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Refraction of Light: Laws of Refraction, Snell's Law, Refractive Index, Total Internal Reflection, Critical Angle, Optical Fibre, and Complete CBSE Guide

A comprehensive guide to refraction of light for CBSE Class 10 and Class 12 Physics — laws of refraction, Snell's law and derivation, refractive index (absolute and relative), refraction through a glass slab, real vs apparent depth, total internal reflection, critical angle, applications of total internal reflection (optical fibre, mirage, diamonds), and five exam-ready solved problems.
1 August 2026 by
Refraction of Light: Laws of Refraction, Snell's Law, Refractive Index, Total Internal Reflection, Critical Angle, Optical Fibre, and Complete CBSE Guide
Krishan Kant
● CBSE Class 10 & Class 12 Physics — Optics: Refraction of Light

Have you ever noticed how a straw appears to bend when placed in a glass of water? Or why a swimming pool looks shallower than it actually is? Or why a diamond sparkles with such extraordinary brilliance? All of these everyday phenomena are caused by the same physical principle: refraction of light — the bending of light as it passes from one transparent medium to another.

Refraction occurs because light travels at different speeds in different media. In vacuum, light travels at c = 3×10&sup8; m/s — the fastest possible speed in the universe. In glass, it slows to about 2×10&sup8; m/s; in water, to 2.25×10&sup8; m/s. This change in speed causes the light ray to change direction at the boundary between two media, a behaviour mathematically described by Snell’s Law.

Refraction is not just a curiosity — it is the operating principle behind eyeglasses, cameras, telescopes, microscopes, optical fibres, endoscopes, and medical imaging. Every lens ever made uses refraction to focus or diverge light. For CBSE Class 10 Physics Chapter 10 (Light — Reflection and Refraction) and CBSE Class 12 Physics Chapter 9 (Ray Optics and Optical Instruments), refraction is a core topic appearing in every board examination. This guide covers all key subtopics with derivations, diagrams, applications, and five CBSE exam-ready solved problems.

Snell’s Law of Refraction
n₁ sinθ₁ = n₂ sinθ₂
The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for a given pair of media — equal to the ratio of their refractive indices.
n₁
Refractive index of medium 1 (incident)
θ₁
Angle of incidence (from normal)
n₂
Refractive index of medium 2 (refracted)
θ₂
Angle of refraction (from normal)

1. Laws of Refraction

When a ray of light passes from one transparent medium to another, it changes both speed and direction (unless it strikes the surface at 90°, i.e., normal incidence). There are two laws that govern refraction:

First Law of Refraction
Coplanarity of Rays
All rays in one plane
The incident ray, the refracted ray, and the normal to the interface at the point of incidence all lie in the same plane. This plane is called the plane of incidence. This law is simply a statement about geometry and has no formula — it tells us refraction is a 2D phenomenon.
Second Law of Refraction
Snell’s Law
sinθ₁ / sinθ₂ = n₂₁ = constant
For a given pair of media and a given colour of light, the ratio of the sine of the angle of incidence (θ₁) to the sine of the angle of refraction (θ₂) is a constant, called the relative refractive index (n₂₁) of the second medium with respect to the first.

Equivalently: n₁ sinθ₁ = n₂ sinθ₂
Named after Dutch mathematician Willebrord Snell (1621), though Descartes published it independently.

2. Refractive Index — Absolute and Relative

▶ Absolute Refractive Index (n)
n = c / v = Speed of light in vacuum / Speed of light in medium
Definition: The absolute refractive index of a medium is the ratio of the speed of light in vacuum to the speed of light in that medium.

n = c/v, where c = 3×10&sup8; m/s (speed of light in vacuum), v = speed of light in the medium.

