L-22: Wave Phenomena and Light
Physics — Class 12 · NIOS Code 312 · Module 6 · Source: 312_Physics_Eng_Lesson22.pdf
Wave Phenomena and Light
Rectilinear propagation cannot explain superposition, bending around corners, or energy redistribution. Huygens' wave theory, Young's interference experiment, diffraction, and polarisation together established that light is a transverse wave. This lesson completes Module 6 (Optics).
NIOS objectives: Huygens' principle; interference and diffraction; single-slit diffraction; polarisation and Brewster's law.
22.1 Huygens' Principle
A wavefront is the locus of all points vibrating in the same phase at an instant. Point source → spherical wavefronts; line source → cylindrical; distant source → plane wavefront over small regions. A ray is perpendicular to the wavefront; a collection of rays forms a beam.
Huygens' principle states:
- Every point on a wavefront acts as a source of secondary disturbances (wavelets).
- The new wavefront at a later instant is the forward common envelope of all secondary wavelets.
- In an isotropic medium, energy spreads equally in all directions.
- Wavefronts do not travel backward.
22.1.1 Propagation of Waves
From known shape, position, direction and speed of a wavefront, its later position is found by drawing arcs of radius r = vT from points on the initial wavefront and drawing the tangent envelope. This describes wave motion geometrically.
Intext 22.1: Wavefront ⊥ direction of propagation. Ratio of wavelet radii at t = 3 s and t = 6 s = ½.
22.2 Interference of Light
Interference is redistribution of energy due to superposition of waves from two coherent sources (same frequency, same amplitude, constant phase difference, close together). Thomas Young demonstrated this in 1802.
22.2.1 Young's Double Slit Experiment
Monochromatic light through pinhole S divides at slits S₁ and S₂ (equidistant from S). In-phase waves superpose on screen C → alternate bright and dark fringes.
- Constructive interference: crests meet crests (or troughs meet troughs) → bright fringe. Phase difference δ = 0, 2π, 4π, … = 2nπ.
- Destructive interference: crest meets trough → dark fringe. δ = π, 3π, … = (2n+1)π.
a = amplitude of each wave
δ = phase difference between the two waves
Max I = 4a² (constructive); Min I = 0 (destructive)
One wavelength path difference = 2π phase difference
Constructive when Δ = nλ
Destructive when Δ = (2n+1)λ/2
D = slit-to-screen distance
x = fringe position from central maximum
Small-angle approximation: sin θ ≈ tan θ ≈ x/D
Central bright fringe at n = 0
β ∝ λ and D; β ∝ 1/d
Energy conserved: dark fringes lose energy reappearing at bright fringes (max 4a² vs average 2a²)
Coherent vs incoherent: Two independent bulbs cannot produce stable fringes — random phase changes wash out the pattern. Sodium lamp through single pinhole gives coherence.
22.3 Diffraction of Light
Light bends around edges of obstacles or narrow apertures — violates strict rectilinear propagation. Observable when obstacle/aperture size ≈ wavelength (~10⁻⁶ m) or screen distance ≫ aperture size.
22.3.1 Diffraction at a Single Slit
Monochromatic plane wavefront incident on narrow slit (width a). Huygens' wavelets from all points in the slit superpose on screen.
- Principal maximum: centre O — all wavelet pairs arrive in phase; brightest; width twice secondary maxima.
- Minima: path difference between waves from extreme edges = nλ → zero intensity.
- Secondary maxima: between minima; only a fraction of wavefront contributes → much weaker than principal peak.
Interference vs diffraction: Interference = superposition from two separate coherent sources; diffraction = superposition of wavelets from different portions of the same wavefront.
22.4 Polarisation of Light
Interference and diffraction prove wave nature; polarisation proves light is a transverse wave (vibrations perpendicular to propagation). Longitudinal waves (e.g. sound in air) cannot be polarised.
Unpolarised light has vibrations in all planes perpendicular to propagation. A polaroid (dichroic crystals in nitrocellulose) transmits one plane and absorbs the perpendicular component → plane (linearly) polarised light.
When unpolarised light reflects from a transparent surface, reflected light is partially polarised. At the polarising angle (Brewster angle ip), reflected ray is completely plane polarised and reflected + transmitted rays are perpendicular.
Air–water: ip = 53° (Sun 37° above horizon → polarised reflection)
Example: ip = 60° → μ = 1.73; μ = 1.42 → ip ≈ 54°
Applications: polaroid sunglasses (reduce glare), camera filters, polarimeters in sugar industry. Two polaroids with transmission axes at 90° block light completely.
