← Physics (312) · Class 12

L-21: Dispersion and Scattering of Light

Physics — Class 12 · NIOS Code 312 · Module 6 · Source: 312_Physics_Eng_Lesson21.pdf

Dispersion and Scattering of Light

When white light passes through a prism, colour bands appear — this is dispersion, distinct from ordinary refraction. Scattering explains why the sky is blue and the Sun looks red near the horizon. This lesson is part of Module 6 (Optics).

NIOS objectives: dispersion; prism deviation and μ; wavelength dependence; primary/secondary rainbows; scattering applications; Raman effect.

21.1 Dispersion of Light

Visible light is a small part of the EM spectrum. Sunlight contains seven wavelengths (VIBGYOR). In a dispersive medium, different wavelengths travel at different speeds → different μ → separation of colours. Air/vacuum are nearly non-dispersive for visible light.

μ depends on nature of material and wavelength λ. Spectral dispersive power: Δμ/Δλ.

21.1.1 Dispersion through a Prism

A glass slab does not show clear dispersion (emergent rays stay parallel to incident). A prism widely separates colours on a screen — forming a spectrum. Violet bends most; red least.

Fig 21.2 — Dispersion by a Prism A white in R V δV > δR · μV > μR · VIBGYOR spectrum
Fig 21.2 — White light splits into spectrum; violet deviated more than red

21.1.2 Angle of Deviation

δ = (i + e) − (r₁ + r₂)  |  i + e = A + δ
δ = angle of deviation
A = refracting angle of prism
i, e = angles of incidence and emergence
r₁ + r₂ = A

Minimum deviation (δm): e = i; r₁ = r₂ = A/2; ray passes symmetrically parallel to base.

μ = sin((A + δm)/2) / sin(A/2)
Used to find refractive index experimentally
δm differs for each colour → dispersion
Emergent beam brightest at minimum deviation
δ = (μ − 1) A  (small-angle prism)
Valid when A, i, r are small
Since μV > μR, we have δV > δR
μ increases as wavelength decreases

21.1.3 Angular Dispersion and Dispersive Power

ω = (δV − δR) / δY = Δμ / μ
ω = dispersive power (dimensionless)
Angular dispersion = δV − δR
δY ≈ mean deviation for yellow

Rainbow Formation

Sunlight dispersed by water droplets in air. With Sun behind observer, coloured arcs appear.

  • Primary rainbow: 2 refractions + 1 internal reflection. Minimum deviation ~137°29′ → cone ~42° at eye. Outer edge red, inner violet (VIBGYOR).
  • Secondary rainbow: 2 refractions + 2 internal reflections. Red inner, violet outer — colours reversed. Fainter; lies above primary. Dark band (Alexander's dark band) between them.
Fig 21.5–21.7 — Primary & Secondary Rainbow drop Sun rays to eye ~42° primary secondary (fainter) Sun behind observer · dispersion in water drops
Fig 21.5–21.7 — Primary bow: 1 reflection; secondary: 2 reflections, reversed colours

21.2 Scattering of Light in Atmosphere

21.2.1 Rayleigh Scattering

Interaction of radiation with particles much smaller than λ. Two-step process: absorption then re-emission in all directions (not reflection). Particle size must be < λ for wavelength-dependent scattering.

I ∝ 1 / λ⁴  (Rayleigh's law)
I = intensity of scattered light
Shorter λ scattered much more intensely
Blue scattered ~6× more than red
Large droplets scatter all wavelengths equally → white
Fig 21.9 — Blue Sky & Red Sun Sun blue scattered → sky blue red reaches eye at horizon long path at sunrise/sunset removes blue → Sun appears red
Fig 21.9 — Rayleigh scattering: short wavelengths deflected; red transmitted at low Sun
  • Blue sky: air molecules scatter blue/violet more; eye less sensitive to violet → sky looks blue.
  • White clouds: water droplets > λ scatter all colours equally.
  • Dense clouds black: absorb sunlight rather than transmit.
  • Red Sun at sunrise/sunset: long atmospheric path removes blue by scattering.
  • Space: no scattering particles → sky black to astronauts.
  • Deep blue after rain: dust removed; purer Rayleigh scattering.

21.2.2 Raman Effect

When light scatters from transparent solids, liquids, or gases, the scattered radiation may have frequency greater or less than incident frequency (discovered by C.V. Raman, 1928; Nobel Prize 1930).

  • No energy exchange: frequency unchanged (Rayleigh scattering).
  • Stokes lines: scattered frequency less than incident (light loses energy to substance).
  • Anti-Stokes lines: scattered frequency greater than incident (substance in excited state gives energy to light).

Raman spectrum reveals molecular structure; analogue of Compton effect for X-rays.

