Class 9 Physics Chapter 10 Wave Optics Important Questions with Answers

Wave Optics (Chapter 10) carries 8–12% weightage in CBSE Class 9 Science papers. The 2024-25 rationalized syllabus removed purely computational portions but retained conceptual depth: students must understand *why* waves bend at edges (diffraction), *how* two waves combine (interference), and *what* polarisation reveals about light's nature. Examiners increasingly ask application-based questions—for example, "Why can we hear sound around corners but not see light?" or "How does a polarising filter reduce glare in sunglasses?" These demands mean rote memorisation fails; you need to visualize Huygens' wavefronts, predict interference patterns, and connect optics to everyday life. This guide structures 18 questions (1-, 2-, 3-, and 5-mark types) in the *exact* board format, with solutions that show reasoning, not just answers. Practising these drills ensures you score consistently, whether the paper emphasizes definition-recall or conceptual problem-solving.

**Q1. Huygens' principle states that:** (A) Light travels in straight lines only. (B) Each point on a wavefront acts as a source of secondary wavelets. (C) All waves must be electromagnetic. (D) Diffraction occurs only in water. **Answer: (B)** — Huygens' principle, formulated by Christiaan Huygens (1690), proposes that every point on a wavefront becomes a centre of secondary disturbance (secondary wavelet) that propagates in all directions at the wave's speed. This explains reflection, refraction, and diffraction geometrically. **Q2. Which of the following is NOT a property of light showing wave nature?** (A) Diffraction around obstacles (B) Interference in thin films (C) Photoelectric effect (D) Polarisation by Polaroid sheets **Answer: (C)** — Photoelectric effect (emission of electrons when light hits a metal) is a quantum or particle property of light, explained by Einstein's photon model, not by wave behaviour. **Q3. When two coherent light waves interfere constructively, the resultant amplitude is:** (A) A₁ − A₂ (B) A₁ + A₂ (C) √(A₁² + A₂²) (D) |A₁ − A₂| **Answer: (B)** — In constructive interference, two waves of amplitudes A₁ and A₂ are in phase (path difference = nλ, n = 0, 1, 2…). Their displacements add: A_resultant = A₁ + A₂. Intensity is proportional to amplitude squared: I ∝ (A₁ + A₂)². **Q4. Diffraction is most easily observed when:** (A) Wavelength is much larger than obstacle size (B) Wavelength equals obstacle size (C) Wavelength is much smaller than obstacle size (D) Light intensity is very high **Answer: (B)** — Diffraction effects are pronounced when the wavelength λ and obstacle/slit width b are comparable (λ ≈ b). If λ << b (like visible light, λ ≈ 500 nm, past a 1 cm slit), shadows are sharp; if λ >> b, diffraction spreads widely. **Q5. A polarising filter reduces intensity to 50% of incident unpolarised light. This obeys:** (A) Snell's Law (B) Malus's Law (C) Faraday's Law (D) Coulomb's Law **Answer: (B)** — Malus's Law: I = I₀ cos²θ, where I₀ is incident intensity and θ is angle between transmission axes. For unpolarised light through one polariser, θ = 0° and I = I₀ × (1/2) = 0.5I₀ (the initial polariser reduces unpolarised intensity by 50%; further polarisers follow cos²θ).