WAVE PHENOMENA & LIGHT — KEY POINTS
=====================================
Huygens : wavelets → forward envelope
Wavefront : locus of same-phase points; ray ⊥ wavefront
Coherent sources: same f, λ, amplitude; constant phase diff.
Bright fringe : Δ = nλ ; Dark: Δ = (n+½)λ
Fringe width : β = λD/d
Intensity : I ∝ 4a² cos²(δ/2)
Diffraction : bending at narrow apertures; same wavefront
Single slit : principal max brightest; 2× width of others
Polarisation : light is transverse wave
Brewster's law : tan iₚ = μ ; OR ⊥ OT at iₚ
Quick Revision
- Huygens: every wavefront point emits secondary wavelets; tangent = new wavefront.
- Young's slits need coherent monochromatic sources for stable fringes.
- β = λD/d — increase D or λ, or decrease d, to widen fringes.
- Energy is redistributed in interference, not destroyed.
- Diffraction needs aperture size ~ λ; interference uses two separate sources.
- Polarisation impossible for longitudinal waves; Brewster angle gives complete polarisation.
Q1. According to Huygens' principle, the new position of a wavefront is obtained by:
Q2. The direction of propagation of a wave is:
Q3. In Young's double slit experiment, bright fringes occur when path difference is:
Q4. The fringe width in Young's experiment is given by:
Q5. Two independent incandescent bulbs cannot produce a stable interference pattern because they are:
Q6. Diffraction of light demonstrates that light:
Q7. In single-slit diffraction, the central maximum is:
Q8. Polarisation of light proves that light waves are:
Q9. Brewster's law states that at the polarising angle:
Q10. For a material of refractive index 1.42, the polarising angle is approximately:
PYQ — Previous Year Questions
Extracted from NIOS Physics (312) board exam papers in your PDF. Chapter L22 — Wave Phenomena and Light only. Use Model Answer for marking points; Explanation for concept clarity.
Chapter 22 — Wave Phenomena and Light (L22)
12 questions · Sections A & B · Sources: 312/TUS/104A, 312/MAY/204A–C, 68/ESS/1-312-A
Section A — Objective (1 mark)
PYQ1. A single-slit diffraction pattern is obtained using a beam of red light. When red light is replaced by blue light: (A) the diffraction pattern disappears (B) there is no change (C) the diffraction fringes become narrower and get crowded together (D) the diffraction fringes become broader and move further apart
Model Answer
(C) — fringes become narrower and crowded.
Blue light has shorter λ → fringe width ∝ λ → smaller angular spread.
Explanation
Single-slit central maximum angular width ≈ 2λ/a. Shorter λ tightens the pattern (L22 §22.3).
PYQ2. If the angle of maximum polarization on the surface of a medium is P, the velocity of light v in the medium is given by (c = speed in vacuum): (A) tan(v/c) = P (B) cot(v/c) = P (C) sec(v/c) = P (D) cosec(v/c) = P
Model Answer
Brewster: μ = tan P and μ = c/v
⇒ v/c = 1/tan P = cot P → (B)
Explanation
At polarising angle, reflected ray is completely polarized. μ = tan ip links refractive index to Brewster angle (L22 §22.4).
PYQ3. Two waves having amplitudes in the ratio 3 : 5 produce interference. The ratio of maximum to minimum intensity in the interference pattern will be (A) 5 : 3 (B) 16 : 1 (C) 25 : 9 (D) 17 : 8
Model Answer
Imax ∝ (3+5)² = 64; Imin ∝ (3−5)² = 4
Ratio = 16 : 1 → (B)
Explanation
Intensity ∝ (amplitude)². For superposition: Imax/(Imin) = (A₁+A₂)²/(A₁−A₂)² (L22 §22.2).
PYQ4. Two waves having intensities in the ratio 1 : 4 produce interference. The ratio of maximum to minimum intensity in the interference pattern will be (A) 5 : 3 (B) 9 : 1 (C) 25 : 9 (D) 4 : 1
Model Answer
Amplitude ratio √1 : √4 = 1 : 2
Imax/Imin = (1+2)²/(1−2)² = 9/1 → (B) 9 : 1
Explanation
Convert intensity ratio to amplitude ratio first, then apply interference formula.
PYQ5. Read the passage: “Phenomena like interference, diffraction and polarization show that light has a wave nature.” Interference is a phenomenon exhibited by: (A) waves only (B) particles only (C) waves as well as particles (D) neither waves nor particles
Model Answer
(A) waves only
Explanation
Interference requires superposition of waves with fixed phase relation — a wave phenomenon (L22 §22.2). Matter waves also interfere, but the board option here is “waves only” in classical optics context.