Key properties:
• n is always ≥ 1 (light is always slower in a medium than in vacuum).
• For vacuum: n = 1 (exactly). For air: n ≈ 1.0003 ≈ 1 (treated as 1 in most problems).
• Optically denser medium: higher n (e.g., glass n ≈ 1.5, diamond n = 2.42).
• Optically rarer medium: lower n (e.g., air n ≈ 1).
• When light goes from rarer to denser medium (e.g., air to glass): it bends toward the normal (θ₂ < θ₁).
• When light goes from denser to rarer medium (e.g., glass to air): it bends away from the normal (θ₂ > θ₁).
MediumRefractive Index (n)Speed of LightOptical Density
Vacuum1.00003.00 × 10&sup8; m/sReference
Air1.0003 ≈ 12.999 × 10&sup8; m/sVery low
Water (liquid)1.332.26 × 10&sup8; m/sLow
Crown Glass1.521.97 × 10&sup8; m/sMedium
Flint Glass1.701.76 × 10&sup8; m/sHigh
Diamond2.421.24 × 10&sup8; m/sVery high
Ice1.312.29 × 10&sup8; m/sLow
Relative Refractive Index
n₂₁ = n₂/n₁ = v₁/v₂ = sinθ₁/sinθ₂
n₂₁ = refractive index of medium 2 with respect to medium 1
n₁ = absolute RI of medium 1  |  n₂ = absolute RI of medium 2
v₁ = speed in medium 1  |  v₂ = speed in medium 2
Reversibility: n₁₂ = 1/n₂₁ (RI from medium 2 to medium 1 = reciprocal of RI from 1 to 2)

3. Refraction Through a Glass Slab — Lateral Displacement

💡 Ray Diagram: Refraction Through a Rectangular Glass Slab

Air (n=1) N1 (Normal at entry surface) \ | Incident\ | ray \ i | \ | ===============\=|========================= Entry surface Glass (n=1.5) \|r |\ Refracted ray inside glass | | ================ |===\===================== Exit surface Air (n=1) | \ e | N2 (Normal at exit surface) \ Emergent ray Key: i = angle of incidence, r = angle of refraction (inside glass) e = angle of emergence = i (emergent ray is parallel to incident ray) Lateral displacement d = t sin(i-r)/cos(r) [t = thickness of slab]
Key results for a glass slab:
• The incident ray and the emergent ray are parallel to each other (same angle, same direction), but laterally displaced.
• The emergent ray is shifted sideways from the original direction — this sideways shift is called lateral displacement (d).
• Lateral displacement: d = t sin(i−r)/cos(r), where t = thickness of slab, i = angle of incidence, r = angle of refraction.
• Lateral displacement increases with: (1) increasing angle of incidence, (2) increasing slab thickness, (3) increasing refractive index of glass.
• When i = 0 (normal incidence): no bending, no lateral displacement (ray passes straight through).

4. Real and Apparent Depth

🔍 Real Depth vs Apparent Depth
n = Real Depth / Apparent Depth  ⇒  Apparent Depth = Real Depth / n
Why does a pool look shallower? When light travels from water (denser, n=1.33) to air (rarer, n=1), it bends away from the normal. Our eyes trace the refracted rays back in straight lines (ignoring refraction), so the object appears to be at a shallower depth than it actually is.

Formula: n (of denser medium, viewed from rarer) = Real Depth / Apparent Depth
⇒ Apparent Depth = Real Depth / n
⇒ Shift (apparent rise) = Real Depth − Apparent Depth = Real Depth × (1 − 1/n)

Examples:
• A swimming pool 2 m deep appears (2/1.33) ≈ 1.5 m deep when viewed from above.
• A coin at the bottom of a water-filled vessel appears closer to the surface.
• A fish swimming 3 m below the water surface appears to be at 3/1.33 ≈ 2.25 m depth.
• Stars appear higher in the sky than their actual position (atmospheric refraction: apparent shift upward at horizon ≈ 0.5°, roughly equal to the Moon’s angular diameter).

5. Total Internal Reflection (TIR) and Critical Angle

⚡ Total Internal Reflection (TIR)

sin C = 1/n   ⇒   n = 1/sin C
Conditions for TIR:
(1) Light must travel from a denser medium to a rarer medium (e.g., glass to air, water to air).
(2) The angle of incidence must be greater than the critical angle (i > C).

Critical Angle (C): The angle of incidence in the denser medium for which the angle of refraction in the rarer medium is exactly 90° (refracted ray grazes the surface). When i > C, no refraction occurs — all light is reflected back into the denser medium (Total Internal Reflection).

Derivation of critical angle: At critical angle C, θ₂ = 90° (refracted ray along surface).
Using Snell’s law: n sin C = 1 × sin 90° = 1 ⇒ sin C = 1/n ⇒ C = sin²(1/n)

Critical angles for common materials:
• Water (n=1.33): C = sin²(1/1.33) = 48.8°
• Crown Glass (n=1.52): C = sin²(1/1.52) = 41.1°
• Diamond (n=2.42): C = sin²(1/2.42) = 24.4° — very small C means light bounces internally many times → extraordinary sparkle!