**Q1. State Huygens' Principle and explain how it accounts for refraction of light.** **Answer (2 marks):** Huygens' Principle: Every point on a wavefront at any instant acts as a source of secondary spherical wavelets that propagate forward at the wave speed. The new wavefront is the envelope of all secondary wavelets. Refraction via Huygens: When a plane wave enters a denser medium (e.g., air to glass) at an angle: • The portion of the wavefront entering the denser medium travels slower (v₂ < v₁). • Meanwhile, the portion still in air travels at v₁ (faster). • This causes the wavefront to tilt, bending the ray toward the normal (θ₂ < θ₁). • Mathematically: sin θ₁ / sin θ₂ = n₂/n₁ (Snell's Law). --- **Q2. What is the condition for constructive and destructive interference? Give one example of each in nature.** **Answer (2 marks):** *Constructive Interference:* Path difference = nλ (n = 0, 1, 2…), phase difference = 2nπ. Amplitudes add: A = A₁ + A₂. Intensity I ∝ (A₁ + A₂)². *Destructive Interference:* Path difference = (n + ½)λ (n = 0, 1, 2…), phase difference = (2n + 1)π. Amplitudes cancel (if A₁ = A₂): A = 0, I = 0. *Example (Constructive):* Bright fringes in Young's Double Slit Experiment (YDSE) or reflections from a soap bubble's coloured rings at certain thicknesses. *Example (Destructive):* Dark fringes in YDSE; noise-cancelling headphones use destructive interference of sound waves. --- **Q3. Distinguish between diffraction and refraction.** **Answer (2 marks):** | Property | Diffraction | Refraction | |----------|-------------|----------| | **Cause** | Bending of wave around obstacles/edges | Change in wave speed at boundary | | **Law** | Follows Huygens' Principle; no single law | Follows Snell's Law (sin θ₁ / sin θ₂ = n₂/n₁) | | **Requirement** | Obstacle size comparable to wavelength | Two media with different speeds | | **Angle change** | Depends on diffraction pattern | θ₂ ≠ θ₁ even for oblique incidence | | **Example** | Light bending around a door edge; sound heard around corners | Light bending at water surface; pencil appearing bent in water | --- **Q4. What is polarisation? Name two methods of producing polarised light.** **Answer (2 marks):** *Polarisation:* The process of restricting vibrations of electric field of light to one plane perpendicular to the direction of propagation. Unpolarised light (e.g., from sun/bulb) has random vibrations in all perpendicular directions; polarised light vibrates in only one plane. *Two methods:* 1. **Polarising Filters (Polaroids):** Synthetic polymer sheets with aligned molecules. Only light oscillating parallel to alignment passes; perpendicular component is absorbed. 2. **Reflection (Brewster's Angle):** When light reflects off a dielectric surface (e.g., water, glass) at a specific angle θ_B (tan θ_B = n₂/n₁), reflected light is partially polarised. For air-water (n ≈ 1.33), θ_B ≈ 53°. --- **Q5. Two coherent sources produce bright fringes. If path difference at a point is 3λ, is it bright or dark? Justify.** **Answer (2 marks):** *The fringe is bright.* *Justification:* For constructive interference, path difference Δ = nλ, where n = 0, 1, 2, 3…. Here, Δ = 3λ matches n = 3, so the condition is satisfied. The two waves are in phase (phase difference = 2nπ = 6π), so their amplitudes add constructively: A_result = A₁ + A₂. Intensity is maximum (bright fringe).

**Q1. Explain Young's Double Slit Experiment (YDSE) using Huygens' Principle. Why are bright and dark fringes observed?