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State Board · Class 12 · Physics

Lesson 7: Wave Optics

7 concepts · 56 practice questions with answers · Free to read

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Nature of light, wavefronts and Huygens' principle

From daily life

A stone in a still pond makes circular ripples; light also spreads as wavefronts.

Simple definition

A wavefront is a surface of points in the same phase; Huygens' principle tells how the next wavefront is formed.

Wavefronts and Huygens secondary waveletsc twavefront (t = 0)new wavefront

In the figure : The left wavefront → secondary wavelets → the new wavefront at the right (distance c t).

Rule 1: Wavefront — A surface of points in the same phase. Types : spherical (point source), cylindrical (line source), plane (distant source). Rays are perpendicular to the wavefront.

  • Easy : Sunlight reaching Earth is almost a plane wavefront.
  • Medium : Spherical wavefronts around a bulb.
  • Exam (CET) : Rays of a plane wavefront are? (1) parallel (2) diverging (3) converging (4) circular → answer 1)

Rule 2: Huygens' principle — Every point on a wavefront is a new secondary source; the common tangent (envelope) of their wavelets after time t is the new wavefront. Distance = v t.

  • Easy : v = 3 × 10⁸ m/s, t = 2 ns ⇒ the wavefront moves 0.6 m.
  • Medium : Wavelets considered only in the forward direction (no back-wave).
  • Exam (CET) : The new wavefront is? (1) the tangent envelope (2) the centre of the wavelets (3) half (4) none → answer 1)

Rule 3: Light in a medium — n = c ÷ v; in a medium f stays the same, λ' = λ ÷ n, v = c ÷ n. Light is an EM wave, transverse.

  • Easy : n = 1.5 : v = 3 × 10⁸ ÷ 1.5 = 2 × 10⁸ m/s.
  • Medium : 600 nm light from air into glass : λ' = 600 ÷ 1.5 = 400 nm.
  • Exam (CET) : Water (n = 4/3) : v = ? (1) 2.25 × 10⁸ (2) 4 × 10⁸ (3) 3 × 10⁸ (4) 1.5 × 10⁸ → answer 1). Trick : 3 ÷ (4/3) = 9/4.

Derivation : λ' = λ ÷ n

  • Step 1 : The source frequency f stays the same in any medium.
  • Step 2 : In air c = f λ; in the medium v = f λ'.
  • Step 3 : Dividing : λ' ÷ λ = v ÷ c = 1 ÷ n.
  • Step 4 : λ' = λ ÷ n.
Remember

n = c/v; λ' = λ/n; f unchanged; wavefront = tangent envelope.

Common mistake / CET trap

In a medium λ and v change, not f. Rays are perpendicular to the wavefront, not parallel.

Shortcut

Memory clue : "in a medium λ shrinks" (λ' = λ/n).

Spot the difference

QuantityAirMedium (n)
Speedcc ÷ n
Wavelengthλλ ÷ n
Frequencyff (same)

Practice questions

Easy
  1. 1When is a wavefront plane?
  2. 2What does n = c ÷ v mean?
  3. 3Does f change in a medium?
Medium
  1. 4n = 2 : v?
  2. 5λ = 500 nm, n = 1.25 : λ'?
  3. 6t = 5 ns, v = 2 × 10⁸ : distance?
Challenging
  1. 7State Huygens' principle.
  2. 8f = 5 × 10¹⁴ Hz, n = 1.5 : λ' (c = 3 × 10⁸)?
Show answers (try them yourself first)
  1. 1When the source is very far
  2. 2Speed in vacuum ÷ speed in the medium
  3. 3No
  4. 4v = 1.5 × 10⁸ m/s
  5. 5λ' = 400 nm
  6. 6d = 1 m
  7. 7Each point is a secondary source; the tangent envelope is the new wavefront
  8. 8λ = 600 nm ⇒ λ' = 400 nm

Reflection and refraction by Huygens' principle

From daily life

A stick in water looks bent; light changes direction at the boundary.

