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Ray Optics • Critical Angle • Optical Fibre

Total Internal Reflection

TIR • Critical Angle • Mirage • Looming • Diamond Brilliance • Prism Applications • Optical Fibre • Numerical Aperture • Bulb Under Water • Numericals • PYQs

Theory45° PrismBulb Under WaterOptical FibreNumericalsCase Studies

1–6. Total Internal Reflection, Conditions and Critical Angle

Introduction to Total Internal Reflection
Total internal reflection is complete reflection of light back into a denser medium when it tries to pass into a rarer medium at an angle greater than the critical angle.
Refraction vs Total Internal Reflection
In ordinary refraction, part of the ray enters the second medium. In TIR, no refracted ray emerges; all light reflects back.
Condition 1
Light must travel from optically denser medium to optically rarer medium.
Condition 2
Angle of incidence in the denser medium must be greater than the critical angle.
Critical Angle
Critical angle C is the angle of incidence in denser medium for which angle of refraction in rarer medium is 90°.
Special Cases
For water-air, sin C = 1/μwater. For glass-air, sin C = 1/μglass. For medium 1 to medium 2, sin C = n₂/n₁.
sin C = n₂/n₁For air outside: sin C = 1/μ
Cr = 90°At critical angle, refracted ray travels along the interface.

Critical angle: refracted ray grazes the surface.

Air, n₂Glass / Water, n₁ > n₂Normalcritical ray grazes surfacei > CReflected ray stays in denser mediumTotal internal reflection: no refracted ray emerges.

When i > C, total internal reflection occurs.

5. Derivation of Critical Angle Formula

Step-by-step derivation

  1. Apply Snell’s law at denser-rarer interface.
  2. Let denser medium have refractive index n₁ and rarer medium n₂.
  3. At critical angle, i = C and r = 90°.
  4. n₁ sin C = n₂ sin 90°.
  5. Since sin 90° = 1, sin C = n₂/n₁.

Important results

n₁ sin C = n₂sin C = n₂/n₁If n₂ = 1, sin C = 1/n₁

7–13. Optical Path, Nature, Mirage, Looming, Diamond, Prism and Fibre Principle

Optical Path Concept
Optical path length is μ times geometrical path length. It measures phase delay due to the medium.
Total Internal Reflection in Nature
Mirage, looming and sparkle of diamond involve refraction plus total internal reflection.
Mirage
Hot air near ground is optically rarer, so rays bend and may undergo TIR-like turning, making virtual water-like images.
Looming
Cold dense air near surface bends rays downward, so distant objects appear lifted.
Diamond Brilliance
Diamond has high refractive index and small critical angle, so light suffers many internal reflections.
Prism Applications
Right angled prisms use TIR for 90° deviation, 180° reversal and retroreflection.
Optical Fibre Principle
Light is guided through a denser core by repeated total internal reflection at core-cladding boundary.
Fibre Requirement
Core refractive index must be greater than cladding refractive index.
Exam Clue
TIR is lossless compared with ordinary mirror reflection in ideal prism questions.

14–17. Numerical Aperture, Acceptance Angle, Optical Fibre Construction and Applications

Acceptance coneCore n₁, cladding n₂NA = μ₀ sin iₘₐₓ = √(n₁² - n₂²)

Optical fibre acceptance cone and guided ray.

Numerical Aperture

Numerical aperture measures the light-gathering ability of an optical fibre.

NA = μ₀ sin iₘₐₓNA = √(n₁² - n₂²)

Derivation

  1. At core-cladding boundary, limiting ray strikes at critical angle.
  2. sin C = n₂/n₁.
  3. Using geometry inside fibre and Snell’s law at entrance face, μ₀ sin iₘₐₓ = √(n₁² - n₂²).
  4. Thus NA = μ₀ sin iₘₐₓ.

Acceptance Angle

The maximum angle with the fibre axis for which entering light remains guided by TIR.

Optical Fibre Construction

An optical fibre contains a high refractive index core, lower refractive index cladding and protective outer jacket. TIR takes place at the core-cladding boundary.

Optical Fibre Applications

Communication, fibre internet, medical endoscopy, sensors and decorative lighting.

Special Section 1: 45°–45°–90° Right Angled Isosceles Prism

For glass-air, the critical angle is about 42°. Inside a right angled isosceles prism, a 45° incidence at the glass-air face produces total internal reflection. The diagrams below show actual object rays, TIR points, normals and image orientation.

Case 1: 45°–45°–90° prism, one TIR, deviation = 90°

Objectupright45°–45°–90° glass prism90°45°45°incident ray enters normallynormali = 45° > CTIR pointemergent rayFinal imageimage rotated by 90°One TIR at the hypotenuse turns the normally entering ray through 90°.

Case 1: one TIR, final ray is perpendicular to the original direction.

  1. The ray starts from the object and enters the first face normally, so it does not bend at entry.
  2. It strikes the hypotenuse at 45°.
  3. For glass-air, C ≈ 42°, so 45° > C and TIR occurs.
  4. The emergent ray is at 90° to the incident direction; the image orientation is correspondingly rotated.
NEET/JEE Question for prism case 1. Why does TIR occur at 45° inside glass?

For glass-air, C ≈ 42°. Since internal incidence is 45° and 45° > C, TIR occurs.

Answer: 45° is greater than critical angle.

