Corpuscular model vs Wave model | Historical background
History
💡 Corpuscular Model (Newton)
Descartes (1637) gave corpuscular model. Developed by Newton in OPTICKS. Predicted: if ray bends towards normal on refraction, speed in second medium is greater. Later proved WRONG.
Wave Theory
🌊 Huygens Wave Theory (1678)
Dutch physicist Huygens proposed wave model. Predicted: if ray bends towards normal, speed is less in second medium. Confirmed by Foucault (1850) experimentally. Also explains reflection & refraction.
Key Event
🔬 Young's Interference (1801)
Thomas Young's double-slit experiment firmly established light as a wave. Wavelength of visible light is very small (~400–750 nm). Maxwell later showed light is an electromagnetic wave.
🖼️[Historical timeline: Descartes 1637 → Huygens 1678 → Young 1801 → Maxwell 1855 → Foucault 1850]
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🎯 Practice MCQs — Section 10.1
Q1. According to the corpuscular model, when light refracts and bends towards the normal, its speed in the second medium is: Easy
A) Greater than in the first medium
B) Less than in the first medium
C) Equal to the first medium
D) Zero
The corpuscular model predicted that if refracted ray bends towards normal, speed increases in second medium. This was later proved WRONG by Foucault's experiment (1850). The wave theory correctly predicts speed decreases. Option A.
Q2. Thomas Young's interference experiment was performed in the year: Easy
A) 1637
B) 1678
C) 1801
D) 1850
Thomas Young performed his famous double-slit interference experiment in 1801, firmly establishing the wave nature of light. Option C.
Q3. The wavelength of visible light is in the range of: Easy
A) 1 nm – 100 nm
B) 400 nm – 750 nm
C) 1 mm – 10 mm
D) 750 nm – 1500 nm
Visible light wavelength range is 400 nm (violet) to 750 nm (red). Yellow light ≈ 589 nm. Option B.
〰️
10.2 Huygens Principle
Wavefront | Secondary wavelets | Geometrical construction
Definition
🌐 Wavefront
A wavefront is a surface of constant phase. All points on a wavefront vibrate in phase. Energy travels perpendicular to the wavefront (along rays). Point source → spherical wavefront. Far away → plane wavefront.
Principle
🔵 Huygens Principle
Every point on a wavefront is a source of secondary wavelets. These secondary wavelets spread out in all directions with the speed of the wave. The common tangent (envelope) of all secondary wavelets gives the new wavefront at a later time.
Note
⬅️ Backwave Problem
Huygens' construction also predicts a backward wave (backwave). Huygens assumed amplitude of secondary wavelets is maximum forward and zero backward (adhoc assumption). More rigorous justification comes from wave theory.
Huygens Construction Steps
1. Know wavefront at t = 0 (say F₁F₂).
2. Draw spheres of radius vτ from each point on F₁F₂.
3. Draw common forward tangent = new wavefront G₁G₂ at time t = τ.
4. Rays are always perpendicular to wavefronts.
🖼️[NCERT Fig. 10.1(a): Spherical wavefront from point source | Fig. 10.1(b): Plane wave far from source | Fig. 10.2: Huygens construction — F₁F₂ → G₁G₂]
Add wavefront diagrams here
🎯 Practice MCQs — Section 10.2
Q4. According to Huygens principle, every point on a wavefront acts as: Easy
A) A point of zero amplitude
B) A source of secondary wavelets
C) A point of destructive interference
D) A reflecting surface
Huygens' Principle states every point on the wavefront is a source of secondary disturbance (secondary wavelets) that spread out in all directions with the speed of the wave. Option B.
Q5. When a point source is very far away, its wavefront becomes: Easy
A) Spherical
B) Cylindrical
C) Plane
D) Parabolic
At a very large distance from a point source, a small portion of the spherical wavefront appears flat (plane). Thus, parallel rays from distant stars are approximated as plane waves. Option C.
Q6. The direction of energy propagation of a wave is: Moderate
A) Parallel to the wavefront
B) At 45° to the wavefront
C) Perpendicular to the wavefront
D) Along the wavefront
Energy of a wave always travels in a direction perpendicular to the wavefront. This direction is what we call the "ray" in geometrical optics. Option C.