Simple definition

Huygens' principle proves the laws of reflection and refraction.

Huygens : refraction — the wavefront bendsirn = sin i ÷ sin r= v₁ ÷ v₂n₂n₁

In the figure : i = angle of incidence, r = angle of refraction; in the denser medium the ray moves towards the normal.

Rule 1: Law of reflection — Proved by Huygens : i = r; the incident ray, reflected ray and normal are in one plane. In reflection v, λ, f stay the same.

  • Easy : i = 35° ⇒ r = 35°; angle between the rays 70°.
  • Medium : A ray incident along the normal returns along itself.
  • Exam (CET) : If the angle of incidence is 0°, r = ? (1) 0° (2) 90° (3) 45° (4) 180° → answer 1)

Rule 2: Snell's law — sin i ÷ sin r = n₂ ÷ n₁ = v₁ ÷ v₂ = λ₁ ÷ λ₂. Rarer → denser : the ray bends towards the normal; denser → rarer : away from it.

  • Easy : Air → glass (1.5), i = 30° : sin r = 0.5 ÷ 1.5 ⇒ r = 19.47°.
  • Medium : i = 45°, r = 30° : n = sin 45° ÷ sin 30° = 1.414; v = 2.12 × 10⁸ m/s.
  • Exam (CET) : i = 60°, n = √3 : sin r = (√3/2) ÷ √3 = 1/2 ⇒ r = 30°. (Options : 1) 30° 2) 45° 3) 60° 4) 15° → answer 1)

Rule 3: What changes in refraction — In refraction f stays; v and λ are divided by n. One part of the wavefront reaches the interface earlier, so the wavefront turns.

  • Easy : Air → glass : v smaller, λ smaller, f same.
  • Medium : Glass → air : v larger, λ larger.
  • Exam (CET) : Which does not change in refraction? (1) v (2) λ (3) f (4) direction → answer 3)

Derivation : Snell's law (Huygens)

  • Step 1 : A plane wavefront AB meets the surface at angle i; the end A arrives first.
  • Step 2 : B needs time t to reach the surface : BC = v₁ t. In the same t the wavelet from A travels AD = v₂ t in medium 2.
  • Step 3 : In triangle ABC : sin i = BC ÷ AC; in ADC : sin r = AD ÷ AC.
  • Step 4 : Dividing : sin i ÷ sin r = v₁ ÷ v₂ = n₂ ÷ n₁.
Remember

i = r; sin i/sin r = n₂/n₁ = v₁/v₂ = λ₁/λ₂; f same.

Common mistake / CET trap

Measure angles from the normal, not from the surface. Do not invert n₂/n₁.

Shortcut

Memory clue : "denser = slower, shorter λ, bends towards the normal".

Spot the difference

ChangeReflectionRefraction
fsamesame
v, λsamechange
Directioni = rSnell

Practice questions

Easy
  1. 1Law of reflection?
  2. 2Snell's law?
  3. 3f in refraction?
Medium
  1. 4i = 60°, r = 30° : n?
  2. 5n₁ = 1, n₂ = 2, i = 30° : r?
  3. 6n = 1.5 : v₁ ÷ v₂ (air → glass)?
Challenging
  1. 7In glass λ = 400 nm, n = 1.5 : λ in air?
  2. 8The ray bends towards the normal : what kind of medium?
Show answers (try them yourself first)
  1. 1i = r
  2. 2sin i ÷ sin r = n₂ ÷ n₁
  3. 3Same
  4. 4n = 0.866 ÷ 0.5 = 1.73
  5. 5sin r = 0.25 ⇒ r = 14.5°
  6. 61.5
  7. 7λ = 600 nm
  8. 8A denser medium

Polarisation and Malus' law

From daily life

Polaroid sunglasses reduce the glare from the road.

Simple definition

Polarisation confines the vibrations of light to one direction.

Polariser and analyser : I = I₀ cos² θpolariseranalyserunpolarised I₀I₀ ÷ 2I₀ cos²θ ÷ 2θ

In the figure : Unpolarised light is halved by the polariser; the analyser reduces it by cos²θ.