Case 2: 45°–45°–90° prism, two TIRs, deviation = 180°

Object ↑Ray enters normally through the hypotenuseincident rayFirst TIRSecond TIRemergent ray opposite to incident raynormali = 45° at both reflecting facesImage ↓orientation reversed by 180°Two successive TIRs reverse the direction of the ray; the emergent ray travels back opposite to the incident ray.

Case 2: two TIRs produce 180° deviation and reversed image orientation.

  1. The object ray enters normally through the hypotenuse.
  2. It meets the first reflecting face at 45° and undergoes TIR.
  3. It meets the second reflecting face at 45° and undergoes TIR again.
  4. The final emergent ray is opposite to the incident direction; image orientation is reversed by 180°.
NEET/JEE Question for prism case 2. Why does two-TIR prism reverse the direction?

Two reflections at mutually perpendicular faces reverse both transverse components of the ray direction, so the emergent ray is opposite to the incident ray.

Answer: Two perpendicular TIR reflections give 180° deviation.

Case 3: Retroreflector case

ObjectIncident ray →First TIRSecond TIR← Emergent rayincident and emergent rays are parallel but oppositei = 45° > C at both TIR pointsImageRetroreflector: returns light toward source.

Case 3: ray returns parallel and opposite to the incident direction.

  1. The incident ray from the object enters the prism face and travels inside.
  2. It undergoes two TIR events at the two glass-air faces.
  3. The emergent ray is parallel to the incident ray but travels in the opposite direction.
  4. This is why this arrangement is called a retroreflector.
NEET/JEE Question for prism case 3. What makes this arrangement a retroreflector?

The ray returns parallel to its incident path but in the opposite direction after two TIR events.

Answer: Emergent ray is parallel and opposite to incident ray.

Special Section 2: Bulb Placed Under Water

Only rays from the bulb that strike the water surface with incidence less than or equal to the critical angle can emerge. These rays form a circular bright region on the surface.

Bulb at depth hbright circle radius rCr = h tan C, Area = πh² tan²C

Circular region through which light emerges from water.

Derivation

  1. Let bulb be at depth h below water surface.
  2. For water-air boundary, limiting ray emerges at critical angle C.
  3. sin C = 1/μ.
  4. The boundary of emergent light makes a cone of semi-angle C.
  5. Radius of bright circle: r = h tan C.
  6. Area of circular bright patch: A = πr².
r = h tan CA = πh² tan²C

Useful Form

sin C = 1/μtan²C = sin²C / (1 - sin²C) = 1/(μ² - 1)A = πh²/(μ² - 1)
Bulb Numerical 1. A bulb is 2 m under water of μ = 1.33. Find area of surface through which light emerges.

Given: h = 2 m, μ = 1.33

A = πh²/(μ² - 1)
  1. A = π(2)² / (1.33² - 1)
  2. A = 16.34 m²
Final Answer: A = 16.34 m²
Bulb Numerical 2. A bulb is 3 m under water of μ = 1.33. Find area of surface through which light emerges.

Given: h = 3 m, μ = 1.33

A = πh²/(μ² - 1)
  1. A = π(3)² / (1.33² - 1)
  2. A = 36.77 m²
Final Answer: A = 36.77 m²
Bulb Numerical 3. A bulb is 5 m under water of μ = 1.5. Find area of surface through which light emerges.

Given: h = 5 m, μ = 1.5

A = πh²/(μ² - 1)
  1. A = π(5)² / (1.5² - 1)
  2. A = 62.83 m²
Final Answer: A = 62.83 m²

18. Formula Revision Sheet

Critical anglesin C = n₂/n₁
Air outsidesin C = 1/μ
TIR conditioni > C
Optical pathOPL = μx
Numerical apertureNA = μ₀ sin iₘₐₓ
Fibre NANA = √(n₁² - n₂²)
Bulb radiusr = h tan C
Bulb areaA = πh²/(μ² - 1)

19. Quick Revision Notes

TIR Conditions
  • Denser to rarer medium.
  • i greater than C.
  • No refracted ray.
Applications
  • Mirage
  • Diamond sparkle
  • Prisms
  • Optical fibre
Must Remember
  • sin C = n₂/n₁
  • NA = √(n₁² - n₂²)
  • A = πh²/(μ² - 1)

Numerical Problems

Solved questions for CBSE, NEET, JEE Main, JEE Advanced, IB Physics, IGCSE, ICSE, A-Level and British Curriculum.

CBSE

CBSE Numerical 1. Find critical angle from glass n₁=1.5 to water n₂=1.33.
sin C = n₂/n₁
  1. sin C = 1.33/1.5
  2. C = 62.46°
Final Answer: C = 62.46°
CBSE Numerical 2. Fibre has n₁=1.48, n₂=1.46. Find numerical aperture in air.
NA = √(n₁² - n₂²)
  1. NA = √(1.48² - 1.46²)
  2. NA = 0.242
Final Answer: NA = 0.242
CBSE Numerical 3. Bulb is 4 m below water surface, μ=1.33. Find bright area.
A = πh²/(μ² - 1)
  1. A = π(4)²/(1.33² - 1)
  2. A = 65.37 m²
Final Answer: 65.37 m²
CBSE Numerical 4. For glass-air prism, will a ray at 45° undergo TIR if μ=1.5?
sin C = 1/μ
  1. C = sin⁻¹(1/1.5) = 41.81°
  2. 45° > 41.81°
  3. Therefore TIR occurs.
Final Answer: Yes, TIR occurs.
CBSE Numerical 5. Find critical angle for glass-air if μ = 1.5.
sin C = 1/μ
  1. sin C = 1/1.5
  2. C = sin⁻¹(0.667)
  3. C = 41.81°
Final Answer: C = 41.81°

