (λ = 650 nm) passes through two slits separated by 0.5 mm onto a screen 2 m away, bright fringes appear. If the student uses a different light source with λ = 325 nm (same slit separation and screen distance), how does the new fringe width compare? (A) Half of the original (B) Twice the original (C) Same as the original (D) Cannot be determined **Answer:** (A) Half of the original — Δx = λD/d. New Δx = (325/650) × original = 0.5 × original; halved wavelength → halved fringe spacing. **Question 25 (Reasoning):** **Statement:** In an interference experiment, two light waves with amplitudes A₁ = 2 units and A₂ = 3 units interfere constructively at a point. The resulting amplitude is 5 units. **Why is this statement true?** (A) Because amplitudes always add in phase (B) Because A₁ + A₂ = 2 + 3 = 5 for constructive interference (C) Because the intensity is maximum (D) None; the resulting amplitude should be √(A₁² + A₂²) **Answer:** (B) Because A₁ + A₂ = 2 + 3 = 5 for constructive interference — When two waves are in phase (Δφ = 0), amplitudes add directly; A_resultant = A₁ + A₂ (not vector sum). **Question 26 (Assertion-Reason):** **Assertion (A):** Brewster's angle for glass (n = 1.5) is approximately 56.3°, and at this angle, the reflected light is completely polarised perpendicular to the plane of incidence. **Reason (R):** At Brewster's angle, the refracted and reflected rays are perpendicular to each other, and reflected light has only the perpendicular component. (A) Both A and R are true; R is the correct explanation of A (B) Both A and R are true; R is NOT the correct explanation (C) A is true; R is false (D) A is false; R is true **Answer:** (A) Both A and R are true; R is the correct explanation of A — tan(θ_B) = n₂/n₁ = 1.5/1 → θ_B ≈ 56.3°. At this angle, reflected and refracted rays are perpendicular; the incident wave's parallel component is refracted away, leaving only perpendicular vibrations in the reflected ray. **Question 27 (Critical Analysis):** Three polaroid sheets are arranged in sequence: the first has a transmission axis at 0°, the second at 45°, and the third at 90°. Unpolarised light of intensity I₀ passes through. What is the final intensity? (A) 0 (B) I₀/8 (C) I₀/4 (D) I₀/16 **Answer:** (B) I₀/8 — After 1st: I = I₀/2 (0° axis). After 2nd: I = (I₀/2)·cos²(45°) = I₀/4. After 3rd: I = (I₀/4)·cos²(45°) = I₀/8 (Malus' law applied twice). **Question 28 (Assertion-Reason):** **Assertion (A):** Diffraction maxima in a diffraction grating are sharper and brighter than in a double-slit interference pattern. **Reason (R):** A diffraction grating has many more slits than a double-slit arrangement, leading to stronger constructive interference and steeper intensity variation near maxima. (A) Both A and R are true; R is the correct explanation of A (B) Both A and R are true; R is NOT the correct explanation (C) A is true; R is false (D) Both A and R are false **Answer:** (A) Both A and R are true; R is the correct explanation of A — Grating: ~1000 slits; double-slit: 2 slits. More sources coherently contributing → sharper maxima (width ∝ 1/N) and higher intensity (∝ N²). **Question 29 (Conceptual):** When sound waves (wavelength ≈ 0.34 m at 1000 Hz) pass through a doorway (≈ 1 m wide) and light waves (λ ≈ 500 nm) pass through the same doorway, which diffracts more noticeably, and why? (A) Sound, because its wavelength is larger relative to the doorway size (B) Light, because it has higher frequency (C) Neither; diffraction is independent of wavelength (D) Light, because it travels faster **Answer:** (A) Sound, because its wavelength is larger relative to the doorway size — Diffraction effect ≈ λ/b. For sound: 0.34/1 = 0.34; for light: 500×10⁻⁹/1 ≈ negligible. Sound diffracts noticeably; light does not. **Question 30 (Assertion-Reason):** **Assertion (A):** When observing Newton's rings, the central spot is always dark, regardless of the wavelength of light used. **Reason (R):** At the centre, the air gap between the curved surface and the flat plate is zero, causing a path difference of zero, leading to destructive interference due to a phase change of π at the glass-air interface. (A) Both A and R are true; R is the correct explanation of A (B) Both A and R are true; R is NOT the correct explanation (C) A is true; R is false (D) A is false; R is true **Answer:** (A) Both A and R are true; R is the correct explanation of A — Central path difference = 0, but reflection from the glass-air interface introduces a phase change of π (≈ λ/2 path difference equivalent), making it a destructive interference condition → dark spot.