🔄
10.3 Refraction & Reflection Using Huygens Principle
Deriving Snell's Law | Deriving Law of Reflection | Wavelength change
Refraction
💧 Snell's Law from Huygens
Plane wave AB incident at angle i on interface PP'. In time τ: BC = v₁τ (medium 1), AE = v₂τ (medium 2). sin i / sin r = v₁/v₂ = n₂/n₁ → n₁ sin i = n₂ sin r
Key Result
📉 What Changes on Refraction?
When light goes from medium 1 to denser medium 2 (v₁ > v₂): ✅ Frequency ν remains the SAME ❌ Speed decreases (v₂ < v₁) ❌ Wavelength decreases (λ₂ < λ₁) v/λ = constant = ν
Reflection
🪞 Law of Reflection
Using Huygens construction: BC = AE = vτ. Triangles EAC and BAC are congruent → angles i and r are equal. This proves ∠i = ∠r (Law of reflection).
Snell's Law derived from Huygens
sin i / sin r = v₁/v₂ = n₂/n₁ → n₁ sin i = n₂ sin r
🖼️[NCERT Fig. 10.4: Refraction of plane wave at interface | Fig. 10.5: Refraction at rarer medium | Fig. 10.6: Reflection of plane wave | Fig. 10.7(a,b,c): Prism, convex lens, concave mirror wavefronts]
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🎯 Practice MCQs — Section 10.3
Q7. Monochromatic light of wavelength 589 nm passes from air into water (n = 1.33). Its wavelength in water is: Moderate
A) 783.4 nm
B) 443 nm
C) 589 nm
D) 300 nm
λ₂ = λ₁/n = 589/1.33 ≈ 443 nm. Speed and wavelength decrease on entering denser medium; frequency stays the same. (NCERT Ex 10.1) Option B.
Q8. Speed of light in glass (n = 1.5) is: Easy
A) 3 × 10⁸ m/s
B) 1.5 × 10⁸ m/s
C) 2 × 10⁸ m/s
D) 4 × 10⁸ m/s
v = c/n = (3×10⁸)/1.5 = 2×10⁸ m/s. (NCERT Ex 10.3) Option C.
Q9. When light travels from rarer to denser medium, which quantity remains unchanged? Easy
A) Speed
B) Wavelength
C) Frequency
D) Amplitude
Frequency ν = v/λ remains unchanged when light travels from one medium to another. Speed and wavelength both change but their ratio (frequency) stays constant. Option C.
🔗
10.4 Coherent and Incoherent Addition of Waves
Superposition | Constructive & Destructive Interference | Intensity formula
Superposition
➕ Superposition Principle
Resultant displacement at any point = vector sum of displacements from each wave. Forms the basis of all interference phenomena. Coherent sources maintain a constant phase difference.
I = 4I₀ cos²(φ/2)
φ = phase difference between waves; I₀ = intensity from each source
Constructive: Path diff = nλ → I = 4I₀
Destructive: Path diff = (n+½)λ → I = 0
Incoherent Sources (time-averaged)
I = 2I₀ (intensities simply add)
🖼️[NCERT Fig. 10.8: Two coherent sources in water | Fig. 10.9: Constructive (Q) and Destructive (R) interference | Fig. 10.10: Locus of points S₁P-S₂P = 0,±λ,±2λ]
Add interference diagrams here
🎯 Practice MCQs — Section 10.4
Q10. Two coherent sources each of intensity I₀ interfere. The maximum resultant intensity is: Easy
A) I₀
B) 2I₀
C) 4I₀
D) I₀/2
At constructive interference, amplitude = 2a. Intensity ∝ amplitude² → I = (2a)² ∝ 4I₀. Option C.
Q11. Two incoherent sources each of intensity I₀ illuminate a wall. The resultant intensity is: Easy
A) 4I₀
B) 2I₀
C) I₀
D) 0
For incoherent sources, phase difference changes randomly. Time-averaged intensity = I₀ + I₀ = 2I₀. No interference fringes are formed. Option B.