What happens at different angles (denser to rarer):
• i < C: Refraction occurs (some reflection, mostly refraction).
• i = C: Refracted ray grazes the surface (θ₂ = 90°).
• i > C: Total Internal Reflection (no refraction; all light reflects back inside denser medium).

6. Applications of Total Internal Reflection

💡
Optical Fibres (Fibre Optics)
Light is passed into a glass core (high n) surrounded by a cladding (lower n). The angle of incidence at the core-cladding boundary always exceeds the critical angle, causing TIR. Light bounces along the fibre with virtually no loss, enabling high-speed internet, telephone cables, medical endoscopes, and industrial inspection cameras.
🔮
Diamonds & Gemstones
Diamond has n = 2.42 and critical angle C = 24.4°. Light entering a well-cut diamond undergoes multiple TIR bounces before exiting through the top faces at specific angles, creating the intense sparkle and fire effect. Diamond cutters design facets to maximise TIR and brilliance.
🌵
Mirage (Hot Road Mirage)
Hot air near a road surface has lower density (lower n) than cooler air above. Light from the sky undergoes TIR in the hot air layer near the ground and reaches our eyes from below the horizon — appearing as a reflective puddle of water on the road (a mirage). Seen in deserts and on hot tarmac roads.
🏫
Medical Endoscopes
Endoscopes use fibre optic bundles to transmit images from inside the body (stomach, lungs, intestines) to a screen outside. Light travels by TIR through thousands of thin glass fibres. Doctors can diagnose internal conditions without surgery.
📷
Periscopes with Prisms
45° right-angle prisms replace mirrors in high-quality periscopes and binoculars. Light hits the hypotenuse face of a glass prism at 45° (well above C ≈ 41° for glass), undergoing TIR. Prisms give brighter images than silvered mirrors (which lose 10–15% light) and don’t tarnish.
🔌
High-Speed Internet (Fibre Broadband)
Modern broadband internet uses optical fibre cables running under oceans and across countries. Data is encoded as pulses of light (laser beams) that travel by TIR with losses as low as 0.2 dB/km — far superior to copper cables. A single optical fibre can carry millions of phone calls or internet sessions simultaneously.

7. Real-Life Examples of Refraction

🍽
Spoon in a glass of water
A spoon appears bent at the water surface. Light from the spoon below water refracts as it passes from water to air, bending away from the normal. Our brain traces the refracted rays back in straight lines, making the spoon look bent at the surface.
Early sunrise / late sunset
We see the Sun about 2 minutes before it physically rises above the horizon, and 2 minutes after it actually sets, due to atmospheric refraction. Light bends as it passes through layers of atmosphere with increasing density, curving around Earth’s curvature.
👁
Eyeglasses and contact lenses
Corrective lenses use refraction to bend light before it enters the eye, compensating for defects like myopia (concave lens, diverges light) or hypermetropia (convex lens, converges light). The curvature of the lens determines the correction provided.
🌈
Rainbow formation
Sunlight entering spherical raindrops is refracted, internally reflected, and refracted again on exit. Dispersion (different n for different wavelengths) separates colours: red (least refracted, outer arc, n≈1.331) to violet (most refracted, inner arc, n≈1.343).
🔯
Twinkling of stars
Stars twinkle (scintillate) because starlight passes through many layers of turbulent atmosphere with varying refractive indices. The continuously changing n bends light randomly, making the apparent position of the star shift slightly moment to moment — seen as twinkling. Planets don’t twinkle because they are extended sources.
🚿
Camera and telescope lenses
Camera lenses use multiple elements of different glass types (different n) to focus light precisely on the sensor, correct chromatic aberration, and minimise distortion. The refraction at each curved glass surface contributes to image formation — basic principle of all refracting optical instruments.