Rule 1: Polarisation — Evidence that light is a transverse wave. A polariser passes vibrations along one direction only. Unpolarised light of intensity I₀ emerges with half : I₀ ÷ 2.

  • Easy : I₀ = 80 W/m² ⇒ 40 W/m² after the polariser.
  • Medium : Sound cannot be polarised (longitudinal).
  • Exam (CET) : Unpolarised light through a polariser : intensity? (1) I₀/2 (2) I₀ (3) I₀/4 (4) 0 → answer 1)

Rule 2: Malus' law — Polarised light (I₀) through an analyser at angle θ : I = I₀ cos² θ. θ = 0° ⇒ I₀; 90° ⇒ 0; 45° ⇒ I₀ ÷ 2.

  • Easy : I₀ = 40, θ = 60° ⇒ I = 40 × ¼ = 10.
  • Medium : θ = 30° ⇒ I = ¾ I₀.
  • Exam (CET) : θ = 45° : I ÷ I₀ = ? (1) 1/2 (2) 1/4 (3) 1/√2 (4) 1 → answer 1). Trick : cos²45° = ½.

Rule 3: Two or three polaroids — Unpolarised I₀ → P₁ (I₀/2) → P₂ at θ ⇒ (I₀/2) cos²θ. Two crossed (90°) polaroids block light; a third at 45° between them gives I₀/8.

  • Easy : I₀ = 80, θ = 60° ⇒ (80/2) × ¼ = 10.
  • Medium : Crossed polaroids plus one at 45° between : I₀/2 × ½ × ½ = I₀/8.
  • Exam (CET) : Through two crossed polaroids? (1) 0 (2) I₀/2 (3) I₀/4 (4) I₀ → answer 1)

Derivation : Malus' law

  • Step 1 : The field E₀ makes angle θ with the analyser axis; only the component E₀ cos θ passes.
  • Step 2 : Intensity ∝ (field)².
  • Step 3 : I = k (E₀ cos θ)² = k E₀² cos²θ.
  • Step 4 : k E₀² = I₀ ⇒ I = I₀ cos²θ.
Remember

Unpolarised → I₀/2; I = I₀cos²θ; crossed polaroids ⇒ 0.

Common mistake / CET trap

Always half at the first polariser, not cos². θ is the angle between the axes.

Shortcut

Memory clue : "half at the first, cos² at the next".

Spot the difference

CaseθI
Parallel axes0°I₀
45°45°I₀/2
Crossed axes90°0

Practice questions

Easy
  1. 1Polarisation is evidence of?
  2. 2Malus' law?
  3. 3θ = 90° : I?
Medium
  1. 4I₀ = 100, θ = 30° : I?
  2. 5Unpolarised 100 → P₁ → P₂ at 60° : I?
  3. 6I = I₀ ÷ 4 : θ?
Challenging
  1. 7Three polaroids at 0°, 45°, 90° : unpolarised I₀ → I?
  2. 8Why can sound not be polarised?
Show answers (try them yourself first)
  1. 1light is a transverse wave
  2. 2I = I₀ cos²θ
  3. 3Zero
  4. 4I = 75
  5. 5I = 50 × 0.25 = 12.5
  6. 6cos²θ = ¼ ⇒ θ = 60°
  7. 7I₀/8
  8. 8Sound is longitudinal

Brewster's law and polarisation by scattering

From daily life

The sun's glare off water is strongly polarised at a particular angle, so polaroid cuts it.

Simple definition

The Brewster angle is the angle of incidence at which the reflected light is completely polarised.

Brewster angle : reflected and refracted rays are perpendiculari_pi_prtan i_p = n90°

In the figure : At i_p the reflected (red) and refracted (green) rays are 90° apart.

Rule 1: Brewster's law — At a particular angle of incidence (i_p) the reflected light is completely polarised : tan i_p = n. Then i_p + r = 90° (reflected ⊥ refracted).