Q12. At a point P, two coherent waves have path difference 2.5λ. The resultant intensity is: Moderate
A) 4I₀
B) 2I₀
C) I₀
D) 0
Path difference = 2.5λ = (2 + ½)λ → destructive interference → I = 0. This is the condition (n + ½)λ with n = 2. Option D.
Single source S → two slits S₁ and S₂ (separation d) → screen at distance D. S₁ and S₂ are coherent (derived from same source). Spherical waves from S₁ and S₂ interfere on screen GG'.
Bright Fringes
💡 Maxima (Bright Fringes)
Path diff = nλ Position: xₙ = nλD/d n = 0, ±1, ±2,... n=0 is central bright fringe. Fringes are equally spaced on both sides.
Dark Fringes
🌑 Minima (Dark Fringes)
Path diff = (n+½)λ Position: xₙ = (n+½)λD/d n = 0, ±1, ±2,... Fringe width β = λD/d (same for bright & dark fringes).
Fringe width: β = λD/d (spacing between consecutive bright or dark fringes)
Intensity: I = 4I₀ cos²(φ/2) where φ = (2π/λ)Δx
🖼️[NCERT Fig. 10.11: Incoherent sources — no fringes | Fig. 10.12(a,b): Young's setup with coherent sources | Fig. 10.13: Computed fringe pattern]
Add YDSE setup and fringe pattern diagrams here
🎯 Practice MCQs — Section 10.5
Q13. In YDSE, d = 0.28 mm, D = 1.4 m, 4th bright fringe is at 1.2 cm from centre. Wavelength of light is: Hard
A) 400 nm
B) 520 nm
C) 589 nm
D) 600 nm
xₙ = nλD/d → λ = xₙd/(nD) = (1.2×10⁻²×0.28×10⁻³)/(4×1.4) = (3.36×10⁻⁶)/(5.6) = 6×10⁻⁷ m = 600 nm. (NCERT Ex 10.4) Option D.
Q14. In YDSE with λ = 600 nm, D = 1 m, d = 1 mm, the fringe width is: Moderate
A) 0.3 mm
B) 0.6 mm
C) 1.2 mm
D) 6 mm
β = λD/d = (600×10⁻⁹×1)/(1×10⁻³) = 6×10⁻⁴ m = 0.6 mm. Option B.
Q15. In YDSE using light of wavelength λ, intensity at path difference λ is K. Intensity at path difference λ/3 is: Hard
A) K
B) K/2
C) K/4
D) 3K/4
At Δx = λ: φ = 2π → I = 4I₀cos²(π) = 4I₀×1 = 4I₀ = K → I₀ = K/4. At Δx = λ/3: φ = 2π/3 → I = 4I₀cos²(π/3) = 4(K/4)×(1/2)² = K×(1/4) = K/4. (NCERT Ex 10.5) Option C.
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10.6 Diffraction
Single slit diffraction | Central maximum | Secondary maxima | Minima
Concept
〰️ What is Diffraction?
Bending of light around obstacles/slits. A general wave property. For light, effects are significant when slit width ≈ wavelength. Colors on a CD are due to diffraction. Limits resolution of optical instruments.
Single Slit
🔲 Single Slit Pattern
Slit of width a, wavelength λ, screen at D. Minima: sin θ = nλ/a (n = ±1, ±2,...) Central max: at θ = 0 (widest, brightest) Secondary maxima at θ ≈ (n+½)λ/a (weaker)
Key Difference
⚡ Interference vs Diffraction
No strict physical difference (Feynman). Interference: few sources (2 slits). Diffraction: large number of sources (continuous slit). Double-slit pattern = interference × single-slit diffraction envelope.
Single Slit Diffraction
Minima condition: a sin θ = nλ (n = ±1, ±2,...)
or: sin θ = nλ/a
Note: In YDSE minima, path diff = (n+½)λ. In single slit, minima at asinθ = nλ — different!