         DISPERSION & SCATTERING — KEY POINTS
         =====================================
    Dispersion      :  μ depends on λ; prism splits VIBGYOR
    Min deviation   :  μ = sin((A+δm)/2)/sin(A/2)
    Small prism     :  δ = (μ−1)A ;  δV > δR
    Dispersive power:  ω = Δμ/μ
    Primary rainbow :  2 refractions + 1 reflection (~42°)
    Secondary bow   :  2 refractions + 2 reflections (reversed)
    Rayleigh scatter:  I ∝ 1/λ⁴
    Raman effect    :  Stokes & anti-Stokes frequency shifts

Quick Revision

  • Dispersion = splitting by wavelength-dependent μ; prism not glass slab.
  • Violet deviated most; red least; δV > δR.
  • Minimum deviation gives μ from prism geometry.
  • Primary rainbow: red outside; secondary: colours reversed, fainter.
  • Rayleigh: I ∝ 1/λ⁴ explains blue sky and red Sun.
  • Clouds white (large droplets); Raman effect shifts scattered frequency.
20 cards · click any card to flip
Dispersion of light
Splitting of white light into constituent colours/wavelengths because μ varies with λ in a dispersive medium. Different from simple refraction.
Dispersive vs non-dispersive medium
Dispersive: different λ travel at different speeds (glass prism). Non-dispersive: all visible λ same speed (air, vacuum).
Why prism, not glass slab?
Prism widely separates colours on screen; slab emergent rays stay parallel to incident — colours do not stay separated.
VIBGYOR order in prism
Violet deviated most (μV highest); red least. δV > δR because shorter λ has higher μ in glass.
Angle of deviation relation
δ = (i+e) − A; or i + e = A + δ. r₁ + r₂ = A. δ minimum when e = i and ray symmetric in prism.
Refractive index from prism
μ = sin((A+δm)/2)/sin(A/2) at minimum deviation. δm depends on colour/wavelength.
Small-angle prism formula
δ = (μ−1)A. Useful when A, i, r are small. Shows deviation proportional to μ−1.
Dispersive power
ω = (δV−δR)/δY = Δμ/μ. Measures how strongly material separates colours. Spectral dispersive power: Δμ/Δλ.
Primary rainbow
2 refractions + 1 internal reflection in water drop. ~42° cone at eye. Red outer, violet inner (VIBGYOR).
Secondary rainbow
2 refractions + 2 internal reflections. Fainter; above primary. Red inner, violet outer — colours reversed.
Rayleigh scattering law
I ∝ 1/λ⁴. Applies when scatterer size < λ. Blue scattered most; red least. Not ordinary reflection.
Why is the sky blue?
Air molecules scatter short wavelengths more (Rayleigh). Blue/violet scattered across sky; eye sees predominantly blue.
Why are clouds white?
Water droplets larger than λ scatter all visible wavelengths with nearly equal intensity → white.
Red Sun at sunrise/sunset
Long atmospheric path scatters away blue/violet (90° scattering). Remaining light rich in red reaches observer.
Sky from space
No air molecules or dust at high altitude → no scattering → sky appears black to astronauts.
Raman effect
Scattered light frequency may differ from incident. Stokes lines (lower ν); anti-Stokes (higher ν). Nobel 1930 to C.V. Raman.
Stokes vs anti-Stokes lines
Stokes: scattered frequency less than incident (energy to substance). Anti-Stokes: greater (excited substance gives energy to light).
Monochromatic vs polychromatic
Monochromatic: single wavelength/colour. Polychromatic: many wavelengths (e.g. sunlight). Spectrum shows separated colours.
μ and wavelength
μ = λa/λm = c/v. Higher μ for shorter λ in glass. Causes dispersion and different deviations per colour.
Does dispersion depend on prism size?
No — depends on material (μ vs λ) and angle A, not physical size. Separation needs adequate path in prism shape.

Q1. Dispersion of light occurs because:

Q2. In a prism spectrum, which colour is deviated the most?

Q3. For a small-angle prism, the angle of deviation is:

Q4. At minimum deviation in a prism:

Q5. A primary rainbow is formed by:

Q6. Compared to the primary rainbow, the secondary rainbow has:

Q7. According to Rayleigh's law, intensity of scattered light varies as:

Q8. The blue colour of the sky is mainly due to:

Q9. Clouds appear white because:

Q10. In the Raman effect, anti-Stokes lines have frequency:

δ = (i + e) − (r₁ + r₂)
i + e = A + δ
r₁ + r₂ = A
μ = sin((A + δ_m)/2) / sin(A/2)
δ_m: e = i, r₁ = r₂ = A/2
δ = (μ − 1) A
ω = (δ_V − δ_R)/δ_Y = Δμ/μ
Angular dispersion = δ_V − δ_R
μ = c/v = λ_a/λ_m
I ∝ 1/λ⁴
Δμ/Δλ — spectral dispersive power

1. Formulas & Definitions

Full Ch 21 study guide — prism dispersion, deviation, dispersive power, Rayleigh scattering, and rainbows.