  • Easy : Glass n = 1.5 ⇒ i_p = tan⁻¹ 1.5 = 56.31°.
  • Medium : Water n = 4/3 ⇒ i_p = 53.13°.
  • Exam (CET) : n = √3 ⇒ i_p = ? (1) 60° (2) 30° (3) 45° (4) 90° → answer 1). Trick : tan 60° = √3.

Rule 2: Consequences — The refracted angle r = 90° − i_p; the reflected light is completely polarised, with vibrations perpendicular to the plane of incidence; the refracted light is partially polarised.

  • Easy : i_p = 57° ⇒ r = 33°.
  • Medium : Glare off a lake is cut by polaroid.
  • Exam (CET) : At i_p the refracted light is? (1) partially polarised (2) completely polarised (3) unpolarised (4) absent → answer 1)

Rule 3: Polarisation by scattering — Sunlight scattered by air molecules (the sky) is completely polarised at 90°; the sky is blue as short wavelengths scatter more (∝ 1 ÷ λ⁴).

  • Easy : Blue (450 nm) vs red (650 nm) : (650 ÷ 450)⁴ = 4.35 times more scattering.
  • Medium : The sky is red at sunrise (long path).
  • Exam (CET) : Scattering ∝ ? (1) 1/λ⁴ (2) λ (3) 1/λ (4) λ⁴ → answer 1)

Derivation : tan i_p = n

  • Step 1 : Snell : n = sin i_p ÷ sin r.
  • Step 2 : Brewster condition : reflected ⊥ refracted ⇒ i_p + r = 90° ⇒ sin r = cos i_p.
  • Step 3 : n = sin i_p ÷ cos i_p.
  • Step 4 : n = tan i_p.
Remember

tan i_p = n; i_p + r = 90°; scattering ∝ 1/λ⁴.

Common mistake / CET trap

At i_p only the reflected ray is completely polarised, not the refracted. It is tan i_p = n, not sin.

Shortcut

Memory clue : "tan = n; the two rays at a right angle".

Spot the difference

Materialni_p
Glass1.556.3°
Water1.3353.1°
Diamond2.4267.5°

Practice questions

Easy
  1. 1Brewster's law?
  2. 2i_p + r = ?
  3. 3Why is the sky blue?
Medium
  1. 4n = 1 : i_p?
  2. 5i_p = 53° : n? (tan 53° = 1.33)
  3. 6i_p = 60° : r?
Challenging
  1. 7n = 1.73 : i_p and r?
  2. 8(700 ÷ 400)⁴ = ?
Show answers (try them yourself first)
  1. 1tan i_p = n
  2. 290°
  3. 3Shorter wavelengths scatter more
  4. 4i_p = 45°
  5. 5n = 1.33
  6. 6r = 30°
  7. 7i_p = 60°, r = 30°
  8. 89.38

Interference and Young's double-slit experiment

From daily life

The colours on a soap bubble are due to interference of light.

Simple definition

Young's experiment : light from one source passes through two close slits and forms bright and dark stripes on a screen.

Young's double-slit : d, D and fringe width βS₁S₂dDββ = λ D ÷ d

In the figure : Rays from S₁ and S₂ make bright or dark stripes on the screen according to the path difference.

Live figure

Interference and Young's double-slit experiment

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Open in the app

Rule 1: Coherent sources — Sources with a constant phase difference and equal f. Two slits S₁, S₂ lit by one source give coherent sources. The slits must be close (small d) and narrow.

  • Easy : Two separate bulbs are not coherent.
  • Medium : A laser : an excellent coherent source.
  • Exam (CET) : Needed for a lasting interference pattern? (1) coherence (2) high speed (3) different f (4) polarisation → answer 1)

Rule 2: Fringe width — Bright fringes : x = n λ D ÷ d; dark : x = (n + ½) λ D ÷ d. β = λ D ÷ d (uniform). Equally spaced stripes on the screen.