NEET

NEET Numerical 1. Find critical angle from glass n₁=1.5 to water n₂=1.33.
sin C = n₂/n₁
  1. sin C = 1.33/1.5
  2. C = 62.46°
Final Answer: C = 62.46°
NEET Numerical 2. Fibre has n₁=1.48, n₂=1.46. Find numerical aperture in air.
NA = √(n₁² - n₂²)
  1. NA = √(1.48² - 1.46²)
  2. NA = 0.242
Final Answer: NA = 0.242
NEET Numerical 3. Bulb is 4 m below water surface, μ=1.33. Find bright area.
A = πh²/(μ² - 1)
  1. A = π(4)²/(1.33² - 1)
  2. A = 65.37 m²
Final Answer: 65.37 m²
NEET Numerical 4. For glass-air prism, will a ray at 45° undergo TIR if μ=1.5?
sin C = 1/μ
  1. C = sin⁻¹(1/1.5) = 41.81°
  2. 45° > 41.81°
  3. Therefore TIR occurs.
Final Answer: Yes, TIR occurs.
NEET Numerical 5. Find critical angle for glass-air if μ = 1.5.
sin C = 1/μ
  1. sin C = 1/1.5
  2. C = sin⁻¹(0.667)
  3. C = 41.81°
Final Answer: C = 41.81°
NEET Numerical 6. Find critical angle from glass n₁=1.5 to water n₂=1.33.
sin C = n₂/n₁
  1. sin C = 1.33/1.5
  2. C = 62.46°
Final Answer: C = 62.46°

JEE Main

JEE Main Numerical 1. Find critical angle from glass n₁=1.5 to water n₂=1.33.
sin C = n₂/n₁
  1. sin C = 1.33/1.5
  2. C = 62.46°
Final Answer: C = 62.46°
JEE Main Numerical 2. Fibre has n₁=1.48, n₂=1.46. Find numerical aperture in air.
NA = √(n₁² - n₂²)
  1. NA = √(1.48² - 1.46²)
  2. NA = 0.242
Final Answer: NA = 0.242
JEE Main Numerical 3. Bulb is 4 m below water surface, μ=1.33. Find bright area.
A = πh²/(μ² - 1)
  1. A = π(4)²/(1.33² - 1)
  2. A = 65.37 m²
Final Answer: 65.37 m²
JEE Main Numerical 4. For glass-air prism, will a ray at 45° undergo TIR if μ=1.5?
sin C = 1/μ
  1. C = sin⁻¹(1/1.5) = 41.81°
  2. 45° > 41.81°
  3. Therefore TIR occurs.
Final Answer: Yes, TIR occurs.
JEE Main Numerical 5. Find critical angle for glass-air if μ = 1.5.
sin C = 1/μ
  1. sin C = 1/1.5
  2. C = sin⁻¹(0.667)
  3. C = 41.81°
Final Answer: C = 41.81°
JEE Main Numerical 6. Find critical angle from glass n₁=1.5 to water n₂=1.33.
sin C = n₂/n₁
  1. sin C = 1.33/1.5
  2. C = 62.46°
Final Answer: C = 62.46°

JEE Advanced

JEE Advanced Numerical 1. Find critical angle from glass n₁=1.5 to water n₂=1.33.
sin C = n₂/n₁
  1. sin C = 1.33/1.5
  2. C = 62.46°
Final Answer: C = 62.46°
JEE Advanced Numerical 2. Fibre has n₁=1.48, n₂=1.46. Find numerical aperture in air.
NA = √(n₁² - n₂²)
  1. NA = √(1.48² - 1.46²)
  2. NA = 0.242
Final Answer: NA = 0.242
JEE Advanced Numerical 3. Bulb is 4 m below water surface, μ=1.33. Find bright area.
A = πh²/(μ² - 1)
  1. A = π(4)²/(1.33² - 1)
  2. A = 65.37 m²
Final Answer: 65.37 m²
JEE Advanced Numerical 4. For glass-air prism, will a ray at 45° undergo TIR if μ=1.5?
sin C = 1/μ
  1. C = sin⁻¹(1/1.5) = 41.81°
  2. 45° > 41.81°
  3. Therefore TIR occurs.
Final Answer: Yes, TIR occurs.
JEE Advanced Numerical 5. Find critical angle for glass-air if μ = 1.5.
sin C = 1/μ
  1. sin C = 1/1.5
  2. C = sin⁻¹(0.667)
  3. C = 41.81°
Final Answer: C = 41.81°
JEE Advanced Numerical 6. Find critical angle from glass n₁=1.5 to water n₂=1.33.
sin C = n₂/n₁
  1. sin C = 1.33/1.5
  2. C = 62.46°
Final Answer: C = 62.46°

IB Physics

IB Physics Numerical 1. Find critical angle from glass n₁=1.5 to water n₂=1.33.
sin C = n₂/n₁
  1. sin C = 1.33/1.5
  2. C = 62.46°
Final Answer: C = 62.46°
IB Physics Numerical 2. Fibre has n₁=1.48, n₂=1.46. Find numerical aperture in air.
NA = √(n₁² - n₂²)
  1. NA = √(1.48² - 1.46²)
  2. NA = 0.242
Final Answer: NA = 0.242
IB Physics Numerical 3. Bulb is 4 m below water surface, μ=1.33. Find bright area.
A = πh²/(μ² - 1)
  1. A = π(4)²/(1.33² - 1)
  2. A = 65.37 m²
Final Answer: 65.37 m²
IB Physics Numerical 4. For glass-air prism, will a ray at 45° undergo TIR if μ=1.5?
sin C = 1/μ
  1. C = sin⁻¹(1/1.5) = 41.81°
  2. 45° > 41.81°
  3. Therefore TIR occurs.
Final Answer: Yes, TIR occurs.

