Original practice sets for Electromagnetic Waves and Wave Optics are being prepared. For now, use the chapter notes for concept mastery and revision.
Start with short chapter-wise drills or take the full chapter bank in one run. These questions are original and aligned to JEE-style concept checks and numerical thinking.
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1. Electromagnetic waves are:
Explanation: EM waves are transverse (E and B fields oscillate perpendicular to propagation and to each other) and require no medium — they travel through vacuum at c = 3×10⁸ m/s.
2. Speed of electromagnetic waves in vacuum is:
Explanation: c = 1/√(μ₀ε₀) ≈ 3×10⁸ m/s for all EM waves (radio, microwave, infrared, visible, UV, X-ray, gamma) in vacuum. Frequency and wavelength vary; c = fλ remains constant.
3. For EM waves in vacuum, the relation between speed c, wavelength λ, and frequency f is:
Explanation: c = fλ. All EM waves travel at c in vacuum, but different types have different frequencies and wavelengths: radio waves (long λ, low f) to gamma rays (short λ, high f).
4. In order of increasing frequency, the correct sequence of EM waves is:
Explanation: Frequency increases: Radio (kHz–GHz) 10¹⁹ Hz). Wavelength decreases in the same order.
5. In an electromagnetic wave, the ratio E/B equals:
Explanation: E = cB. The amplitudes of the electric and magnetic fields are related by E₀/B₀ = c = 1/√(μ₀ε₀). This comes from Maxwell's equations for plane waves.
6. Energy density of an EM wave is equally shared between:
Explanation: Average energy density u = ε₀E²(rms) = B²(rms)/μ₀. Both are equal: u_E = u_B. Total u = ε₀E²(rms) = B²(rms)/μ₀.
7. The Poynting vector S = (1/μ₀)(E × B) represents:
Explanation: The Poynting vector S = (E × B)/μ₀ gives the direction and rate of electromagnetic energy transport per unit area (W/m²). Its time average is the intensity I = c ε₀ E₀²/2.
8. Gamma rays are produced by:
Explanation: Gamma rays originate from nuclear energy transitions (e.g., after alpha or beta decay) and have the highest frequency (lowest wavelength) in the EM spectrum — typically >10¹⁹ Hz.
9. X-rays are produced when:
Explanation: X-rays (λ: 0.01–10 nm) arise from Bremsstrahlung (braking radiation when fast electrons decelerate in a target) or characteristic X-rays from inner electron shell transitions in heavy atoms.
10. Microwaves are used in radar and cooking because:
Explanation: Radar: microwave pulses reflect off objects allowing distance and speed measurement. Microwave ovens: 2.45 GHz microwaves are absorbed by water molecules (resonance), heating food from inside out.
11. Ultraviolet radiation is harmful to humans primarily because:
Explanation: UV-B and UV-C radiation has enough energy per photon to break DNA covalent bonds, causing mutations and skin cancer. UV-A also contributes to skin damage. The ozone layer absorbs most UV-C.
12. Which application uses infrared radiation?
Explanation: Infrared (IR) cameras detect heat emitted by objects (night vision, thermal imaging). TV remotes use near-IR (≈950 nm LED) to send signals. IR is also used in fiber optic communications.
13. The displacement current between the plates of a charging capacitor produces:
Explanation: Maxwell's displacement current id = ε₀(dΦE/dt) acts like a real current and produces a magnetic field exactly as conduction current does. This ensures Ampere's law is consistent and is the basis for EM wave propagation.
14. Radiation pressure P on a perfectly absorbing surface for an EM wave of intensity I is:
Explanation: For total absorption: P = I/c. For perfect reflection: P = 2I/c (momentum is reversed). Radiation pressure is the basis for the concept of solar sails in space propulsion.
15. The fact that light can be polarised proves that it is:
Explanation: Polarisation can only occur for transverse waves (oscillations perpendicular to propagation). Longitudinal waves (like sound) cannot be polarised because oscillations are along the propagation direction.
16. Maxwell derived the speed of EM waves in vacuum as c = 1/√(μ₀ε₀). Numerically this gives:
Explanation: 1/√(μ₀ε₀) = 1/√(4π×10⁻⁷ × 8.854×10⁻¹²) ≈ 3×10⁸ m/s. This agreed with the measured speed of light and led Maxwell to conclude that light itself is an electromagnetic wave — one of the greatest unifications in physics.