  • Easy : λ = 600 nm, D = 1.5 m, d = 0.3 mm : β = 3 mm.
  • Medium : Fifth bright fringe : x = 5 β = 15 mm.
  • Exam (CET) : d halved ⇒ β? (1) doubles (2) halves (3) same (4) four times → answer 1). Trick : β ∝ 1/d.

Rule 3: Medium, intensity and derivation — The whole set-up in a medium : β' = β ÷ n. Intensities : I_max ÷ I_min = (√I₁ + √I₂)² ÷ (√I₁ − √I₂)². With white light the centre is white and the edges coloured.

  • Easy : Water n = 4/3 : β' = 3 × 3/4 = 2.25 mm.
  • Medium : I₁ : I₂ = 4 : 1 ⇒ √ = 2 : 1 ⇒ I_max : I_min = 9 : 1.
  • Exam (CET) : Slit intensities 9 : 1 ⇒ I_max : I_min? (1) 4 : 1 (2) 16 : 4 (3) 9 : 1 (4) 10 : 8 → answer 1)

Derivation : β = λ D ÷ d

  • Step 1 : A point P on the screen at distance x from the centre; path difference Δ = S₂P − S₁P.
  • Step 2 : For small angles Δ ≈ d x ÷ D.
  • Step 3 : Bright fringe : Δ = n λ ⇒ x_n = n λ D ÷ d.
  • Step 4 : Spacing of adjacent fringes β = x_{n+1} − x_n = λ D ÷ d.
Remember

β = λD/d; x_n = nλD/d; β ∝ λ, D, 1/d; in a medium β/n.

Common mistake / CET trap

Do not mix d (slit separation) and a (slit width). Use the same units (m). In water it is only λ that changes.

Shortcut

Memory clue : "long λ, long D, small d ⇒ wide stripes".

Spot the difference

QuantityFormulaUnit / dimensions
βλ D ÷ dm; [L]
Bright fringex = nλD/dm
Dark fringex = (n+½)λD/dm
Path differenced x ÷ Dm

Practice questions

Easy
  1. 1What are coherent sources?
  2. 2Formula for β?
  3. 3Increasing D : β?
Medium
  1. 4λ = 500 nm, D = 2 m, d = 1 mm : β?
  2. 5Third bright fringe x for the same?
  3. 6β = 2 mm in air; in water (n = 4/3) β'?
Challenging
  1. 7λ from 600 nm to 400 nm : ratio of β?
  2. 8I₁ : I₂ = 1 : 16 : I_max : I_min?
Show answers (try them yourself first)
  1. 1Sources with constant phase difference and equal f
  2. 2λD/d
  3. 3β increases
  4. 4β = 1 mm
  5. 5x = 3 mm
  6. 6β' = 1.5 mm
  7. 7β ÷ β' = 3 : 2
  8. 8√ = 1 : 4 ⇒ (5)² : (3)² = 25 : 9

Diffraction at a single slit

From daily life

Looking at a lamp through a narrow gap between your fingers shows stripes around the light.

Simple definition

Diffraction is the spreading of light into the geometric shadow as it passes a slit.

Single-slit diffraction : broad central maximumaD2λD ÷ aθ

In the figure : Slit a, screen at distance D; the central band is wide and bright.

Live figure

Diffraction at a single slit

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Open in the app

Rule 1: Diffraction — Light bends around the edge of an obstacle or slit into the shadow. It is noticeable when the slit width a ≈ λ. Smaller a ⇒ more diffraction.

  • Easy : Sound (λ ≈ 1 m) bends through a doorway; light (λ ≈ 5 × 10⁻⁷ m) does not.
  • Medium : Radio waves reach behind buildings.
  • Exam (CET) : If a ≫ λ the diffraction is? (1) negligible (2) large (3) 100 % (4) infinite → answer 1)

Rule 2: Minima and the central maximum — a sin θ = n λ (n = ±1, ±2 …) gives the dark minima. The angular half-width of the central maximum θ ≈ λ ÷ a; its linear width 2 λ D ÷ a.