**Trap 1: Confusing Interference with Diffraction** *Trap option:* 'Diffraction is when two waves overlap and create bright and dark bands.' *Why it's wrong:* This describes interference. Diffraction is the bending of light around edges. *Correct concept:* Diffraction produces single-slit patterns (one central bright band); interference (e.g., double-slit) produces multiple equally-spaced bright fringes. **Trap 2: Wavelength and Frequency Relationships** *Trap option:* 'If wavelength increases, frequency increases.' *Why it's wrong:* They are inversely related (c = λf). If wavelength increases, frequency decreases. *Correct concept:* In a given medium, c is constant, so λ ∝ 1/f. In free space, c ≈ 3 × 10⁸ m/s; in glass (n = 1.5), c reduces to 2 × 10⁸ m/s, but frequency stays the same—wavelength decreases. **Trap 3: Huygens' Principle as 'Just a Formula'** *Trap option:* 'Huygens' principle is used only to derive Snell's law.' *Why it's wrong:* It's far broader. *Correct concept:* Huygens' principle applies to refraction, reflection, diffraction, and polarisation. Every point on a wavefront is a secondary source; the envelope of these secondary wavelets forms the new wavefront. **Trap 4: Polarisation and 'Intensity Loss'** *Trap option:* 'When unpolarised light passes through a polariser, half is absorbed and half is transmitted as useless scattered light.' *Why it's wrong:* Transmitted light is polarised, not scattered—it's coherent and useful. *Correct concept:* Unpolarised light → 50% transmitted (polarised). If a second polariser is at angle θ, transmission = 50% × cos²θ (Malus' law). **Trap 5: Path Difference vs. Phase Difference** *Trap option:* 'Constructive interference occurs when phase difference = π.' *Why it's wrong:* This is destructive interference. *Correct concept:* Constructive: phase difference = 0, 2π, 4π… (path difference = 0, λ, 2λ…). Destructive: phase difference = π, 3π, 5π… (path difference = λ/2, 3λ/2…). **Trap 6: Diffraction Order and Slit Width** *Trap option:* 'Using a wider slit produces more diffraction orders.' *Why it's wrong:* Wider slit reduces diffraction; minima occur at angles determined by b·sinθ = nλ, so larger b → smaller angles → fewer visible orders. *Correct concept:* Narrower slits → more prominent diffraction, more orders visible on a screen. **Trap 7: Brewster's Angle Misconception** *Trap option:* 'At Brewster's angle, all light is reflected; none is refracted.' *Why it's wrong:* Most light is refracted; only reflected light is polarised. *Correct concept:* At θ_B, reflected light is completely polarised perpendicular to the plane of incidence. Refracted light is also partially polarised. No special cancellation of refracted rays occurs. **Trap 8: Young's Double-Slit and Slit Width** *Trap option:* 'In Young's experiment, the fringe width is inversely proportional to slit width (d).' *Why it's wrong:* Only partially true. *Correct concept:* Fringe width Δx = λD/d (inverse proportionality). However, if individual slits are too wide, diffraction effects smear the fringes. Fringes become visible only if each slit's diffraction pattern is broad enough to cover multiple interference fringes. **Trap 9: Coherence and Visibility** *Trap option:* 'Two sources are coherent if they are at the same distance from the screen.' *Why it's wrong:* Spatial proximity doesn't ensure coherence. *Correct concept:* Coherence requires constant phase difference and same frequency. Distance affects path difference (and thus fringe visibility), not coherence itself. Incoherent sources (e.g., two separate light bulbs) never produce stable fringes, regardless of positioning. **Trap 10: Polaroid Intensity Calculation Errors** *Trap option:* 'If two polaroids have axes at 90°, transmitted intensity = I₀/2.' *Why it's wrong:* At 90°, transmission = I₀/2 × cos²(90°) = 0 (complete blocking). *Correct concept:* Use Malus' law: I = I₀·cos²θ. At 90°, θ = 90° → cos²(90°) = 0 → no light emerges. At 45°, half of the already-polarised light passes: cos²(45°) = 0.5.