δ = (i + e) − (r₁ + r₂)

Definition: Angle of deviation — angle between incident and emergent rays through a prism.

Derivation

From geometry of refraction at two faces; total bend equals excess of (i+e) over (r₁+r₂).

Variables

δ = deviation · i, e = incidence, emergence · r₁, r₂ = refraction angles

Why it works

Measures how much light bends; differs for each colour → dispersion.

Historical context

Alternative form: i + e = A + δ when r₁ + r₂ = A.

Deep understanding

Violet has larger δ than red because μ_V > μ_R.

2. Diagrams & Visuals

δ

Color-coded visual · step-by-step breakdown below

  1. Measure i and e at prism faces
  2. Find r₁, r₂ inside prism
  3. δ = (i+e) − (r₁+r₂)
  4. Compare colours for dispersion

3. Solved Examples

Basic

Q: i=60°, e=60°, r₁+r₂=40°.

Solution: δ=80°

Answer: 80°

Intermediate

Q: Violet vs red?

Solution: δ_V > δ_R

Answer: More violet bend

Advanced

Q: Symmetric ray?

Solution: e=i at δ_m

Answer: Min deviation

Exam

Q: Deviation angle?

Solution: δ=(i+e)−(r₁+r₂)

Answer: Sec 21.1.2

i + e = A + δ

Definition: Relation between incidence, emergence, prism angle A, and deviation δ.

Derivation

From prism geometry with r₁ + r₂ = A.

Variables

A = refracting angle of prism

Why it works

Useful when only i, e, A known; links to experimental measurements.

Historical context

Combined with Snell's law at each face for full analysis.

Deep understanding

At minimum deviation: e = i and δ = δ_m.

2. Diagrams & Visuals

i + e = A + δ

Color-coded visual · step-by-step breakdown below

  1. Know prism angle A
  2. Measure i, e, δ
  3. Check i+e = A+δ
  4. Solve for unknown

3. Solved Examples

Basic

Q: A=60°, δ=40°, i=50°.

Solution: e=50°

Answer: e=50°

Intermediate

Q: At δ_m, i=45°, A=60°.

Solution: e=45°

Answer: Symmetric

Advanced

Q: Increase i?

Solution: δ changes

Answer: Not constant

Exam

Q: Prism relation?

Solution: i+e=A+δ

Answer: Sec 21.1.2

r₁ + r₂ = A

Definition: Sum of internal refraction angles equals prism refracting angle.

Derivation

From quadrilateral geometry inside prism (apex angle A).

Variables

r₁ at first face, r₂ at second face

Why it works

Connects external angles to prism geometry; used in μ derivation.

Historical context

At minimum deviation: r₁ = r₂ = A/2.

Deep understanding

Valid for any colour; r values differ per wavelength.

2. Diagrams & Visuals

r₁+r₂ = A

Color-coded visual · step-by-step breakdown below

  1. Identify prism angle A
  2. r₁ at entry face
  3. r₂ at exit face
  4. r₁+r₂ = A always

3. Solved Examples

Basic

Q: A=60°.

Solution: r₁+r₂=60°

Answer: 60°

Intermediate

Q: δ_m case?

Solution: r₁=r₂=30°

Answer: A/2 each

Advanced

Q: Different colour?

Solution: r change, sum=A

Answer: Same A

Exam

Q: Internal angles?

Solution: r₁+r₂=A

Answer: Sec 21.1.2

μ = sin((A + δ_m)/2) / sin(A/2)

Definition: Refractive index from prism at minimum deviation.

Derivation

Apply Snell's law at symmetric configuration e = i, r₁ = r₂ = A/2.

Variables

δ_m = minimum deviation angle · μ for that wavelength

Why it works

Standard experiment to measure μ of prism material.

Historical context

Emergent beam brightest at δ_m; each colour has its own δ_m and μ.

Deep understanding

μ_V > μ_R → different δ_m for violet and red.

2. Diagrams & Visuals

μ = sin((A+δm)/2)/sin(A/2)

Color-coded visual · step-by-step breakdown below

  1. Find prism angle A
  2. Measure δ_m for colour
  3. μ = sin((A+δ_m)/2)/sin(A/2)
  4. Repeat per wavelength

3. Solved Examples

Basic

Q: A=60°, δ_m=38°.