  • Easy : λ = 500 nm, a = 0.1 mm, D = 2 m : θ = 5 × 10⁻³ rad; width = 20 mm.
  • Medium : a halved ⇒ width doubles.
  • Exam (CET) : a doubled : width of the central maximum? (1) halves (2) doubles (3) same (4) four times → answer 1)

Rule 3: Diffraction vs interference — Interference : equally wide stripes of nearly equal intensity (ideal). Diffraction : a wide, brightest central band; the side bands are narrow and dim (secondary maxima about 4.5 %, 1.6 % …).

  • Easy : The first secondary maximum is about 4.5 % of the central one.
  • Medium : Young's experiment in reality combines both.
  • Exam (CET) : The central bright band is? (1) widest and brightest (2) narrowest (3) dim (4) dark → answer 1)

Derivation : a sin θ = n λ

  • Step 1 : Slit AB of width a; split it into two equal halves of a/2.
  • Step 2 : Path difference between a point in the upper half and the matching point in the lower half = (a/2) sin θ.
  • Step 3 : Destructive : (a/2) sin θ = λ/2 ⇒ a sin θ = λ (first minimum).
  • Step 4 : In general a sin θ = n λ; for small θ : θ = nλ ÷ a ⇒ band width 2λD ÷ a.
Remember

a sin θ = nλ; central width 2λD/a; θ ≈ λ/a.

Common mistake / CET trap

A larger a gives a narrower band (inverse). Do not mix with Young's β = λD/d; here it is 2λD/a.

Shortcut

Memory clue : "narrow slit ⇒ wide band".

Spot the difference

PointInterference (Young)Diffraction (single slit)
Stripesequal widthcentral one wider
Intensitynearly equalfalls quickly
Conditiontwo coherent sourcesa ≈ λ

Practice questions

Easy
  1. 1What is diffraction?
  2. 2Condition for minima?
  3. 3Smaller a ⇒ diffraction?
Medium
  1. 4λ = 600 nm, a = 0.2 mm : θ (first minimum)?
  2. 5Same, D = 1 m : width of central band?
  3. 6a doubled ⇒ θ?
Challenging
  1. 7λ = 500 nm : with a = 0.2 mm at a sin θ = 2λ; θ?
  2. 8Red or blue : which diffracts more?
Show answers (try them yourself first)
  1. 1Bending of light around an obstacle
  2. 2a sin θ = nλ
  3. 3More
  4. 4θ = 3 × 10⁻³ rad
  5. 52 × 3 × 10⁻³ × 1 = 6 mm
  6. 6θ halves
  7. 7θ = 2 × 5 × 10⁻⁷ ÷ 2 × 10⁻⁴ = 5 × 10⁻³ rad
  8. 8Red (larger λ)

Resolving power of optical instruments

From daily life

From far away two close headlights look like a single lamp.

Simple definition

Resolving power is an instrument's ability to show two close objects as separate; diffraction limits it.

Rayleigh criterion : diffraction images of two objectsnot resolvedjust resolvedclearly resolvedΔθ = 1.22 λ ÷ D

In the figure : As two images come closer they merge; Δθ = 1.22λ/D is the limit.

Rule 1: Rayleigh's criterion — Two objects are just resolved when the first minimum of one diffraction pattern falls on the central maximum of the other. For a circular aperture the least angle is Δθ = 1.22 λ ÷ D.

  • Easy : Eye D = 2 mm, λ = 550 nm : Δθ = 1.22 × 5.5 × 10⁻⁷ ÷ 2 × 10⁻³ = 3.4 × 10⁻⁴ rad.
  • Medium : From 10 m that angle corresponds to 3.4 mm.
  • Exam (CET) : Δθ ∝ ? (1) λ/D (2) D/λ (3) λD (4) 1/(λD) → answer 1)

Rule 2: Resolving power of a telescope — R = 1 ÷ Δθ = D ÷ (1.22 λ). A bigger objective (D) ⇒ larger R; a smaller λ ⇒ larger R.