17. EM waves are produced by:
Explanation: Accelerating charges radiate electromagnetic energy. An oscillating electric dipole (alternating current in an antenna) creates oscillating E and B fields that propagate outward as EM waves. Constant-velocity charges do not radiate.
18. The visible spectrum spans approximately:
Explanation: Visible light: violet (~400 nm) to red (~700 nm). Memorise VIBGYOR: Violet, Indigo, Blue, Green, Yellow, Orange, Red — increasing wavelength. Below 400 nm is UV; above 700 nm is IR.
19. The intensity of EM radiation from a point source varies with distance r as:
Explanation: Energy spreads over a sphere of area 4πr², so intensity I = P/(4πr²) ∝ 1/r². This is the inverse square law for radiation from a point source.
20. Radio waves are produced by:
Explanation: Radio waves (λ from mm to km) are produced by oscillating electrical circuits with frequencies from MHz to GHz and broadcast through antennae. AM radio: 0.5–1.7 MHz; FM radio: 88–108 MHz; WiFi: 2.4 GHz.
21. For a stable and observable interference pattern, the two sources must be:
Explanation: Incoherent sources (like two independent lamps) have rapidly varying phase differences, so the interference pattern averages out to uniform illumination. Coherent sources (derived from the same source) maintain a fixed phase relationship.
22. Fringe width β in Young's double-slit experiment (slit separation d, screen distance D, wavelength λ) is:
Explanation: β = λD/d. Fringe width increases with larger D (screen farther away) and larger λ (longer wavelength), and decreases with larger d (slits closer together).
23. Constructive interference occurs at a point when path difference Δ equals:
Explanation: When path difference = nλ (whole number of wavelengths), crest meets crest → constructive interference → bright fringe. Phase difference = 2nπ.
24. Destructive interference occurs when path difference Δ equals:
Explanation: When path difference = (2n+1)λ/2 = odd multiple of λ/2, crest meets trough → destructive interference → dark fringe. Phase difference = (2n+1)π.
25. The central maximum in YDSE is located where:
Explanation: At the central fringe (n = 0), path difference = 0 → both sources reach in phase → maximum constructive interference → brightest fringe at the centre.
26. If two coherent sources have intensities I₀ each, the maximum intensity at a bright fringe is:
Explanation: Amplitude adds: A_max = A₁ + A₂ = 2A (when amplitudes are equal). Intensity ∝ A²: I_max = (2A)² = 4A² = 4I₀. Minimum intensity = (A₁ − A₂)² = 0 (for equal amplitudes).
27. In YDSE with white light, the central maximum is:
Explanation: At the centre (path difference = 0), all wavelengths interfere constructively → all colours combine → white central maximum. Adjacent fringes are coloured because different wavelengths have different fringe spacings.
28. YDSE is performed in water (refractive index 4/3). Fringe width compared to air is:
Explanation: In water, λ_water = λ_air/n. Fringe width β = λD/d → β_water = (λ/n)D/d = β_air/n = (3/4)β_air.
29. In YDSE, a glass slab of thickness t and refractive index n is placed over one slit. The central maximum shifts toward:
Explanation: The slab increases the optical path length for that slit by (n−1)t. The central maximum (zero path difference) shifts toward the slab side to compensate for the extra optical path. Shift = (n−1)t × D/d fringe widths.
30. In practice, coherent light sources are obtained by:
Explanation: Two independent sources (even two identical lasers) are not coherent due to random phase jumps. Coherent beams must be derived from the same wavefront — as in YDSE (same slit source), Lloyd's mirror, or interferometers (Michelson, Fabry-Perot).
31. In an interference pattern, energy at dark fringes goes to:
Explanation: Interference does not create or destroy energy — it redistributes it. The energy 'missing' from dark fringes appears as extra energy at the bright fringes. Average intensity = sum of individual intensities, conserving energy.
32. In YDSE with slit separation d = 1 mm and screen at D = 1 m, the 3rd bright fringe is at distance y from centre for λ = 500 nm. y equals:
Explanation: y_n = nλD/d = 3 × 500×10⁻⁹ × 1/(10⁻³) = 3 × 500×10⁻⁶ = 1500×10⁻⁶ = 1.5 mm.
33. Path difference of λ/3 corresponds to a phase difference of:
Explanation: Phase difference φ = (2π/λ) × path difference = (2π/λ) × (λ/3) = 2π/3 radians.