IGCSE

IGCSE Numerical 1. Find critical angle from glass n₁=1.5 to water n₂=1.33.
sin C = n₂/n₁
  1. sin C = 1.33/1.5
  2. C = 62.46°
Final Answer: C = 62.46°
IGCSE Numerical 2. Fibre has n₁=1.48, n₂=1.46. Find numerical aperture in air.
NA = √(n₁² - n₂²)
  1. NA = √(1.48² - 1.46²)
  2. NA = 0.242
Final Answer: NA = 0.242
IGCSE Numerical 3. Bulb is 4 m below water surface, μ=1.33. Find bright area.
A = πh²/(μ² - 1)
  1. A = π(4)²/(1.33² - 1)
  2. A = 65.37 m²
Final Answer: 65.37 m²
IGCSE Numerical 4. For glass-air prism, will a ray at 45° undergo TIR if μ=1.5?
sin C = 1/μ
  1. C = sin⁻¹(1/1.5) = 41.81°
  2. 45° > 41.81°
  3. Therefore TIR occurs.
Final Answer: Yes, TIR occurs.

ICSE

ICSE Numerical 1. Find critical angle from glass n₁=1.5 to water n₂=1.33.
sin C = n₂/n₁
  1. sin C = 1.33/1.5
  2. C = 62.46°
Final Answer: C = 62.46°
ICSE Numerical 2. Fibre has n₁=1.48, n₂=1.46. Find numerical aperture in air.
NA = √(n₁² - n₂²)
  1. NA = √(1.48² - 1.46²)
  2. NA = 0.242
Final Answer: NA = 0.242
ICSE Numerical 3. Bulb is 4 m below water surface, μ=1.33. Find bright area.
A = πh²/(μ² - 1)
  1. A = π(4)²/(1.33² - 1)
  2. A = 65.37 m²
Final Answer: 65.37 m²
ICSE Numerical 4. For glass-air prism, will a ray at 45° undergo TIR if μ=1.5?
sin C = 1/μ
  1. C = sin⁻¹(1/1.5) = 41.81°
  2. 45° > 41.81°
  3. Therefore TIR occurs.
Final Answer: Yes, TIR occurs.

A-Level / British Curriculum

A-Level / British Curriculum Numerical 1. Find critical angle from glass n₁=1.5 to water n₂=1.33.
sin C = n₂/n₁
  1. sin C = 1.33/1.5
  2. C = 62.46°
Final Answer: C = 62.46°
A-Level / British Curriculum Numerical 2. Fibre has n₁=1.48, n₂=1.46. Find numerical aperture in air.
NA = √(n₁² - n₂²)
  1. NA = √(1.48² - 1.46²)
  2. NA = 0.242
Final Answer: NA = 0.242
A-Level / British Curriculum Numerical 3. Bulb is 4 m below water surface, μ=1.33. Find bright area.
A = πh²/(μ² - 1)
  1. A = π(4)²/(1.33² - 1)
  2. A = 65.37 m²
Final Answer: 65.37 m²
A-Level / British Curriculum Numerical 4. For glass-air prism, will a ray at 45° undergo TIR if μ=1.5?
sin C = 1/μ
  1. C = sin⁻¹(1/1.5) = 41.81°
  2. 45° > 41.81°
  3. Therefore TIR occurs.
Final Answer: Yes, TIR occurs.
A-Level / British Curriculum Numerical 5. Find critical angle for glass-air if μ = 1.5.
sin C = 1/μ
  1. sin C = 1/1.5
  2. C = sin⁻¹(0.667)
  3. C = 41.81°
Final Answer: C = 41.81°

PYQ Section

CBSE PYQs

CBSE PYQs 1. State two conditions for TIR.

Light must travel from denser to rarer medium, and i must be greater than C.

Answer: Light must travel from denser to rarer medium, and i must be greater than C.
CBSE PYQs 2. Define critical angle.

Critical angle is angle in denser medium for which refracted angle in rarer medium is 90°.

Answer: Critical angle is angle in denser medium for which refracted angle in rarer medium is 90°.
CBSE PYQs 3. Why does diamond sparkle?

Small critical angle causes multiple total internal reflections.

Answer: Small critical angle causes multiple total internal reflections.
CBSE PYQs 4. Why are prisms better than plane mirrors in periscopes?

TIR in prisms is nearly lossless and produces brighter image.

Answer: TIR in prisms is nearly lossless and produces brighter image.
CBSE PYQs 5. What is numerical aperture?

NA is μ₀ sin of acceptance angle and measures light-gathering ability.