Solution: μ≈1.52

Answer: ~1.52 glass

Intermediate

Q: Violet δ_m larger?

Solution: μ_V higher

Answer: Shorter λ

Advanced

Q: Yellow light μ?

Solution: Between μ_R and μ_V

Answer: Middle

Exam

Q: Prism μ formula?

Solution: sin((A+δm)/2)/sin(A/2)

Answer: Sec 21.1.2

δ_m: e = i, r₁ = r₂ = A/2

Definition: Conditions for minimum deviation through a prism.

Derivation

Symmetric ray path parallel to base; deviation is smallest.

Variables

δ_m = minimum δ · emergent beam brightest here

Why it works

Simplifies experiment; enables μ formula above.

Historical context

Only one angle i gives δ_m for given colour.

Deep understanding

Ray passes symmetrically — equal bending at both faces.

2. Diagrams & Visuals

symmetric path

Color-coded visual · step-by-step breakdown below

  1. Vary angle of incidence
  2. Find smallest δ = δ_m
  3. At δ_m: e = i
  4. r₁ = r₂ = A/2

3. Solved Examples

Basic

Q: A=60° at δ_m.

Solution: r₁=r₂=30°

Answer: 30° each

Intermediate

Q: e ≠ i?

Solution: δ > δ_m

Answer: Not minimum

Advanced

Q: Use for μ?

Solution: Yes — standard method

Answer: Experiment

Exam

Q: Min deviation condition?

Solution: e=i, r=A/2

Answer: Sec 21.1.2

δ = (μ − 1) A

Definition: Approximate deviation for thin/small-angle prism.

Derivation

Small A, i, r → deviation proportional to (μ−1) and prism angle.

Variables

Valid when angles are small (radians)

Why it works

Quick estimate; shows δ ∝ (μ−1) and ∝ A.

Historical context

Since μ_V > μ_R, δ_V > δ_R — explains colour separation.

Deep understanding

μ increases as wavelength λ decreases in glass.

2. Diagrams & Visuals

δ = (μ−1)A

Color-coded visual · step-by-step breakdown below

  1. Confirm small angles
  2. Find μ for colour
  3. δ ≈ (μ−1)A
  4. Compare colours

3. Solved Examples

Basic

Q: μ=1.5, A=4°.

Solution: δ≈2°

Answer:

Intermediate

Q: Double A?

Solution: δ doubles

Answer: ∝ A

Advanced

Q: μ=1 (vacuum)?

Solution: δ=0

Answer: No deviation

Exam

Q: Thin prism?

Solution: δ=(μ−1)A

Answer: Sec 21.1.2

ω = (δ_V − δ_R)/δ_Y = Δμ/μ

Definition: Dispersive power — how strongly a material separates colours.

Derivation

Ratio of angular dispersion to mean deviation for yellow.

Variables

ω dimensionless · δ_Y ≈ mean deviation · Δμ = μ_V − μ_R

Why it works

Compare dispersive ability of crown vs flint glass in spectrometers.

Historical context

Spectral dispersive power also: Δμ/Δλ.

Deep understanding

Higher ω → wider spectrum for same prism.

2. Diagrams & Visuals

ω = Δδ/δ_Y

Color-coded visual · step-by-step breakdown below

  1. Measure δ_V and δ_R
  2. Find mean δ_Y
  3. ω = (δ_V−δ_R)/δ_Y
  4. Or ω = Δμ/μ

3. Solved Examples

Basic

Q: δ_V−δ_R=2°, δ_Y=40°.

Solution: ω=0.05

Answer: 0.05

Intermediate

Q: Flint vs crown?

Solution: Flint ω higher

Answer: More dispersion

Advanced

Q: ω units?

Solution: Dimensionless

Answer: Ratio

Exam

Q: Dispersive power?

Solution: Δμ/μ

Answer: Sec 21.1.3

Angular dispersion = δ_V − δ_R

Definition: Difference in deviation between violet and red components.

Derivation

Each colour has different μ and hence different δ.

Variables

δ_V > δ_R always in glass prism

Why it works

Direct measure of colour spread on screen.

Historical context

Used in numerator of dispersive power ω.

Deep understanding

Prism separates white light into VIBGYOR spectrum.

2. Diagrams & Visuals

δ_V − δ_R

Color-coded visual · step-by-step breakdown below

  1. Deviate white light through prism
  2. Measure δ for violet and red
  3. Angular dispersion = δ_V − δ_R
  4. Larger → wider spectrum

3. Solved Examples

Basic

Q: δ_V=42°, δ_R=40°.

Solution: Dispersion=2°

Answer:

Intermediate

Q: Larger A?

Solution: Dispersion increases

Answer: Wider spread

Advanced

Q: Air prism?