34. Fresnel biprism is used to obtain two coherent sources because:
Explanation: The prism's two halves refract the incoming wavefront from a single slit in slightly different directions. These two refracted beams overlap and interfere, as if coming from two virtual coherent sources.
35. In YDSE, the maximum order of the bright fringe that can be observed depends on:
Explanation: Path difference = d sinθ. Maximum sinθ = 1 → maximum path difference = d. Maximum order: n_max = d/λ. For example, if d = 0.5 mm and λ = 500 nm: n_max = 0.5×10⁻³/500×10⁻⁹ = 1000.
36. Two coherent waves of intensities I₁ and I₂ with phase difference φ interfere. Resultant intensity is:
Explanation: I = I₁ + I₂ + 2√(I₁I₂) cosφ. For I₁ = I₂ = I₀: I = 2I₀(1 + cosφ) = 4I₀cos²(φ/2). Maximum (φ = 0) = 4I₀; minimum (φ = π) = 0.
37. Diffraction of light is the phenomenon of:
Explanation: Diffraction occurs when light encounters an obstacle or aperture whose size is comparable to the wavelength. The wave bends into the 'shadow' region — a purely wave phenomenon.
38. In single-slit diffraction (slit width a), the first minimum occurs when:
Explanation: For single-slit diffraction, minima occur at a sinθ = mλ where m = ±1, ±2, ... The central maximum is widest (between the ±1st minima). Contrast with double-slit maxima at d sinθ = nλ.
39. Width of the central maximum in single-slit diffraction increases when:
Explanation: Angular half-width of central max = λ/a. Narrower slit → larger diffraction → wider central maximum. This is why a pinhole gives a very spread diffraction pattern.
40. A beam of plane polarised light (intensity I₀) passes through an analyser at angle θ to the polariser axis. Transmitted intensity is:
Explanation: Malus's law: I = I₀ cos²θ. At θ = 0° (analyser parallel to polariser): I = I₀ (full transmission). At θ = 90°: I = 0 (complete extinction).
41. At Brewster's angle θ_B, reflected light is completely polarised. tan θ_B equals:
Explanation: Brewster's law: tan θ_B = n. At this angle, the reflected and refracted rays are perpendicular. The reflected ray contains only the polarisation component perpendicular to the plane of incidence.
42. When unpolarised light passes through a single polariser, the transmitted intensity is:
Explanation: Unpolarised light has all polarisation directions equally. On average, cos²θ averaged over all θ = 1/2. So I_transmitted = I₀/2. The polariser creates plane-polarised light at half the original intensity.
43. Diffraction grating condition for maxima (grating spacing d, angle θ, wavelength λ) is:
Explanation: For a diffraction grating with N slits separated by d: constructive interference at d sinθ = nλ (n = 0, ±1, ±2, ...). More slits → sharper maxima. Used for precision wavelength measurement.
44. Rayleigh's criterion for resolution of two closely spaced point sources states they are just resolved when:
Explanation: Rayleigh limit: minimum resolvable angle θ_min = 1.22λ/D for a circular aperture of diameter D. A larger aperture (telescope mirror, eye pupil) resolves finer detail.
45. The sky appears blue because:
Explanation: Rayleigh scattering intensity ∝ 1/λ⁴. Blue light (λ ≈ 450 nm) is scattered ≈5–6× more than red light (λ ≈ 700 nm). Scattered blue light reaches us from all directions — the sky appears blue. At sunset, the direct path is longer, blue is scattered away, and red/orange dominates.
46. Brewster's angle for a glass–air interface (n = 1.5) is approximately:
Explanation: tan θ_B = n = 1.5 → θ_B = tan⁻¹(1.5) ≈ 56.3°.
47. Double refraction (birefringence) occurs in:
Explanation: Birefringent crystals have two refractive indices (ordinary and extraordinary). An incident beam splits into two rays polarised perpendicular to each other. Calcite is the classic example, used to make wave plates and polarising prisms.
48. Polaroid sunglasses reduce glare from horizontal surfaces because:
Explanation: Glare from horizontal surfaces (water, road) is predominantly horizontally polarised (near Brewster's angle). Polaroids oriented vertically block this polarisation, greatly reducing glare while transmitting vertically polarised components.
49. A single slit of width 0.2 mm is illuminated by λ = 500 nm. Angle of first diffraction minimum is:
Explanation: a sinθ = λ → sinθ = λ/a = 500×10⁻⁹/0.2×10⁻³ = 2.5×10⁻³. For small angles sinθ ≈ θ = 2.5 mrad.