Section A — Short Answer (2 marks)
PYQ6. Fill in the blanks: (a) In Young's double-slit experiment, if the separation between the slits is tripled, the fringe width will become _____ times the initial value. (b) With λ = 6000 Å, 99 fringes are seen; with λ = 5500 Å in the same space, the number of fringes will be _____.
Model Answer
(a) 1/3 — β = λD/d; d tripled → β becomes one-third.
(b) 108 — same field width W = nβ; n ∝ 1/λ → n₂ = 99 × 6000/5500 ≈ 108
Explanation
Young's fringe width β = λD/d (L22 §22.2). Fewer fringes when β larger; more fringes when λ decreases at fixed screen area.
PYQ7. Write ‘True’ for correct statement and ‘False’ for incorrect: (a) Superposition of two light waves from two coherent sources produces a fringe pattern in which the central maximum is twice the width of all other maxima. (b) Huygens' wave theory explains various optical phenomena in terms of wavefront.
Model Answer
(a) False — in Young's double-slit interference, bright fringes have equal width; “central max twice as wide” describes single-slit diffraction.
(b) True — Huygens' principle uses secondary wavelets from each point on a wavefront.
Explanation
Do not mix interference (equal-width fringes) with diffraction (broad central peak) (L22 §22.1, §22.3).
Section B — Short Answer (2 marks)
PYQ8. In a single-slit diffraction pattern, what can we say about (a) the relation between the width of the central bright fringe and the widths of the other fringes, and (b) the intensity of various bright fringes?
Model Answer
(a) Central maximum is twice as wide as any secondary maximum.
(b) Intensity of secondary maxima decreases with order (first secondary ≈ 4.5% of central).
Explanation
Single-slit pattern: broad central envelope from width a; side maxima narrower and fainter (L22 §22.3).
PYQ9. The angle of maximum polarisation for a certain medium is 60°. Calculate the refractive index of the medium.
Model Answer
Brewster's law: μ = tan ip
μ = tan 60° = √3 ≈ 1.732
Explanation
At polarising angle, reflected and refracted rays are perpendicular; tan ip = μ (L22 §22.4).
Section B — Short Answer (3 marks)
PYQ10. Write any three distinguishing features of the fringe patterns formed in Young's double-slit experiment and single-slit diffraction.
Model Answer
Any three, e.g.:
- Young: equally spaced fringes; all bright fringes nearly same width/intensity.
- Single slit: central maximum is brightest and twice as wide as others.
- Single slit: secondary maxima decrease rapidly in intensity.
- Young: needs two coherent sources; single slit: one aperture, diffraction envelope.
Explanation
Interference + diffraction together in real double-slit, but board contrast is ideal Young fringes vs pure single-slit pattern (L22 §22.2–22.3).
PYQ11. In a single-slit diffraction experiment, how will the angular width of the central maximum change when: (i) slit width is decreased, (ii) distance between slit and screen is increased, (iii) light of smaller wavelength is used? Explain.
Model Answer
Angular width θ ≈ 2λ/a (a = slit width).
(i) a decreased → θ increases (broader central max).
(ii) D increased → angular width unchanged; linear width on screen increases.
(iii) smaller λ → θ decreases (narrower pattern).
Explanation
Angular spread set by λ and slit width only. Screen distance scales linear size, not angle (L22 §22.3).
PYQ12. State and explain Brewster's law with the help of a diagram. The value of Brewster angle for a transparent medium is different for light of different colours. Give reason.
Model Answer
Brewster's law: When unpolarized light is incident at polarising angle ip, the reflected ray is completely plane-polarized perpendicular to the plane of incidence; reflected and refracted rays are at 90°.
μ = tan ip
Diagram: interface, incident/unpolarized ray, polarized reflected ray, refracted ray at 90° to reflected.
Different colours: μ depends on wavelength (dispersion) → ip = tan⁻¹(μ) differs for each colour.
Explanation
Marking scheme: μ(λ) varies in glass → each colour has its own Brewster angle (L22 §22.4, links to L21 dispersion).
Problem Solving — L22 Wave Phenomena and Light
Six problems spanning this chapter’s NIOS syllabus. Every question is built from the notes and formula sheet: solve with equations first, then read the formal textbook-style write-up, the easy explanation, and the topic in depth (formulas, meaning, exam tips). Explanations open by default.
State Huygens’ principle and one application.
Solution — step by step with formulas
- Every point on a wavefront is a source of secondary spherical wavelets; new front is envelope.
- Explains reflection/refraction laws.