**Pre-Exam Preparation (2–3 weeks before)** 1. *Build conceptual clarity first.* Spend 50% of study time on theory (Huygens' principle, interference conditions, diffraction formulas, Brewster's angle). Use NCERT diagrams and perform mental visualisations (e.g., wavefront progression, secondary wavelets spreading from slit edges). 2. *Learn standard formulas and their domain.* Know when to apply fringe width Δx = λD/d (double-slit only, not single-slit), diffraction minimum b·sinθ = nλ, and Malus' law I = I₀·cos²θ (polarisers only). Mis-applying formulas is a common error. 3. *Solve 15–20 conceptual MCQs per day* (Easy + Medium mix) for 10 days. Track mistakes in a notebook; categorise errors (formula confusion, sign error, units, concept gap). **Exam Day Tactics (30 questions in ~45 minutes)** **Phase 1: Skim & Sort (3–5 minutes)** - Read all 30 MCQ stems (not options yet). Mark 10 'easy/confident' Qs (typical: definition-based, single-step reasoning). - Mark 12–15 'medium' Qs (require formula or 2–3 logic steps). - Mark 3–5 'hard/assertion-reason' Qs (multi-concept or trap-heavy). **Phase 2: Easy MCQs (8–10 minutes)** - Solve marked Easy Qs first. Average time: 1 minute/Q. Examples: 'Name Huygens' principle,' 'Which colour has longest wavelength?' If confident, tick and move; no second-guessing. - Finish all Easy Qs before touching Medium. - **Checkpoint:** After 10 min, you should have 10 Qs done (33% complete). **Phase 3: Medium MCQs (18–20 minutes)** - Tackle marked Medium Qs in order of increasing difficulty within this tier. - *Single-calculation Qs (e.g., 'Find fringe width given λ, D, d'):* Budget 1.5–2 minutes. Recall formula, plug values, check units, tick. - *Two-step reasoning (e.g., 'If wavelength halves, how does fringe width change?'):* Budget 2 min. Write formula relationship on paper if unsure (e.g., Δx ∝ λ). - *Trap options (e.g., 'Which is TRUE: A, B, C, D?'):* Budget 2.5 min. Eliminate 2 obviously wrong options; compare remaining two carefully using conceptual reasoning, not intuition. - **Checkpoint:** After 28 min, aim for 22–24 Qs complete. **Phase 4: Hard / Assertion-Reason MCQs (10–12 minutes)** - These demand care. Assertion-reason Qs have 4 meta-options (A and R both true + R explains; A and R both true but R doesn't explain; A true R false; A false R true). - *Strategy:* Always evaluate A independently first. Is it factually correct per NCERT? YES → move to R. NO → immediately choose (C) or (D). - If A is TRUE, read R carefully for logical causation. Use conceptual reasoning: *Does R logically cause A?* Example: 'Light diffracts because wavelength is comparable to slit size'—YES, causation is direct. Tick (A). - *Time per hard Q:* 2–2.5 min. If stuck after 2 min, mark and return in final review. - **Final Checkpoint:** 38–40 min → 28–30 Qs attempted; 2–5 min left for review. **Phase 5: Review & Blind Spots (2–5 minutes)** - Revisit any marked 'unsure' or 'skipped' Qs. - *For Qs you answered but feel doubtful:* Re-read the stem (not options); does your answer align with NCERT definitions or formulas? If YES, leave it. If NO or vague, change only if you're 90% confident in the new answer. - *For Qs skipped entirely:* If time remains, quickly assess: Is it a definition Q (guess intelligently by elimination) or a calculation Q (mark and skip if no time)? Definitions and assertion-reason Qs have higher success rates under time pressure. **Common Time-Saver Techniques** 1. *Eliminate illogical options first.