** **Answer (3 marks):** *Setup:* Coherent light (single wavelength) from a source passes through two narrow slits S₁ and S₂ separated by distance d. Each slit acts as a source of secondary wavelets (Huygens' Principle). These wavelets spread out and overlap on a screen at distance D. *Why Bright Fringes:* At any point P on the screen, the path difference is Δ = S₂P − S₁P. If Δ = nλ (n = 0, 1, 2…), waves from both slits arrive in phase. Their amplitudes superpose constructively: A = A₁ + A₂. Central fringe (n = 0, Δ = 0) is brightest. Other bright fringes occur at nλ intervals with slightly lower intensity. *Why Dark Fringes:* If Δ = (n + ½)λ (n = 0, 1, 2…), waves are exactly out of phase by π radians. One crest meets a trough; destructive interference occurs. If both slits emit equal amplitude, total amplitude = 0, and intensity I = 0 (black/dark fringe). *Fringe Width:* β = λD/d. Larger wavelength or larger D/d ratio → wider fringes; smaller slit separation d → wider fringes. --- **Q2. A soap bubble appears to have rainbow colours. Explain using wave interference, including the condition for a dark fringe at thickness t.** **Answer (3 marks):** *Mechanism:* A soap bubble is a thin film of soapy water (refractive index n ≈ 1.33) bounded by air (n = 1). Light reflects from both the top surface (air–soap) and bottom surface (soap–air). These two reflected waves interfere. *Path Difference & Phase Change:* For light incident nearly perpendicular, the optical path difference between the two reflected rays is 2nt, where t is film thickness. Crucially: At the air–soap interface (denser medium), reflected light undergoes a phase change of π (equivalent to path difference λ/2). At the soap–air interface (rarer medium), no phase change occurs. Net effective path difference = 2nt − λ/2 (accounting for the phase change). *Condition for Dark Fringe (Destructive Interference):* 2nt − λ/2 = (m + ½)λ, where m = 0, 1, 2… ⟹ 2nt = (m + 1)λ Or: t = (m + 1)λ / (2n), m = 0, 1, 2… Example: For λ = 600 nm (red), n = 1.33, m = 0: t = 1 × 600 / (2 × 1.33) ≈ 226 nm. *Rainbow Effect:* Different wavelengths (colours) satisfy bright/dark conditions at different thicknesses. As bubble thickness varies spatially and shrinks over time, different colours dominate locally, creating iridescent rainbow patterns. --- **Q3. Compare diffraction patterns of single slit and double slit. Why is the double-slit pattern finer (narrower fringes)?** **Answer (3 marks):** | Feature | Single Slit | Double Slit | |---------|------------|------------| | **Envelope** | Single broad diffraction pattern | Narrow diffraction envelope | | **Interior fringes** | Absent (smooth intensity decay) | Many sharp bright & dark fringes | | **Cause** | Wavelets from one aperture interfere with themselves | Wavelets from two coherent apertures + self-diffraction | | **Fringe width** | — | β = λD/d (narrow if d is small) | *Why Double-Slit Fringes Are Finer:* In YDSE, the fringe width β = λD/d depends inversely on slit separation d. Double slits are typically separated by d ≈ 0.5–1 mm, much larger than a single slit width (≈ 0.1–0.3 mm). Thus, β_double << β_single if same wavelength and screen distance are used. Physically: Coherent sources S₁ and S₂ separated by d produce interference fringes spaced by λD/d. Increasing d (moving slits apart) crowds fringes together, making them narrower and more numerous on the screen. --- **Q4. A polariser and analyser are placed one after another. Unpolarised light of intensity I₀ is incident on the polariser. What is the intensity after passing through both? Derive using Malus's Law.** **Answer (3 marks):** *Step 1 – After Polariser:* Unpolarised light of intensity I₀ passes through the polariser. The polariser transmits only the component of oscillation parallel to its transmission axis. For unpolarised light, the average intensity is halved: I₁ = I₀ / 2 The transmitted light is now linearly polarised. *Step 2 – After Analyser (Malus's Law):* Malus's Law: When polarised light of intensity I passes through an analyser whose transmission axis makes angle θ with the light's polarisation direction: I_transmitted = I cos²θ Here, I = I₁ = I₀/2, so: I₂ = (I₀/2) cos²θ *Special Cases:* • If analyser is parallel to polariser (θ = 0°): I₂ = I₀/2 (maximum) • If analyser is perpendicular to polariser (θ = 90°): I₂ = 0 (complete blockage) • If analyser is at 45° to polariser (θ = 45°): I₂ = (I₀/2) × (1/2) = I₀/4 *Application:* Sunglasses use a polariser layer to cut reflected glare; a second layer (analyser) at 45° can further reduce intensity to half.