Answer: NA is μ₀ sin of acceptance angle and measures light-gathering ability.

NEET PYQs

NEET PYQs 1. State two conditions for TIR.

Light must travel from denser to rarer medium, and i must be greater than C.

Answer: Light must travel from denser to rarer medium, and i must be greater than C.
NEET PYQs 2. Define critical angle.

Critical angle is angle in denser medium for which refracted angle in rarer medium is 90°.

Answer: Critical angle is angle in denser medium for which refracted angle in rarer medium is 90°.
NEET PYQs 3. Why does diamond sparkle?

Small critical angle causes multiple total internal reflections.

Answer: Small critical angle causes multiple total internal reflections.
NEET PYQs 4. Why are prisms better than plane mirrors in periscopes?

TIR in prisms is nearly lossless and produces brighter image.

Answer: TIR in prisms is nearly lossless and produces brighter image.
NEET PYQs 5. What is numerical aperture?

NA is μ₀ sin of acceptance angle and measures light-gathering ability.

Answer: NA is μ₀ sin of acceptance angle and measures light-gathering ability.

JEE Main PYQs

JEE Main PYQs 1. State two conditions for TIR.

Light must travel from denser to rarer medium, and i must be greater than C.

Answer: Light must travel from denser to rarer medium, and i must be greater than C.
JEE Main PYQs 2. Define critical angle.

Critical angle is angle in denser medium for which refracted angle in rarer medium is 90°.

Answer: Critical angle is angle in denser medium for which refracted angle in rarer medium is 90°.
JEE Main PYQs 3. Why does diamond sparkle?

Small critical angle causes multiple total internal reflections.

Answer: Small critical angle causes multiple total internal reflections.
JEE Main PYQs 4. Why are prisms better than plane mirrors in periscopes?

TIR in prisms is nearly lossless and produces brighter image.

Answer: TIR in prisms is nearly lossless and produces brighter image.
JEE Main PYQs 5. What is numerical aperture?

NA is μ₀ sin of acceptance angle and measures light-gathering ability.

Answer: NA is μ₀ sin of acceptance angle and measures light-gathering ability.

JEE Advanced PYQs

JEE Advanced PYQs 1. State two conditions for TIR.

Light must travel from denser to rarer medium, and i must be greater than C.

Answer: Light must travel from denser to rarer medium, and i must be greater than C.
JEE Advanced PYQs 2. Define critical angle.

Critical angle is angle in denser medium for which refracted angle in rarer medium is 90°.

Answer: Critical angle is angle in denser medium for which refracted angle in rarer medium is 90°.
JEE Advanced PYQs 3. Why does diamond sparkle?

Small critical angle causes multiple total internal reflections.

Answer: Small critical angle causes multiple total internal reflections.
JEE Advanced PYQs 4. Why are prisms better than plane mirrors in periscopes?

TIR in prisms is nearly lossless and produces brighter image.

Answer: TIR in prisms is nearly lossless and produces brighter image.
JEE Advanced PYQs 5. What is numerical aperture?

NA is μ₀ sin of acceptance angle and measures light-gathering ability.

Answer: NA is μ₀ sin of acceptance angle and measures light-gathering ability.

IB Physics Questions

IB Physics Questions 1. State two conditions for TIR.

Light must travel from denser to rarer medium, and i must be greater than C.

Answer: Light must travel from denser to rarer medium, and i must be greater than C.
IB Physics Questions 2. Define critical angle.

Critical angle is angle in denser medium for which refracted angle in rarer medium is 90°.

Answer: Critical angle is angle in denser medium for which refracted angle in rarer medium is 90°.
IB Physics Questions 3. Why does diamond sparkle?

Small critical angle causes multiple total internal reflections.

Answer: Small critical angle causes multiple total internal reflections.
IB Physics Questions 4. Why are prisms better than plane mirrors in periscopes?

TIR in prisms is nearly lossless and produces brighter image.

Answer: TIR in prisms is nearly lossless and produces brighter image.
IB Physics Questions 5. What is numerical aperture?

NA is μ₀ sin of acceptance angle and measures light-gathering ability.

Answer: NA is μ₀ sin of acceptance angle and measures light-gathering ability.

IGCSE Questions

IGCSE Questions 1. State two conditions for TIR.

Light must travel from denser to rarer medium, and i must be greater than C.

Answer: Light must travel from denser to rarer medium, and i must be greater than C.
IGCSE Questions 2. Define critical angle.

Critical angle is angle in denser medium for which refracted angle in rarer medium is 90°.

Answer: Critical angle is angle in denser medium for which refracted angle in rarer medium is 90°.
IGCSE Questions 3. Why does diamond sparkle?

Small critical angle causes multiple total internal reflections.

Answer: Small critical angle causes multiple total internal reflections.
IGCSE Questions 4. Why are prisms better than plane mirrors in periscopes?

TIR in prisms is nearly lossless and produces brighter image.

Answer: TIR in prisms is nearly lossless and produces brighter image.
IGCSE Questions 5. What is numerical aperture?

NA is μ₀ sin of acceptance angle and measures light-gathering ability.

Answer: NA is μ₀ sin of acceptance angle and measures light-gathering ability.

A-Level Questions

A-Level Questions 1. State two conditions for TIR.

Light must travel from denser to rarer medium, and i must be greater than C.

Answer: Light must travel from denser to rarer medium, and i must be greater than C.
A-Level Questions 2. Define critical angle.