50. A telescope with objective diameter 5 cm is used with λ = 550 nm. Minimum angular separation (Rayleigh) it can resolve is:
Explanation: θ = 1.22λ/D = 1.22 × 550×10⁻⁹/0.05 = 1.22 × 1.1×10⁻⁵ = 1.342×10⁻⁵ rad ≈ 13.4 μrad.
51. In YDSE (d = 0.5 mm, D = 1 m, λ = 500 nm), the number of complete fringes in 1 cm on the screen is:
Explanation: β = λD/d = 500×10⁻⁹ × 1/0.5×10⁻³ = 10⁻³ m = 1 mm. Number of fringes in 1 cm = 10 mm / 1 mm = 10.
52. Optical path length through a glass slab of thickness t and refractive index n is:
Explanation: Optical path = n × geometric path = nt. The extra optical path introduced compared to air is (n−1)t. This is the basis for thin-film interference and YDSE slab problems.
53. Light scattered by air molecules at 90° to the incident direction is:
Explanation: Scattered light at exactly 90° to the incident direction (Rayleigh scattering) is completely plane polarised. This can be verified by viewing the sky at 90° to the sun through a polaroid, which shows extinction at one orientation.
54. For a thin film of thickness t and refractive index n in air, constructive interference in reflected light occurs when (considering phase change at top surface):
Explanation: Light reflecting from the top (air-to-film) gets a π phase change (half-wavelength shift). Light from the bottom (film-to-air) does not. Net extra optical path = 2nt. For constructive interference in reflected light: 2nt = (m+½)λ (the phase difference from the extra path must cancel the π shift from reflection).
55. Compared to a double slit, a diffraction grating with many slits produces:
Explanation: All N slits of a grating separated by d produce maxima at the same angles d sinθ = nλ. But with N slits, the maxima are N times narrower (sharper) and N² times brighter. Gratings are used for high-resolution spectroscopy.
56. Intensity after two crossed polaroids (transmission axes at 90°) is:
Explanation: First polaroid transmits I₀/2 with polarisation at 0°. Second polaroid at 90° → Malus: I = (I₀/2) cos²90° = 0.
57. Huygens' principle states that every point on a wavefront acts as:
Explanation: Each point on a wavefront is a source of spherical secondary wavelets. The next wavefront is the common tangent (envelope) to all secondary wavelets. Huygens' principle explains reflection, refraction, and diffraction geometrically.
58. Unpolarised light of intensity I₀ passes through a polariser, then a polaroid at 45°, then one at 90° to the first. Final intensity is:
Explanation: Step 1: after polariser → I₀/2. Step 2: Malus at 45° → (I₀/2)cos²45° = (I₀/2)(1/2) = I₀/4. Step 3: at 90° to polariser (= 45° to middle polaroid) → (I₀/4)cos²45° = (I₀/4)(1/2) = I₀/8.
59. A sodium lamp emits two closely spaced wavelengths 589.0 nm and 589.6 nm. They produce interference fringes that disappear and reappear. Fringes disappear (minima of visibility) when:
Explanation: When n₁λ₁ = (n₁ + ½)λ₂, bright fringe of λ₁ overlaps dark fringe of λ₂ → fringes disappear (minimum visibility). This occurs at path difference Δ = λ₁λ₂/(2|λ₁−λ₂|) ≈ λ̄²/(2Δλ). This phenomenon is used to measure wavelength separation (coherence length measurement).
60. In Lloyd's mirror experiment, the fringe at the edge (zero path difference) is:
Explanation: The reflected ray undergoes a phase change of π (half wavelength) on reflection from the denser glass mirror. At zero path difference, the direct and reflected beams are π out of phase → destructive interference → dark fringe at the edge. This confirms the phase change on reflection.
61. In a Michelson interferometer, moving one mirror by λ/2 causes the fringe pattern to shift by:
Explanation: Moving mirror by λ/2 changes the path length of that arm by 2×(λ/2) = λ → change in path difference = λ → shift of exactly one fringe. Used to measure wavelengths and test optical components with sub-wavelength precision.
62. A thin wedge-shaped air film forms between two glass plates meeting at one edge. Observed in reflected monochromatic light, the fringe nearest the apex (thin edge) is:
Explanation: At the apex (zero thickness), ray 1 reflects from the top surface of the air film (glass to air → no phase change) and ray 2 from the bottom (air to glass → π phase change). At zero thickness, path difference = 0 but phase difference = π → dark fringe at the edge. Each subsequent dark fringe forms where 2t = nλ.