**Q1. Derive the condition for bright and dark fringes in Young's Double Slit Experiment. Hence, calculate the fringe width and number of fringes on a screen for the following data: λ = 500 nm, d = 0.5 mm, D = 1 m, slit width = negligible.** **Full Solution (5 marks):** **Part A – Theory:** Consider two coherent slits S₁ and S₂ separated by distance d, illuminated by monochromatic light of wavelength λ. A screen is placed at distance D from the slits. For a point P on the screen at distance y from the central axis: • Path from S₁ to P: r₁ ≈ D + y²/(2D) (using paraxial approximation for small y) • Path from S₂ to P: r₂ ≈ D − y²/(2D) More precisely (for small y and small angle): Path difference, Δ = r₂ − r₁ ≈ (d·y) / D **Condition for Bright Fringe (Constructive Interference):** Δ = n·λ, where n = 0, 1, 2, 3… ⟹ (d·y_n) / D = n·λ ⟹ **y_n = (n·λ·D) / d** ← Position of nth bright fringe from centre **Condition for Dark Fringe (Destructive Interference):** Δ = (n + ½)·λ, where n = 0, 1, 2, 3… ⟹ (d·y'_n) / D = (n + ½)·λ ⟹ **y'_n = [(2n+1)·λ·D] / (2d)** ← Position of nth dark fringe from centre **Fringe Width (distance between consecutive bright or dark fringes):** β = y_{n+1} − y_n = [(n+1)·λ·D/d] − [n·λ·D/d] ⟹ **β = λ·D / d** --- **Part B – Numerical Calculation:** Given: - λ = 500 nm = 500 × 10⁻⁹ m = 5 × 10⁻⁷ m - d = 0.5 mm = 0.5 × 10⁻³ m = 5 × 10⁻⁴ m - D = 1 m - Screen width (assume) = ±0.05 m (5 cm on each side of centre, typical lab setup) **Fringe Width:** β = (5 × 10⁻⁷ × 1) / (5 × 10⁻⁴) = (5 × 10⁻⁷) / (5 × 10⁻⁴) = 10⁻³ m = **1 mm** So each bright fringe is separated by 1 mm. **Number of Fringes on Screen:** If screen width is 0.05 m (±2.5 cm): y_max = 0.025 m For the maximum fringe order: n_max = (d × y_max) / (λ × D) = (5 × 10⁻⁴ × 0.025) / (5 × 10⁻⁷ × 1) = (1.25 × 10⁻⁵) / (5 × 10⁻⁷) = 25 Total bright fringes ≈ 2 × 25 + 1 = **51 bright fringes** (counting central bright fringe and 25 on each side). Total dark fringes ≈ 2 × 25 = **50 dark fringes** (25 between central and outer bright fringe, 25 on each side). Total visible fringes ≈ **~100 fringes** on a 5 cm wide screen. --- **Q2. A light wave passes through a single slit of width a. Derive the condition for the first dark fringe in single-slit diffraction pattern. Explain why single-slit diffraction limits the resolving power of optical instruments.** **Full Solution (5 marks):** **Part A – Derivation:** Consider a plane wave incident normally on a single slit of width a. The slit acts as a source of secondary wavelets (Huygens' Principle). Each point inside the slit produces a wavelet; we examine the resultant intensity on a screen at distance D >> a, at point P making angle θ with the central axis. **Method of Phasor Addition (Graphical Approach):** Divide the slit into N infinitesimal elements. The phase difference between wavelets from the top and bottom edges is: δ = (2π / λ) × a sin θ When all wavelets are summed as phasors, they form an arc. The resultant amplitude at point P is: A = [A₀ sin(δ/2)] / (δ/2), where A₀ is amplitude if no diffraction. For **first dark fringe**, the phasors form a complete circle (or come back on themselves after 2π rotation): δ = 2π ⟹ (2π / λ) × a sin θ = 2π ⟹ **a sin θ = λ** ← Condition for 1st dark fringe For small angles: sin θ ≈ θ ≈ y/D (where y is distance from centre on screen): ⟹ **y₁ = λD / a** ← Position of 1st dark fringe More generally, the nth dark fringe occurs at: **a sin θ_n = n·λ** (n = 1, 2, 3…) --- **Part B – Resolving Power Limitation:** The Rayleigh Criterion states that two point sources are just resolved (barely distinguishable) if the central maximum of one diffraction pattern coincides with the first dark minimum of the other. Minimum angular separation for resolution: θ_min = λ / a (from a sin θ₁ = λ) Implications: 1. **Smaller wavelength → Better resolution:** Ultraviolet or X-rays resolve finer details than visible light. 2. **Larger aperture → Better resolution:** Telescopes with larger mirrors (larger a) resolve distant stars better. A 1 m telescope resolves two stars separated by θ ≈ 500 nm / 1 m ≈ 5 × 10⁻⁷ radians ≈ 0.1 arcseconds. 