Solution: Negligible dispersion

Answer: Non-dispersive

Exam

Q: Angular dispersion?

Solution: δ_V−δ_R

Answer: Sec 21.1.3

μ = c/v = λ_a/λ_m

Definition: Refractive index in terms of speed and wavelength.

Derivation

Light slows in denser medium; wavelength in medium shrinks.

Variables

λ_a in air/vacuum · λ_m in medium · v = speed in medium

Why it works

Links dispersion to wavelength — shorter λ → higher μ in glass.

Historical context

μ depends on material and λ; causes VIBGYOR separation.

Deep understanding

Dispersion does not depend on prism size — only material and A.

2. Diagrams & Visuals

μ = c/v = λ_a/λ_m

Color-coded visual · step-by-step breakdown below

  1. Speed c in vacuum, v in medium
  2. μ = c/v
  3. Or compare wavelengths
  4. μ > 1 for glass

3. Solved Examples

Basic

Q: v=c/1.5.

Solution: μ=1.5

Answer: 1.5

Intermediate

Q: Shorter λ in glass?

Solution: μ higher

Answer: Dispersion

Advanced

Q: Vacuum?

Solution: μ=1

Answer: Reference

Exam

Q: μ definition?

Solution: c/v

Answer: Optics link

I ∝ 1/λ⁴

Definition: Rayleigh scattering law — intensity of scattered light vs wavelength.

Derivation

Applies when scatterer size ≪ λ; elastic scattering by small particles.

Variables

I = scattered intensity · λ = wavelength of incident light

Why it works

Explains blue sky, red sunset, and preferential scattering of short waves.

Historical context

Not reflection — absorb then re-emit in all directions.

Deep understanding

Blue scattered ~6× more than red (λ ratio 700/400 to 4th power). Large droplets → all λ equally → white clouds.

2. Diagrams & Visuals

I ∝ 1/λ⁴

Color-coded visual · step-by-step breakdown below

  1. Scatterer size < λ
  2. Shorter λ → much higher I
  3. I ∝ 1/λ⁴
  4. Apply to sky/sunset

3. Solved Examples

Basic

Q: Blue vs red λ halved?

Solution: I ~16× higher for blue

Answer: λ⁴ effect

Intermediate

Q: Cloud droplets large?

Solution: All λ equal

Answer: White cloud

Advanced

Q: Space sky?

Solution: No scatterers

Answer: Black sky

Exam

Q: Rayleigh law?

Solution: I∝1/λ⁴

Answer: Sec 21.2.1

Δμ/Δλ — spectral dispersive power

Definition: Rate of change of refractive index with wavelength.

Derivation

From dispersion curve μ(λ); steeper slope → stronger colour separation.

Variables

Δμ = change in μ · Δλ = change in λ

Why it works

Quantifies how μ varies across the visible spectrum.

Historical context

Related to angular dispersive power ω = Δμ/μ.

Deep understanding

μ increases as λ decreases — violet bends most in prism.

2. Diagrams & Visuals

μ vs λ curve

Color-coded visual · step-by-step breakdown below

  1. Plot μ for different λ
  2. Find slope Δμ/Δλ
  3. Steeper → more dispersion
  4. Material property

3. Solved Examples

Basic

Q: μ changes 0.02 over 100 nm.

Solution: Δμ/Δλ=2×10⁻⁴/nm

Answer: Slope

Intermediate

Q: Flint glass?

Solution: Larger Δμ/Δλ

Answer: High dispersion

Advanced

Q: Air for visible?

Solution: ≈0

Answer: Non-dispersive

Exam

Q: Spectral dispersive power?

Solution: Δμ/Δλ

Answer: Sec 21.1

5. Special Features & Extras

Complete study guide for Dispersion and Scattering of Light.

Exam Tips & Tricks

  • Dispersion ≠ refraction — splitting by λ-dependent μ, not just bending.
  • Prism not slab — slab keeps emergent rays parallel; prism separates colours.
  • VIBGYOR: violet deviated most (μ_V highest); red least.
  • μ from prism: use minimum deviation formula with measured δ_m.
  • Rayleigh: I ∝ 1/λ⁴ — blue sky, red Sun at horizon.
  • Clouds white: droplets > λ scatter all colours equally.
  • Rainbows: primary (1 reflection, red outside); secondary (2 reflections, reversed).