🖼️[NCERT Fig. 10.14: Single slit geometry — slit LN, point P on screen | Fig. 10.15: Intensity pattern with central max and secondary maxima]
Add single slit diffraction diagrams here
🎯 Practice MCQs — Section 10.6
Q16. In single slit diffraction (slit width a = 0.1 mm, λ = 500 nm), the angle of the first minimum is: Moderate
A) 0.005°
B) 0.287°
C) 0.573°
D) 5°
sin θ = λ/a = (500×10⁻⁹)/(0.1×10⁻³) = 5×10⁻³. θ = sin⁻¹(0.005) ≈ 0.287°. Option B.
Q17. In single slit diffraction, the width of central maximum is doubled if slit width a is: Moderate
A) Doubled
B) Halved
C) Quadrupled
D) Unchanged
Width of central max ∝ λD/a ∝ 1/a. So to double the width, slit width must be halved. Option B.
Q18. Colours seen on a CD disc are due to: Easy
A) Refraction
B) Reflection
C) Diffraction
D) Polarisation
The tracks on a CD act as a diffraction grating. Different wavelengths (colours) of white light are diffracted at different angles, producing rainbow-like colours. Option C.
↕️
10.7 Polarisation
Transverse waves | Polaroid | Malus' Law | Applications
Key Concept
↕️ Polarisation
Light waves are transverse — electric field is perpendicular to direction of propagation. Unpolarised light: electric field oscillates in all directions randomly. Linearly polarised light: oscillates in one plane only.
Polaroid
🕶️ Polaroid Sheet
Contains aligned long-chain molecules. Electric field parallel to molecules is absorbed. Perpendicular component (pass-axis direction) is transmitted. Unpolarised → Polaroid → Intensity becomes I₀/2
Malus' Law
📐 Malus' Law
When polarised light (intensity I₀) passes through a polaroid at angle θ to pass-axis: I = I₀ cos²θ θ = 0° → I = I₀ (max) θ = 90° → I = 0 (zero, crossed polaroids)
Malus' Law
I = I₀ cos²θ
θ = angle between pass-axis of analyser and plane of polarisation of incident light
Unpolarised Light through Single Polaroid
Intensity transmitted = I₀/2 (regardless of orientation — rotating polaroid has no effect)
Three Polaroids (P₁, P₂, P₃ where P₁⊥P₃)
I = (I₀/4) sin²2θ where θ = angle of P₂ with P₁. Maximum when θ = 45°.
🖼️[NCERT Fig. 10.17: Transverse wave (y-polarised) | Fig. 10.18(a): Two polaroids at various angles — intensity varies | Fig. 10.18(b): Electric vector component transmitted]
Add polarisation diagrams here
🎯 Practice MCQs — Section 10.7
Q19. Unpolarised light of intensity I₀ passes through a polaroid. Transmitted intensity is: Easy
A) I₀
B) I₀/2
C) I₀/4
D) 2I₀
A polaroid transmits only one component of the electric field. For unpolarised light, the transmitted intensity is half the incident intensity = I₀/2. Option B.
Q20. Polarised light of intensity I₀ is incident on a polaroid making angle 60° with pass-axis. Transmitted intensity is: Moderate
A) I₀
B) I₀/2
C) I₀/4
D) 3I₀/4
By Malus' Law: I = I₀ cos²θ = I₀ cos²60° = I₀ × (0.5)² = I₀/4. Option C.
Q21. Polarisation phenomenon proves that light waves are: Easy
A) Longitudinal
B) Transverse
C) Scalar
D) Mechanical
Polarisation is a property exclusive to transverse waves. Sound waves (longitudinal) cannot be polarised. The fact that light can be polarised proves it is a transverse wave. Option B.
📋
Chapter 10 — Key Summary
All important formulas for Wave Optics — AAI ATC exam quick reference
🌊 Huygens Principle
Each point on wavefront = source of secondary wavelets. New wavefront = envelope of secondary wavelets. Rays ⊥ wavefront.
🔄 Refraction (Wave)
sin i/sin r = v₁/v₂ = n₂/n₁ λ₁/λ₂ = v₁/v₂ Frequency unchanged on refraction
🔗 Interference
I = 4I₀cos²(φ/2) Constructive: Δx = nλ → I = 4I₀ Destructive: Δx=(n+½)λ → I=0