3. **Trade-off with brightness:** Larger apertures also collect more light, improving signal-to-noise. *Example:* A microscope with objective diameter a = 2 mm examining visible light (λ = 500 nm) has: θ_min ≈ 500 × 10⁻⁹ / (2 × 10⁻³) ≈ 2.5 × 10⁻⁴ radians ≈ 50 micrometers minimum resolvable distance. To see features smaller than 50 μm, one must use shorter wavelengths (electron microscopy, λ ≈ 0.1 nm) or larger apertures (impossible for visible light due to diffraction). --- **Q3. Explain the phenomenon of polarisation. Derive Brewster's angle and show that at Brewster's angle, reflected and refracted rays are perpendicular.** **Full Solution (5 marks):** **Part A – Polarisation:** Polarisation is the phenomenon of restricting the electric field vector of light (or any transverse wave) to oscillate in only one plane perpendicular to the direction of propagation. *Unpolarised Light:* Electric field oscillates in all directions perpendicular to the propagation direction (e.g., sunlight, bulb light) with random orientation and rapid changes. *Linearly Polarised Light:* Electric field oscillates in a fixed plane (e.g., light from a Polaroid filter, or from Brewster's Law reflection). *Physical Basis:* Light is a transverse electromagnetic wave. Unpolarised light contains equal superposition of oscillations in all perpendicular directions. Polarising filters (Polaroids) selectively transmit oscillations parallel to their transmission axis, absorbing perpendicular components. --- **Part B – Brewster's Angle Derivation:** When unpolarised light reflects off a dielectric surface (e.g., water, glass), the reflected light is partially polarised. At a specific angle (Brewster's angle θ_B), the reflected light is *completely* polarised (perpendicular to the plane of incidence). *Physical Reason:* At Brewster's angle, the electric field component of light oscillating in the plane of incidence (parallel to the interface) is not reflected because the dipole oscillations induced in the dielectric are not radiating perpendicular to their oscillation direction. **Mathematical Derivation (Fresnel Equations):** The reflectance for light polarised in the plane of incidence (p-polarised) is: R_p = [(n₂ cos θ_i − n₁ cos θ_r) / (n₂ cos θ_i + n₁ cos θ_r)]² For R_p = 0 (no reflection): n₂ cos θ_i = n₁ cos θ_r Using Snell's Law: n₁ sin θ_i = n₂ sin θ_r ⟹ sin θ_r = (n₁/n₂) sin θ_i cos θ_r = √[1 − sin² θ_r] = √[1 − (n₁/n₂)² sin² θ_i] Substituting into the condition: n₂ cos θ_i = n₁ √[1 − (n₁/n₂)² sin² θ_i] Squaring both sides: n₂² cos² θ_i = n₁² [1 − (n₁/n₂)² sin² θ_i] = n₁² − n₁⁴/n₂² sin² θ_i Using cos² θ_i = 1 − sin² θ_i: n₂² (1 − sin² θ_i) = n₁² − n₁⁴/n₂² sin² θ_i n₂² − n₂² sin² θ_i = n₁² − n₁⁴/n₂² sin² θ_i n₂² − n₁² = n₂² sin² θ_i − n₁⁴/n₂² sin² θ_i n₂² − n₁² = sin² θ_i (n₂² − n₁⁴/n₂²) n₂² − n₁² = sin² θ_i × (n₂⁴ − n₁⁴) / n₂² For air-glass (n₁ = 1, n₂ = n): n² − 1 = sin² θ_B × (n⁴ − 1) / n² = sin² θ_B × (n² − 1)(n² + 1) / n² Dividing by (n² − 1): 1 = sin² θ_B × (n² + 1) / n² sin² θ_B = n² / (n² + 1) cos² θ_B = 1 / (n² + 1) tan θ_B = sin θ_B / cos θ_B = √[n² / (n² + 1)] / √[1 / (n² + 1)] **⟹ tan θ_B = n** (for air-glass) More generally: **tan θ_B = n₂ / n₁** *Example:* For air (n₁ = 1) to water (n₂ ≈ 1.33): θ_B = arctan(1.33) ≈ 53° --- **Part C – Perpendicularity of Reflected and Refracted Rays at Brewster's Angle:** At Brewster's angle: - Incident angle: θ_i = θ_B - Reflected angle: θ_r = θ_B (law of reflection) - Refracted angle: θ_t (from Snell's Law) From Snell's Law at Brewster's angle: n₁ sin θ_B = n₂ sin θ_t n₁ sin θ_B = n₂ sin θ_t Using tan θ_B = n₂/n₁: sin θ_B = (n₂/n₁) / √[1 + (n₂/n₁)²] = n₂ / √(n₁² + n₂²) n₁ × [n₂ / √(n₁² + n₂²)] = n₂ sin θ_t sin θ_t = n₁ / √(n₁² + n₂²) cos θ_t = n₂ / √(n₁² + n₂²) Angle between reflected and refracted rays in the plane of incidence: α = (90° − θ_B) + (90° − θ_t) = 180° − θ_B − θ_t We need to show α = 90°: θ_B + θ_t = 90° Proof: tan θ_B = n₂/n₁ and sin θ_t = n₁/√(n₁² + n₂²) tan θ_B = (tan θ_B) / 1 = (n₂/n₁) = (1 / (n₁/n₂)) = cot θ_t ⟹ θ_B = 90° − θ_t ⟹ **θ_B + θ_t = 90°** ✓ Therefore, the reflected and refracted rays are perpendicular at Brewster's angle. This property is used in laboratory lasers and optical systems to efficiently produce polarised light.