Common Student Mistakes

  • Using glass slab instead of prism for dispersion questions
  • Forgetting δ_V > δ_R because μ_V > μ_R
  • Applying Rayleigh law when particle size ≫ λ (clouds)
  • Confusing dispersion with scattering (prism vs sky)
  • Using wrong δ in μ formula — must be δ_m at minimum deviation
  • Thinking prism size alone causes dispersion (it's material + A)

Memory Aids & Mnemonics

VIBGYOR in prism: "Violet bends Violently, Red Relaxes"
Min deviation: "Symmetric ray — e equals i, r halves A"
Rayleigh: "Short waves scatter Strongly — λ to the fourth"
Rainbow: "Primary: 1 bounce, Red out · Secondary: 2 bounces, Reversed"

Which Formula When?

  • Deviation through prism? → δ = (i+e)−(r₁+r₂) or i+e=A+δ
  • Measure μ experimentally? → μ = sin((A+δ_m)/2)/sin(A/2)
  • Thin prism estimate? → δ = (μ−1)A
  • How much colour spread? → ω = (δ_V−δ_R)/δ_Y
  • Blue sky / red sunset? → I ∝ 1/λ⁴
  • μ vs wavelength link? → μ = c/v = λ_a/λ_m
  • Raman effect? → frequency shift (Stokes/anti-Stokes) — not λ⁴ law

QUICK REFERENCE — Ch 21 Dispersion & Scattering

δ = (i + e) − (r₁ + r₂)i + e = A + δr₁ + r₂ = Aμ = sin((A + δ_m)/2) / sin(A/2)δ_m: e = i, r₁ = r₂ = A/2δ = (μ − 1) Aω = (δ_V − δ_R)/δ_Y = Δμ/μAngular dispersion = δ_V − δ_Rμ = c/v = λ_a/λ_mI ∝ 1/λ⁴Δμ/Δλ — spectral dispersive power

Key facts: μ_V > μ_R · δ_V > δ_R · air ≈ non-dispersive · scatterer < λ for Rayleigh

Rainbows: primary ~42° (red outer) · secondary fainter, reversed, above primary

Tip: For prism numericals, always check if question specifies minimum deviation before using μ formula.

PYQ — Previous Year Questions

Extracted from NIOS Physics (312) board exam papers in your PDF. Chapter L21 — Dispersion and Scattering of Light only. Use Model Answer for marking points; Explanation for concept clarity.

Chapter 21 — Dispersion and Scattering of Light (L21)

12 questions · Sections A & B · Sources: 312/TUS/104A, 312/MAY/204A, 312/MAY/204B, 68/ESS/1-312-A

Section A — Objective (1 mark)

PYQ1. In the sky after rains, rainbow is formed due to the phenomenon of (A) interference (B) diffraction (C) polarization (D) dispersion

1 mark · Section A Q2 · 312/TUS/104A

Model Answer

(D) dispersion

Rainbow forms when sunlight undergoes refraction, internal reflection and again refraction in water droplets — different wavelengths deviate by different amounts (dispersion).

Explanation

Primary/secondary rainbows are dispersion phenomena in spherical drops (L21 §21.2). Interference/diffraction/polarization are wave phenomena but not the main cause of the coloured bow.

PYQ2. If we make identical prisms of different types of glass, the broadest spectrum is formed by a (A) soda glass prism (B) crown glass prism (C) flint glass prism (D) quartz glass prism

1 mark · Section A Q12 · 312/MAY/204A

Model Answer

(C) flint glass prism

Flint glass has the highest dispersive power among common glasses → greatest angular separation of colours → broadest spectrum.

Explanation

Dispersive power ω = (μV − μR)/(μY − 1). Flint glass has larger μ variation with λ than crown/soda/quartz (L21 §21.1).

PYQ3. Which of the following colour of white light deviates the most when passes through a prism? (A) Red (B) Violet (C) Yellow (D) Green

1 mark · Section A Q12 · 68/ESS/1-312-A

Model Answer

(B) Violet

Shortest wavelength → highest refractive index in glass → largest angle of deviation.

Explanation

μ increases toward violet end of spectrum; δV > δR. Order in spectrum: VIBGYOR from base of prism outward (L21 §21.1).

PYQ4. The refracting angle of a prism is 30′ and its refractive index is 1.6. Calculate the deviation caused by the prism. (A) 28′ (B) 8′ (C) 30′ (D) 18′

1 mark · Section A Q14 · 68/ESS/1-312-A

Model Answer

Thin-prism formula: δ = (μ − 1) A

δ = (1.6 − 1) × 30′ = 0.6 × 30′ = 18′(D)

Explanation

For small refracting angle, δ ≈ (μ−1)A (L21 §21.1). 30′ = 0.5°; same result in arcminutes.

PYQ5. Read the passage: “Dispersion of light indicates that white light is composed of seven wavelength ranges corresponding to the seven colours of the rainbow.” The phenomenon responsible for the formation of rainbow is: (A) interference (B) diffraction (C) dispersion (D) polarization

1 mark · Section A Q19 (i) · 68/ESS/1-312-A (passage)

Model Answer

(C) dispersion

Explanation

Passage links white light splitting into spectral colours with rainbow formation — both are dispersion (L21 §21.2).