63. Newton's rings show a dark spot at the centre in reflected light because:
Explanation: At the point of contact (t = 0): ray from top of air film (no phase change) and ray from bottom plate (π phase change at air-to-glass). Path difference = 0 but net phase difference = π → dark central spot in reflected light.
64. In YDSE, the source S is moved a distance y perpendicular to the line joining the slits. The central maximum shifts:
Explanation: When S moves up, the path from S to S₁ (lower slit) increases relative to S₂ (upper slit). S₂ now has the shorter total path to the original central max location. The fringe pattern shifts downward (opposite to S's upward motion) to restore equal total path lengths.
65. The intensity ratio of the 1st secondary maximum to the central maximum in a single-slit diffraction pattern is approximately:
Explanation: First secondary maximum occurs at approximately a sinθ ≈ 3λ/2. The intensity ratio is I₁/I₀ ≈ 4/(9π²) ≈ 0.045 ≈ 1/22. Secondary maxima are much weaker than the central maximum.
66. Two coherent sources with a constant phase difference of π (antiphase) are placed at the slit positions in YDSE. The central fringe is:
Explanation: At the centre, geometric path difference = 0. But since sources are antiphase (phase difference π), the actual phase difference at centre = 0 + π = π → destructive interference → dark central fringe. The entire pattern shifts by half a fringe width.
67. A wave of frequency 3×10¹⁴ Hz belongs to which part of the EM spectrum?
Explanation: λ = c/f = 3×10⁸/3×10¹⁴ = 10⁻⁶ m = 1000 nm. This is at the boundary of infrared and red visible light — approximately the near-IR/red region.
68. In YDSE, if slit separation is halved and screen distance is doubled, fringe width:
Explanation: β = λD/d. New β = λ(2D)/(d/2) = 4λD/d = 4β. Fringe width quadruples.
69. Optical activity means a substance can:
Explanation: Optically active substances (sugars, amino acids) rotate the plane of plane-polarised light. Dextrorotatory (+) rotates clockwise, levorotatory (−) anticlockwise. Used in saccharimetry to determine sugar concentration.
70. Two thin glass plates of different thickness are placed in the two slits of a YDSE. The central fringe shifts. The shift depends on:
Explanation: Each slab adds (n−1)t of extra optical path to that beam. The net shift = difference in extra optical paths = (n₁−1)t₁ − (n₂−1)t₂. The fringe pattern shifts by this/β fringe widths.
71. In single-slit diffraction, making the slit very wide (a >> λ) causes:
Explanation: When a >> λ, θ_min = λ/a → 0, so the central maximum narrows to essentially a geometric beam with sharp edges — diffraction is negligible. This is why large objects cast sharp shadows in ordinary light.
72. Unpolarised light (I₀) passes through polariser P₁, then P₂ (at 30° to P₁), then P₃ (at 90° to P₁). Final intensity is:
Explanation: After P₁: I₀/2. After P₂ (30°): (I₀/2)cos²30° = (I₀/2)(3/4) = 3I₀/8. After P₃ (at 90° to P₁ = 60° to P₂): (3I₀/8)cos²60° = (3I₀/8)(1/4) = 3I₀/32.
73. In YDSE with white light, a bright fringe of wavelength λ₁ at order n₁ coincides with a bright fringe of λ₂ at order n₂ when:
Explanation: Position of bright fringe: y = nλD/d. For two wavelengths to coincide: n₁λ₁D/d = n₂λ₂D/d → n₁λ₁ = n₂λ₂. This gives missing orders and spectral overlaps in gratings.
74. Radius of nth bright Newton's ring (radius of curvature R, wavelength λ, thin air film) in reflected light is:
Explanation: For bright rings in reflected light: 2t = (n+½)λ (accounting for the π phase change). For a plano-convex lens: t = r²/(2R). So r² = (n+½)λR → r_n = √((n+½)λR).
75. Dispersive power of a diffraction grating (dθ/dλ) for order n and grating spacing d is:
Explanation: From d sinθ = nλ, differentiating: d cosθ dθ = n dλ → dθ/dλ = n/(d cosθ). Higher order n and smaller d → higher dispersive power → better wavelength separation per unit length.