* E.g., 'wavelength of light ≈ 100 nm' is wrong for visible light (400–700 nm); skip option immediately. 2. *Use dimensional analysis.* If a formula has Δx (distance), units should be mm, m, or μm—not Hz or nm². 3. *Relative reasoning for unknowns.* If you forget Brewster's angle value, recall: 'As refractive index increases, angle increases.' Glass (n = 1.5) → ~56°; diamond (n = 2.4) → ~68°. Use this to eliminate extreme options. 4. *Flag, don't dwell.* Mark a Q as 'return if time' and move on within 1 min if you're stuck. Dwelling costs time and demoralises. **Post-Exam Analysis** - After each practice test or school exam, review wrong Qs within 24 hours. Was the error conceptual, formula-based, or careless (e.g., unit conversion)? Log it. - For repeated mistake types, revisit theory and solve 3–5 similar Qs before the next test. Start a 3-day free trial at cbsetutor.ai to get real-time feedback on your Wave Optics practice MCQs and identify weak spots algorithmically.

**Huygens' Principle** - Every point on a wavefront acts as a source of secondary wavelets expanding in all directions at the wave speed. - The envelope of secondary wavelets forms the new wavefront. - Application: Derives laws of reflection, refraction, diffraction, and polarisation. **Interference (Coherent Light Sources)** - *Constructive:* Path difference = nλ (n = 0, 1, 2…); Amplitude A_net = A₁ + A₂; Intensity I_max = (A₁ + A₂)². - *Destructive:* Path difference = (2n+1)λ/2 (n = 0, 1, 2…); Amplitude A_net = |A₁ − A₂|; Intensity I_min = (A₁ − A₂)². - *Young's double-slit:* Fringe width Δx = λD/d (D = screen distance, d = slit separation). - *Condition for visibility:* Each slit's diffraction pattern must be broad enough to encompass multiple fringes. **Diffraction (Single Source, Edge/Aperture Effects)** - *Single-slit minima:* b·sinθ = nλ (n = 1, 2, 3…; b = slit width). - *Central maximum width:* ≈ 2λD/b (broader for smaller b or larger λ). - *Diffraction grating:* d·sinθ = nλ (d = grating spacing); maxima are sharp and bright (inversely proportional to number of slits). - *Key insight:* Diffraction is prominent when λ ≥ obstacle size. **Polarisation** - *Unpolarised light:* Vibrations in all perpendicular directions to propagation; intensity = I₀. - *After 1st polariser:* Intensity = I₀/2 (only one perpendicular direction transmitted). - *Malus' Law:* After polariser at angle θ to first, I = I₀·cos²θ. - *Brewster's angle:* tan(θ_B) = n₂/n₁ (air-glass: θ_B ≈ 56.3°); reflected light is completely polarised perpendicular to plane of incidence. - *Polaroid application:* Three polarisers at 0°, 45°, 90° → I_final = (I₀/2)·cos²(45°)·cos²(45°) = I₀/8. **Wavelength in Different Media** - In vacuum/air: c = 3 × 10⁸ m/s; λ_air = c/f. - In medium (refractive index n): v = c/n; λ_medium = λ_air/n (frequency f constant). - Visible light: 400 nm (violet) to 700 nm (red) in vacuum. **Common Trap Mistakes to Avoid** - Confusing path difference = nλ (constructive) with path difference = nλ/2 (generic rule). - Forgetting phase change of π (≈ λ/2 path difference) at glass-air reflection in Newton's rings or thin films. - Using diffraction formula for double-slit (not applicable; use interference fringe width). - Mis-applying Malus' law to unpolarised light directly (first pass through polariser, then use Malus' for analyser). - Assuming all diffraction or interference patterns are identical (they're not: single-slit, double-slit, and grating have distinct patterns).