Final answer: Wavefront = envelope of secondary wavelets
Textbook formal language
Huygens construction models propagation of wavefronts in isotropic media.
Working formula set for this problem: (see solution steps). In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Each point on a crest sprays tiny waves; the forward edge is the new crest line.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Huygens’ principle
Diffraction also follows from wavelet interference.
Link to chapter notes (L22 — Huygens’ principle): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: (see solution steps). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write (see solution steps) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
In YDSE, fringe width formula? If λ increases, what happens to β?
Solution — step by step with formulas
- β = λD/d; β increases with λ.
Final answer: β = λD/d; wider fringes for larger λ
Formulas used in this problem
Textbook formal language
Constructive interference when path difference is mλ.
Working formula set for this problem: β = λD/d; path diff = d sinθ. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Fringe spacing grows if light is redder or slits closer or screen farther.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Young’s double slit
Requires coherent sources—slits from one source provide that.
Link to chapter notes (L22 — Young’s double slit): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: β = λD/d; path diff = d sinθ. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write β = λD/d; path diff = d sinθ before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
Why does a narrow slit spread light into a diffraction pattern?
Solution — step by step with formulas
- Wavelets from different parts of slit interfere; minima when path difference conditions met.
Final answer: Interference of wavelets across finite aperture
Formulas used in this problem
Textbook formal language
Diffraction is bending/spreading due to wave nature at obstacles/apertures.
Working formula set for this problem: a sinθ = mλ (minima). In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Light doesn’t stay a sharp beam after a narrow gap—it fans and makes bands.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Single-slit diffraction
Smaller a ⇒ broader central maximum.
Link to chapter notes (L22 — Single-slit diffraction): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: a sinθ = mλ (minima). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write a sinθ = mλ (minima) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
State Rayleigh criterion qualitatively for optical instruments.
Solution — step by step with formulas
- Two point sources just resolved when central max of one coincides with first min of other.
Final answer: Central max meets first min for limit of resolution
Formulas used in this problem
Textbook formal language
Diffraction limits angular resolution of telescopes/microscopes.
Working formula set for this problem: θ ≈ 1.22 λ/D (circular). In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
If two stars’ blur discs overlap too much, you see one blob.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Resolving power idea
Larger aperture D improves resolution.
Link to chapter notes (L22 — Resolving power idea): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: θ ≈ 1.22 λ/D (circular). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write θ ≈ 1.22 λ/D (circular) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
How does a Polaroid produce plane-polarised light from unpolarised light?
Solution — step by step with formulas
- Preferentially transmits E-component along transmission axis; absorbs the perpendicular component.
Final answer: Selects one E-plane
Textbook formal language
Polarisation proves light’s transverse nature.
Working formula set for this problem: (see solution steps). In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
Like a fence that lets only vertical rope waves through.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Polarisation
Malus’s law: I = I₀ cos²φ after analyser.
Link to chapter notes (L22 — Polarisation): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: (see solution steps). In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write (see solution steps) before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).
State Brewster’s law. What is special about reflected light at polarising angle?
Solution — step by step with formulas
- μ = tan i_p; reflected light is completely plane-polarised (perpendicular to plane of incidence).
Final answer: μ = tan i_p; reflected fully polarised
Formulas used in this problem
Textbook formal language
At polarising angle, reflected and refracted rays are perpendicular.
Working formula set for this problem: μ = tan i_p. In the NIOS presentation, physical quantities must be expressed in SI units and the relevant law or definition stated before substitution. Vector quantities require an explicit choice of positive direction; scalar work and energy require attention to sign conventions of the textbook.
Easy language (same idea, plain words)
At a magic angle, glare bounce is polarised—sunglasses cut it.
Read the question once for the story, once for the numbers. Write the formula, plug in values with units, then simplify. If a result looks huge or tiny, re-check powers of ten and whether you used sin/cos of the correct angle.
Topic in depth — Brewster’s law
Used in reducing reflections optically.
Link to chapter notes (L22 — Brewster’s law): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: μ = tan i_p. In multi-step questions, keep a free-body diagram or energy flow sketch before algebra; most errors are missing forces or wrong signs, not hard maths.
Exam tip
Quote the law in one line, then write μ = tan i_p before numbers. Box the final answer with unit. For numericals, keep at least three significant figures until the last step unless the data are coarse.
Common mistakes
- Mixing up scalar and vector quantities (e.g. treating momentum as unsigned).
- Using the wrong sign convention for work/heat/force direction.
- Forgetting to convert units (g↔kg, cm↔m, minutes↔seconds).
- Applying a formula outside its assumptions (e.g. F = ma when mass is not constant).