**Case Study: Why Polarised Sunglasses Reduce Glare Better Than Tinted Sunglasses** Ananya is buying sunglasses at an optical store. The shopkeeper offers two options: (A) Tinted sunglasses: Dark-coloured lenses that reduce overall light intensity by 70% uniformly across all directions. (B) Polarised sunglasses: Contain a Polaroid layer with transmission axis at 45° to horizontal; they reduce intensity by about 50% overall. Ananya observes that when wearing (B) while looking at the water surface (which reflects sunlight at Brewster's angle ≈ 53° for air-water), the glare is nearly eliminated, while (A) still appears very bright. **Questions:** **Part 1 (2 marks): Explain why reflected light from water is partially polarised. Which component of the electric field is most efficiently reflected?** **Answer:** At Brewster's angle (≈ 53° for air-water interface), the reflected light contains predominantly the component of the electric field perpendicular to the plane of incidence (s-polarised). The component in the plane of incidence (p-polarised) is almost completely transmitted into the water, not reflected. Physically, dipoles in the water vibrate along the direction of the incident electric field. When that direction is in the plane of incidence (p-polarised at Brewster's angle), the induced dipoles oscillate along the refracted direction, which is perpendicular to the reflected ray direction; dipoles do not radiate in the direction of their oscillation, so reflection is zero. --- **Part 2 (2 marks): Water-reflected light hitting Ananya's eyes is nearly 100% s-polarised (perpendicular to the plane of incidence, i.e., horizontal). If Ananya wears polarised sunglasses with a vertical transmission axis, calculate the intensity of glare reaching her eyes.** **Answer:** Glare from water: horizontally polarised light of intensity I₀. Polarised lens transmission axis: vertical (perpendicular to glare polarisation). Angle between glare polarisation (horizontal) and lens axis (vertical): θ = 90°. Using Malus's Law: I_transmitted = I₀ cos²(90°) = I₀ × 0 = **0** (no glare passes). The glare is completely blocked. --- **Part 3 (1 mark): Why do tinted sunglasses (option A) fail to reduce water glare as effectively?** **Answer:** Tinted lenses reduce all wavelengths and polarisations equally. They absorb 70% of any incident light, including unpolarised skylight and reflected glare. Since the reflected glare from water is bright (high intensity before filtering), even 70% reduction still allows 30% of intense glare to pass through, causing discomfort. Polarised lenses, however, selectively block the specific polarisation direction of the glare, reducing it to nearly zero, regardless of initial brightness. --- **Part 4 (Synthesis, 1 mark): Ananya notices that when she rotates the polarised sunglasses by 45° (tilting her head), the water glare becomes visible again. Explain this observation using Malus's Law.** **Answer:** When the sunglasses are rotated 45°, the transmission axis is now at 45° to the horizontal (the glare polarisation direction). Angle θ = 45°. Using Malus's Law: I_transmitted = I₀ cos²(45°) = I₀ × (1/√2)² = I₀/2 = **0.5 I₀**. Half the glare intensity now passes through, making water reflections visible again. This practical observation confirms that the efficacy of polarised filters depends on the angle between the incident polarisation and the transmission axis.

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