Critical angle is angle in denser medium for which refracted angle in rarer medium is 90°.

Answer: Critical angle is angle in denser medium for which refracted angle in rarer medium is 90°.
A-Level Questions 3. Why does diamond sparkle?

Small critical angle causes multiple total internal reflections.

Answer: Small critical angle causes multiple total internal reflections.
A-Level Questions 4. Why are prisms better than plane mirrors in periscopes?

TIR in prisms is nearly lossless and produces brighter image.

Answer: TIR in prisms is nearly lossless and produces brighter image.
A-Level Questions 5. What is numerical aperture?

NA is μ₀ sin of acceptance angle and measures light-gathering ability.

Answer: NA is μ₀ sin of acceptance angle and measures light-gathering ability.

Case Study Section

Case Study 1: Mirage in Desert

Passage: Light bends through hot air layers near ground, producing an apparent water-like image.

Air, n₂Glass / Water, n₁ > n₂Normalcritical ray grazes surfacei > CReflected ray stays in denser mediumTotal internal reflection: no refracted ray emerges.

Diagram for Mirage in Desert.

MCQ 1. Which phenomenon is central here?

Total internal reflection is the key phenomenon.

Answer: Total internal reflection.
MCQ 2. Which condition is essential?

The ray must go from denser to rarer medium with i > C.

Answer: Denser to rarer and i > C.
Assertion-Reason. Assertion: TIR gives bright reflection. Reason: Ideally, no light is refracted out.

Both are true and the reason explains the assertion.

Answer: Both true; reason correct.

Case Study 2: Looming in Polar Regions

Passage: Cold dense air layers bend rays downward so distant objects appear lifted.

Air, n₂Glass / Water, n₁ > n₂Normalcritical ray grazes surfacei > CReflected ray stays in denser mediumTotal internal reflection: no refracted ray emerges.

Diagram for Looming in Polar Regions.

MCQ 1. Which phenomenon is central here?

Total internal reflection is the key phenomenon.

Answer: Total internal reflection.
MCQ 2. Which condition is essential?

The ray must go from denser to rarer medium with i > C.

Answer: Denser to rarer and i > C.
Assertion-Reason. Assertion: TIR gives bright reflection. Reason: Ideally, no light is refracted out.

Both are true and the reason explains the assertion.

Answer: Both true; reason correct.

Case Study 3: Diamond Sparkle

Passage: Diamond has high refractive index and low critical angle, producing repeated TIR.

incident white lightTIRTIRbright emergent raysDiamond sparkle is due to repeated internal reflections inside the diamond, not a right-angled prism path.

Diamond sparkle: multiple TIR inside a diamond-shaped medium produces bright emergent rays.

MCQ 1. Which phenomenon is central here?

Total internal reflection is the key phenomenon.

Answer: Total internal reflection.
MCQ 2. Which condition is essential?

The ray must go from denser to rarer medium with i > C.

Answer: Denser to rarer and i > C.
Assertion-Reason. Assertion: TIR gives bright reflection. Reason: Ideally, no light is refracted out.

Both are true and the reason explains the assertion.

Answer: Both true; reason correct.

Case Study 4: Optical Fibre Communication

Passage: Signals travel through core by repeated total internal reflection.

Acceptance coneCore n₁, cladding n₂NA = μ₀ sin iₘₐₓ = √(n₁² - n₂²)

Diagram for Optical Fibre Communication.

MCQ 1. Which phenomenon is central here?

Total internal reflection is the key phenomenon.

Answer: Total internal reflection.
MCQ 2. Which condition is essential?

The ray must go from denser to rarer medium with i > C.

Answer: Denser to rarer and i > C.
Assertion-Reason. Assertion: TIR gives bright reflection. Reason: Ideally, no light is refracted out.

Both are true and the reason explains the assertion.

Answer: Both true; reason correct.

Case Study 5: Medical Endoscopy

Passage: Bundles of optical fibres carry light and images inside the body.

Acceptance coneCore n₁, cladding n₂NA = μ₀ sin iₘₐₓ = √(n₁² - n₂²)

Diagram for Medical Endoscopy.

MCQ 1. Which phenomenon is central here?

Total internal reflection is the key phenomenon.

Answer: Total internal reflection.
MCQ 2. Which condition is essential?

The ray must go from denser to rarer medium with i > C.

Answer: Denser to rarer and i > C.
Assertion-Reason. Assertion: TIR gives bright reflection. Reason: Ideally, no light is refracted out.

Both are true and the reason explains the assertion.

Answer: Both true; reason correct.

Case Study 6: Prism Periscope

Passage: Right angled prisms turn rays by TIR with high brightness.

ObjectFirst prismSecond prismTIR, 90° deviationTIR, 90° deviationFinal imageTwo right-angled prisms give two 90° deviations and carry light from the object to the observer.

Prism periscope: two right-angled prisms, each producing 90° deviation by TIR.

MCQ 1. Which phenomenon is central here?

Total internal reflection is the key phenomenon.

Answer: Total internal reflection.
MCQ 2. Which condition is essential?

The ray must go from denser to rarer medium with i > C.

Answer: Denser to rarer and i > C.
Assertion-Reason. Assertion: TIR gives bright reflection. Reason: Ideally, no light is refracted out.

Both are true and the reason explains the assertion.

Answer: Both true; reason correct.