  • Easy : D = 0.5 m, λ = 550 nm : Δθ = 1.34 × 10⁻⁶ rad.
  • Medium : D doubled ⇒ R doubles.
  • Exam (CET) : In red or blue light is R greater? (1) blue (2) red (3) same (4) none → answer 1)

Rule 3: Resolving power of a microscope — Least separation a = 0.61 λ ÷ (n sin θ) = 0.61 λ ÷ N.A.; R = 1 ÷ a = N.A. ÷ (0.61 λ). An oil-immersion objective (larger n) ⇒ larger N.A.

  • Easy : λ = 500 nm, N.A. = 1.2 : a = 2.54 × 10⁻⁷ m.
  • Medium : Using a short λ (electron microscope) ⇒ a extremely small.
  • Exam (CET) : N.A. doubled ⇒ a? (1) halves (2) doubles (3) same (4) four times → answer 1)

Derivation : telescope R = D ÷ (1.22 λ)

  • Step 1 : For a circular aperture D the first minimum is at sin θ = 1.22 λ ÷ D.
  • Step 2 : Rayleigh : least resolvable angle = angle of the first minimum ⇒ Δθ = 1.22 λ ÷ D (small angle).
  • Step 3 : Resolving power R = 1 ÷ Δθ.
  • Step 4 : R = D ÷ (1.22 λ).
Remember

Δθ = 1.22λ/D; R(telescope) = D/(1.22λ); a = 0.61λ/N.A.

Common mistake / CET trap

For a telescope R is based on angle, for a microscope on distance; do not swap. A smaller λ gives larger R in both.

Shortcut

Memory clue : "big D, small λ = sharp".

Spot the difference

InstrumentLimitR
TelescopeΔθ = 1.22λ/DD/(1.22λ)
Microscopea = 0.61λ/N.A.N.A./(0.61λ)
EyeD ≈ 2 mm—

Practice questions

Easy
  1. 1State Rayleigh's criterion.
  2. 2R of a telescope?
  3. 3N.A. = ?
Medium
  1. 4D = 1 m, λ = 500 nm : Δθ?
  2. 5λ = 600 nm, N.A. = 1.0 : a?
  3. 6D ×4 ⇒ R?
Challenging
  1. 7λ halved ⇒ R of a microscope?
  2. 8Two stars 2 × 10⁻⁷ rad apart; D = 1 m, λ = 500 nm : resolved?
Show answers (try them yourself first)
  1. 1First minimum of one on the central maximum of the other
  2. 2D ÷ 1.22λ
  3. 3n sin θ
  4. 4Δθ = 6.1 × 10⁻⁷ rad
  5. 5a = 3.66 × 10⁻⁷ m
  6. 6R ×4
  7. 7R doubles
  8. 8Limit 6.1 × 10⁻⁷ rad > 2 × 10⁻⁷ ⇒ not resolved

Quick revision of the whole lesson

ConceptKey formula / factUnit / dimensions
Lightn = c/v; λ' = λ/n; f unchangedno unit
Snellsin i/sin r = n₂/n₁ = v₁/v₂—
MalusI = I₀cos²θ; unpolarised → I₀/2W/m²
Brewstertan i_p = n; i_p + r = 90°—
Youngβ = λD/d; x_n = nλD/dm; [L]
Single slita sin θ = nλ; width = 2λD/am
TelescopeΔθ = 1.22λ/D; R = D/1.22λrad
Microscopea = 0.61λ/(n sin θ)m

🎯 Most important points for the exam

  • f unchanged in a medium; λ' = λ/n; v = c/n.
  • Snell : sin i/sin r = n₂/n₁ (proved by Huygens).
  • Malus : I = I₀cos²θ; half at the first polariser.
  • Brewster : tan i_p = n; reflected ⊥ refracted.
  • Young : β = λD/d; β/n in water.
  • Single slit : a sin θ = nλ; central width 2λD/a.
  • Resolution : Δθ = 1.22λ/D; microscope a = 0.61λ/N.A.
Study mantra

Identify the wave effect → find the path difference → right formula → λ, D, d in metres → answer.