Section A — Short Answer (2 marks)

PYQ6. Write ‘True’ for correct statement and ‘False’ for incorrect statement: (a) Angular dispersion for any two colours is independent of the angle of prism. (b) Angular width of primary rainbow is more than the angular width of secondary rainbow.

2 marks · Section A Q25 · 312/TUS/104A

Model Answer

(a) False — angular dispersion (δV − δR) depends on prism angle A; for thin prism δ = (μ−1)A so dispersion ∝ A.

(b) False — secondary rainbow subtends a larger angle (~50°) than primary (~42°) about the antisolar point.

Explanation

Board T/F on dispersion geometry and rainbow cones (L21 §21.1–21.2). Do not confuse colour-band width inside each bow with overall angular radius.

PYQ7. Match Column—I with Column—II: (a) Formation of rainbow — (i) Interference (ii) Diffraction (iii) Scattering (iv) Dispersion; (b) Blue colour of sky — same options.

2 marks · Section A Q26 · 312/MAY/204A

Model Answer

(a) Formation of rainbow → (iv) Dispersion

(b) Blue colour of sky → (iii) Scattering (Rayleigh scattering of sunlight by air molecules)

Explanation

Rainbow = wavelength-dependent refraction in drops. Blue sky = preferential scattering of short λ (I ∝ 1/λ⁴) — L21 §21.2.

PYQ8. Fill in the blanks (attempt any two parts): (i) A ray of light undergoes ______ twice on passing through a prism. (ii) ______ is the most scattered colour. (iii) The deviation through a prism is minimum when angle of incidence equals angle of ______. (iv) According to Rayleigh's law, intensity of scattered light is inversely proportional to the ______ power of wavelength.

2 marks · Section A Q25 · 68/ESS/1-312-A

Model Answer

(i) refraction (at the two refracting faces)

(ii) violet (or blue — shortest visible λ scattered most)

(iii) emergence (e = i at minimum deviation)

(iv) fourth (I ∝ 1/λ⁴)

Explanation

Prism optics + Rayleigh law blanks from ESS paper (L21 §21.1, §21.2.1). At δm, ray is symmetric inside prism: i = e, r₁ = r₂.

PYQ9. Name any two phenomena based on scattering of light.

2 marks · Section B Q30 · 312/TUS/104A

Model Answer

Any two, e.g.:

  • Blue colour of the sky
  • Red colour of the Sun at sunrise/sunset
  • Tyndall effect / reddening at horizon / deep blue sky after rain

Explanation

All arise from Rayleigh (or Tyndall) scattering — preferential deflection of shorter wavelengths by particles smaller than λ (L21 §21.2).

Section B — Short Answer (3 marks)

PYQ10. Draw a ray diagram showing the dispersion through an equiangular triangular glass prism. Write expression for the refractive index of the prism in terms of angle of minimum deviation and angle of prism. Why does the prism disperse rays of different colours at different angles?

3 marks · Section B Q38 · 312/MAY/204B

Model Answer

Ray diagram: White ray incident on prism → emergent fan of coloured rays (VIBGYOR) with violet deviated most.

Refractive index: μ = sin((A + δm)/2) / sin(A/2)

Why dispersion: μ depends on wavelength (μV > μR) → different refraction angles → different deviations for each colour.

Explanation

Equiangular prism: symmetric geometry at minimum deviation gives the standard μ formula (L21 §21.1). Material dispersion causes colour separation.

PYQ11. Write expression for the refractive index of the material of a prism. Reduce the expression for a thin prism of small refracting angle. Calculate the angle of minimum deviation for a thin prism of refractive index 1.6 and refracting angle 1°.

3 marks · Section B Q40 · 68/ESS/1-312-A

Model Answer

General: μ = sin((A + δm)/2) / sin(A/2)

Thin prism (A small): δm = (μ − 1) A

Numerical: δm = (1.6 − 1) × 1° = 0.6°

Explanation

Small-angle reduction uses sin θ ≈ θ (radians) or direct proportionality δ ∝ (μ−1)A in degrees for thin prisms (L21 §21.1).

PYQ12. A glass prism of refracting angle 60° and refractive index 1.5 is completely immersed in water. Calculate the angle of minimum deviation of the prism in this situation. Given: refractive index of water = 1.33, sin⁻¹(0.56) = 34.38°.