Case Study 7: Fibre Internet Systems

Passage: Digital light pulses remain guided over long distances using TIR.

Acceptance coneCore n₁, cladding n₂NA = μ₀ sin iₘₐₓ = √(n₁² - n₂²)

Diagram for Fibre Internet Systems.

MCQ 1. Which phenomenon is central here?

Total internal reflection is the key phenomenon.

Answer: Total internal reflection.
MCQ 2. Which condition is essential?

The ray must go from denser to rarer medium with i > C.

Answer: Denser to rarer and i > C.
Assertion-Reason. Assertion: TIR gives bright reflection. Reason: Ideally, no light is refracted out.

Both are true and the reason explains the assertion.

Answer: Both true; reason correct.

Advanced Question Bank

Conceptual Questions

Conceptual Questions 1. Explain why TIR cannot occur from air to glass.

TIR needs propagation from denser to rarer medium. Air to glass is rarer to denser, so the ray bends towards the normal and cannot totally internally reflect at that boundary.

Answer: Because air to glass is rarer to denser.
Conceptual Questions 2. Explain why TIR cannot occur from air to glass.

TIR needs propagation from denser to rarer medium. Air to glass is rarer to denser, so the ray bends towards the normal and cannot totally internally reflect at that boundary.

Answer: Because air to glass is rarer to denser.
Conceptual Questions 3. Explain why TIR cannot occur from air to glass.

TIR needs propagation from denser to rarer medium. Air to glass is rarer to denser, so the ray bends towards the normal and cannot totally internally reflect at that boundary.

Answer: Because air to glass is rarer to denser.
Conceptual Questions 4. Explain why TIR cannot occur from air to glass.

TIR needs propagation from denser to rarer medium. Air to glass is rarer to denser, so the ray bends towards the normal and cannot totally internally reflect at that boundary.

Answer: Because air to glass is rarer to denser.

Assertion-Reason Questions

Assertion-Reason Questions 1. Explain why TIR cannot occur from air to glass.

TIR needs propagation from denser to rarer medium. Air to glass is rarer to denser, so the ray bends towards the normal and cannot totally internally reflect at that boundary.

Answer: Because air to glass is rarer to denser.
Assertion-Reason Questions 2. Explain why TIR cannot occur from air to glass.

TIR needs propagation from denser to rarer medium. Air to glass is rarer to denser, so the ray bends towards the normal and cannot totally internally reflect at that boundary.

Answer: Because air to glass is rarer to denser.
Assertion-Reason Questions 3. Explain why TIR cannot occur from air to glass.

TIR needs propagation from denser to rarer medium. Air to glass is rarer to denser, so the ray bends towards the normal and cannot totally internally reflect at that boundary.

Answer: Because air to glass is rarer to denser.
Assertion-Reason Questions 4. Explain why TIR cannot occur from air to glass.

TIR needs propagation from denser to rarer medium. Air to glass is rarer to denser, so the ray bends towards the normal and cannot totally internally reflect at that boundary.

Answer: Because air to glass is rarer to denser.

HOTS Questions

HOTS Questions 1. Explain why TIR cannot occur from air to glass.

TIR needs propagation from denser to rarer medium. Air to glass is rarer to denser, so the ray bends towards the normal and cannot totally internally reflect at that boundary.

Answer: Because air to glass is rarer to denser.
HOTS Questions 2. Explain why TIR cannot occur from air to glass.

TIR needs propagation from denser to rarer medium. Air to glass is rarer to denser, so the ray bends towards the normal and cannot totally internally reflect at that boundary.

Answer: Because air to glass is rarer to denser.
HOTS Questions 3. Explain why TIR cannot occur from air to glass.

TIR needs propagation from denser to rarer medium. Air to glass is rarer to denser, so the ray bends towards the normal and cannot totally internally reflect at that boundary.

Answer: Because air to glass is rarer to denser.
HOTS Questions 4. Explain why TIR cannot occur from air to glass.

TIR needs propagation from denser to rarer medium. Air to glass is rarer to denser, so the ray bends towards the normal and cannot totally internally reflect at that boundary.

Answer: Because air to glass is rarer to denser.

Match the Following

Match the Following 1. Match: critical angle, NA, bulb area, prism TIR with formula/result.
  1. Critical angle → sin C = n₂/n₁
  2. NA → √(n₁²-n₂²)
  3. Bulb area → πh²/(μ²-1)
  4. Prism TIR → 45° > C
Answer: Correct matching shown.
Match the Following 2. Match: critical angle, NA, bulb area, prism TIR with formula/result.
  1. Critical angle → sin C = n₂/n₁
  2. NA → √(n₁²-n₂²)
  3. Bulb area → πh²/(μ²-1)
  4. Prism TIR → 45° > C
Answer: Correct matching shown.
Match the Following 3. Match: critical angle, NA, bulb area, prism TIR with formula/result.
  1. Critical angle → sin C = n₂/n₁
  2. NA → √(n₁²-n₂²)
  3. Bulb area → πh²/(μ²-1)
  4. Prism TIR → 45° > C
Answer: Correct matching shown.
Match the Following 4. Match: critical angle, NA, bulb area, prism TIR with formula/result.
  1. Critical angle → sin C = n₂/n₁
  2. NA → √(n₁²-n₂²)
  3. Bulb area → πh²/(μ²-1)
  4. Prism TIR → 45° > C
Answer: Correct matching shown.

Statement-Based Questions

Statement-Based Questions 1. Explain why TIR cannot occur from air to glass.

TIR needs propagation from denser to rarer medium. Air to glass is rarer to denser, so the ray bends towards the normal and cannot totally internally reflect at that boundary.

Answer: Because air to glass is rarer to denser.
Statement-Based Questions 2. Explain why TIR cannot occur from air to glass.

TIR needs propagation from denser to rarer medium. Air to glass is rarer to denser, so the ray bends towards the normal and cannot totally internally reflect at that boundary.

Answer: Because air to glass is rarer to denser.
Statement-Based Questions 3. Explain why TIR cannot occur from air to glass.

TIR needs propagation from denser to rarer medium. Air to glass is rarer to denser, so the ray bends towards the normal and cannot totally internally reflect at that boundary.

Answer: Because air to glass is rarer to denser.
Statement-Based Questions 4. Explain why TIR cannot occur from air to glass.

TIR needs propagation from denser to rarer medium. Air to glass is rarer to denser, so the ray bends towards the normal and cannot totally internally reflect at that boundary.

Answer: Because air to glass is rarer to denser.

Multiple Correct Questions

Multiple Correct Questions 1. Explain why TIR cannot occur from air to glass.

TIR needs propagation from denser to rarer medium. Air to glass is rarer to denser, so the ray bends towards the normal and cannot totally internally reflect at that boundary.

Answer: Because air to glass is rarer to denser.
Multiple Correct Questions 2. Explain why TIR cannot occur from air to glass.

TIR needs propagation from denser to rarer medium. Air to glass is rarer to denser, so the ray bends towards the normal and cannot totally internally reflect at that boundary.

Answer: Because air to glass is rarer to denser.
Multiple Correct Questions 3. Explain why TIR cannot occur from air to glass.

TIR needs propagation from denser to rarer medium. Air to glass is rarer to denser, so the ray bends towards the normal and cannot totally internally reflect at that boundary.

Answer: Because air to glass is rarer to denser.
Multiple Correct Questions 4. Explain why TIR cannot occur from air to glass.

TIR needs propagation from denser to rarer medium. Air to glass is rarer to denser, so the ray bends towards the normal and cannot totally internally reflect at that boundary.

Answer: Because air to glass is rarer to denser.

Integer-Type Questions

Integer-Type Questions 1. If μ=1.5, nearest integer value of critical angle is?
C = sin⁻¹(1/1.5)

C = 41.81°, nearest integer is 42.

Answer: 42
Integer-Type Questions 2. If μ=1.5, nearest integer value of critical angle is?
C = sin⁻¹(1/1.5)

C = 41.81°, nearest integer is 42.

Answer: 42
Integer-Type Questions 3. If μ=1.5, nearest integer value of critical angle is?
C = sin⁻¹(1/1.5)

C = 41.81°, nearest integer is 42.

Answer: 42
Integer-Type Questions 4. If μ=1.5, nearest integer value of critical angle is?
C = sin⁻¹(1/1.5)

C = 41.81°, nearest integer is 42.

Answer: 42

JEE Advanced Style Problems

JEE Advanced Style Problems 1. Explain why TIR cannot occur from air to glass.

TIR needs propagation from denser to rarer medium. Air to glass is rarer to denser, so the ray bends towards the normal and cannot totally internally reflect at that boundary.

Answer: Because air to glass is rarer to denser.
JEE Advanced Style Problems 2. Explain why TIR cannot occur from air to glass.

TIR needs propagation from denser to rarer medium. Air to glass is rarer to denser, so the ray bends towards the normal and cannot totally internally reflect at that boundary.

Answer: Because air to glass is rarer to denser.
JEE Advanced Style Problems 3. Explain why TIR cannot occur from air to glass.

TIR needs propagation from denser to rarer medium. Air to glass is rarer to denser, so the ray bends towards the normal and cannot totally internally reflect at that boundary.

Answer: Because air to glass is rarer to denser.
JEE Advanced Style Problems 4. Explain why TIR cannot occur from air to glass.

TIR needs propagation from denser to rarer medium. Air to glass is rarer to denser, so the ray bends towards the normal and cannot totally internally reflect at that boundary.

Answer: Because air to glass is rarer to denser.

Common Mistakes Students Make

Wrong direction
TIR is possible only from denser to rarer medium.
Using i = C
At i = C the refracted ray grazes; TIR starts for i > C.
Forgetting special case
For air outside, sin C = 1/μ.
NA confusion
NA is not simply n₁; it depends on n₁ and n₂.
Prism angle
In 45° prism, internal incidence is 45° for normal entry cases.
Bulb area
Use A = πh²/(μ² - 1), not πh²/μ².

Exam Tips and Frequently Asked Questions

Tip 1
Always write both TIR conditions before solving conceptual questions.
Tip 2
For prism questions, compare 45° with critical angle.
Tip 3
For bulb under water, draw the limiting cone.
FAQ: Is TIR reflection complete?
In ideal ray optics, yes; all light returns to denser medium.
FAQ: Why optical fibre works?
Core has higher refractive index than cladding, so rays repeatedly undergo TIR.
FAQ: Why diamond shines?
High refractive index gives small critical angle and repeated internal reflections.

Still have doubts in Total Internal Reflection, Optical Fibre or Critical Angle?

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