3 marks · Section B Q41 · 68/ESS/1-312-A (Marking Scheme)

Model Answer

Relative refractive index in water: μ′ = nglass/nwater = 1.5/1.33

μ′ = sin((A + δm)/2) / sin(A/2)

sin((60° + δm)/2) = (1.5/1.33) × sin 30° = 0.564

(60° + δm)/2 = sin⁻¹(0.56) = 34.38°

δm = 8.76° ≈ 8.8°

Explanation

Immersion changes effective μ to nprism/nsurround. Board gives sin⁻¹(0.56) = 34.38° to complete the calculation (L21 §21.1).

Problem Solving — L21 Dispersion and Scattering of 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.

Question 1 of 6Dispersion

Why does a prism disperse white light into a spectrum?

μ depends on λ

Solution — step by step with formulas

  1. μ is larger for violet than red ⇒ greater deviation for violet.

Final answer: Different λ have different μ ⇒ different deviation

Formulas used in this problem

μ depends on λ

Textbook formal language

Dispersion arises because refractive index varies with wavelength.

Working formula set for this problem: μ depends on λ. 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)

Violet bends more than red in glass, so colours split.

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 — Dispersion

Cauchy’s formula approximates μ(λ) for transparent media.

Link to chapter notes (L21 — Dispersion): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: μ depends on λ. 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 depends on λ 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).
Question 2 of 6Rainbow

Outline formation of a primary rainbow (dispersion + reflection).

Solution — step by step with formulas

  1. Sunlight enters raindrop, disperses, reflects internally once, exits—colours at different angles.

Final answer: Dispersion + one TIR-like internal reflection in drops

Textbook formal language

Geometric optics of spherical drops yields angular separation of colours.

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)

Drops act like tiny prisms plus a mirror; red and violet leave at different angles.

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 — Rainbow formation

Secondary rainbow involves two internal reflections.

Link to chapter notes (L21 — Rainbow formation): 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).
Question 3 of 6Scattering

Why is the clear sky blue according to Rayleigh scattering?

I ∝ 1/λ⁴

Solution — step by step with formulas

  1. Shorter λ (blue) scattered more strongly (∝1/λ⁴) by air molecules.

Final answer: Blue scattered more than red

Formulas used in this problem

I ∝ 1/λ⁴

Textbook formal language

Rayleigh scattering intensity varies as inverse fourth power of wavelength for particles ≪ λ.

Working formula set for this problem: I ∝ 1/λ⁴. 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)

Air molecules bounce blue light all over the sky; redder light goes more straight.

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 — Rayleigh scattering

Valid for molecular sizes much smaller than optical wavelengths.

Link to chapter notes (L21 — Rayleigh scattering): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: I ∝ 1/λ⁴. 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 I ∝ 1/λ⁴ 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).
Question 4 of 6Sunset

Explain red appearance of the Sun at sunrise/sunset.

Solution — step by step with formulas

  1. Long path through atmosphere; blue scattered out of line of sight; red/orange remains.

Final answer: Blue removed by long-path scattering

Textbook formal language

Optical path length in atmosphere is maximum near horizon.

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)

Sunlight travels through more air; blue is scattered away, leaving warm colours.

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 — Red sunset

Pollution can enhance red/orange hues.

Link to chapter notes (L21 — Red sunset): 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).
Question 5 of 6μ prism

State the formula for refractive index of prism material in terms of A and δ_m.

δ = i + e − A
μ = sin((A+δ_m)/2)/sin(A/2)

Solution — step by step with formulas

  1. μ = sin((A+δ_m)/2) / sin(A/2).

Final answer: μ = sin((A+δ_m)/2)/sin(A/2)

Formulas used in this problem

δ = i + e − A
μ = sin((A+δ_m)/2)/sin(A/2)

Textbook formal language

Minimum deviation configuration is symmetric; yields standard μ formula.

Working formula set for this problem: δ = i + e − A; μ = sin((A+δ_m)/2)/sin(A/2). 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)

Measure prism angle and least deviation, plug into the sine formula for μ.

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 — Prism deviation

Used in spectrometer experiments.

Link to chapter notes (L21 — Prism deviation): this idea sits with the definitions and worked examples in the detailed notes and formula sheet. Memorise: δ = i + e − A; μ = sin((A+δ_m)/2)/sin(A/2). 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 δ = i + e − A; μ = sin((A+δ_m)/2)/sin(A/2) 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).
Question 6 of 6Complementary

What are complementary colours in the context of mixing light?

Solution — step by step with formulas

  1. Two colours that mix to produce white (e.g. blue + yellow light).

Final answer: Pair that yields white on addition

Textbook formal language

Additive mixing of spectral components can reconstitute white.

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)

If two coloured lights together look white, they complement each other.

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 — Complementary colours

Distinct from subtractive mixing of pigments.

Link to chapter notes (L21 — Complementary colours): 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).