8. Solved Numerical Problems

Q1. A ray of light is incident at an angle of 45° on a glass slab of refractive index 1.5. Find the angle of refraction inside the glass.
Given: θ₁ = 45° (in air, n₁ = 1)  |  n₂ = 1.5 (glass)
Using Snell’s Law: n₁ sinθ₁ = n₂ sinθ₂
1 × sin 45° = 1.5 × sinθ₂
sinθ₂ = sin 45° / 1.5 = 0.7071 / 1.5 = 0.4714
θ₂ = sin²(0.4714)
θ₂ ≈ 28.1° (angle of refraction in glass)
Q2. The refractive index of glass is 1.5. What is the speed of light in glass? (c = 3×10&sup8; m/s)
Given: n = 1.5  |  c = 3×10&sup8; m/s
n = c/v ⇒ v = c/n = (3×10&sup8;) / 1.5
v = 2×10&sup8; m/s = 2,00,000 km/s
Q3. A fish is swimming 1.5 m below the surface of a lake. What is its apparent depth when viewed from directly above? (n of water = 1.33)
Given: Real depth = 1.5 m  |  n = 1.33
Apparent Depth = Real Depth / n = 1.5 / 1.33
Apparent Depth ≈ 1.13 m (the fish appears 0.37 m closer than it actually is)
Q4. Calculate the critical angle for glass-air interface if the refractive index of glass is 1.52.
Given: nᵊᵋᵃᴸᴸ = 1.52  |  nᵃᵄᵅ = 1 (air)
sin C = nᵃᵄᵅ/nᵊᵋᵃᴸᴸ = 1/1.52 = 0.6579
C = sin²(0.6579)
C ≈ 41.1° (for angles > 41.1° inside glass, TIR occurs at glass-air interface)
Q5. A ray of light travels from water (n=1.33) into glass (n=1.5). If the angle of incidence is 30°, find the angle of refraction.
Given: n₁ = 1.33 (water)  |  n₂ = 1.5 (glass)  |  θ₁ = 30°
n₁ sinθ₁ = n₂ sinθ₂
1.33 × sin30° = 1.5 × sinθ₂
1.33 × 0.5 = 1.5 × sinθ₂
sinθ₂ = 0.665 / 1.5 = 0.4433
θ₂ = sin²(0.4433)
θ₂ ≈ 26.3° (ray bends toward normal: going from rarer water to denser glass)

🌎 Complete AJKANT Optics Cluster — 4 Guides

#66
Convex Lens & Focal Length
#67
Concave Mirror & Focal Length
#71
Glass Prism & Refraction
#78
Refraction of Light & Snell’s Law ✓

9. Frequently Asked Questions (FAQ)

Q1. State the two laws of refraction of light.

First Law: The incident ray, the refracted ray, and the normal to the refracting surface at the point of incidence all lie in the same plane.

Second Law (Snell’s Law): For a given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for a given wavelength of light.
sinθ₁/sinθ₂ = n₂₁ = constant (called relative refractive index of medium 2 w.r.t. medium 1)
Equivalently: n₁ sinθ₁ = n₂ sinθ₂ (using absolute refractive indices)

Q2. What is the refractive index of a medium? On what factors does it depend?

Refractive index (n) of a medium is the ratio of the speed of light in vacuum to the speed of light in that medium: n = c/v.

Factors affecting refractive index:
(1) Nature of the medium: Denser materials (more atoms/molecules per unit volume) have higher n (e.g., glass n=1.5, diamond n=2.42).
(2) Wavelength (colour) of light: n is different for different wavelengths. Violet light has higher n than red light in glass (this causes dispersion — prism separates white light into spectrum). This is called dispersion.
(3) Temperature: As temperature increases, density decreases, so n decreases slightly.
(4) Pressure: For gases, higher pressure increases density and hence n.
Note: Refractive index does NOT depend on the angle of incidence (Snell’s law shows n is constant for given media and wavelength).

Q3. What is total internal reflection? State the conditions for TIR.

Total Internal Reflection (TIR): When a light ray travels from a denser medium to a rarer medium and the angle of incidence exceeds the critical angle, no refraction occurs. All the light is completely reflected back into the denser medium. This phenomenon is called total internal reflection.

Conditions for TIR:
(1) Light must travel from denser to rarer medium (e.g., glass to air, water to air, glass to water if nᵊᵋᵃᴸᴸ > nᵏᵃᵗᵃᵅ).
(2) Angle of incidence must be greater than the critical angle: i > C, where sin C = nᵅᵃᵅᵃᵅ/nₒᵃᵗᴸᵃᵅ = 1/n (when rarer medium is air).

Critical Angle: The specific angle of incidence for which the angle of refraction is exactly 90° (refracted ray grazes the boundary surface). sin C = 1/n for glass-air interface.
For crown glass (n=1.52): C = 41.1°
For diamond (n=2.42): C = 24.4° (very small — explains brilliant sparkle!)

Q4. What is optical fibre? Explain its working principle.

Optical fibre is a thin, flexible strand of very pure glass or plastic (diameter ≈ 10 μm for single-mode, 50–100 μm for multi-mode) used to transmit light signals over long distances with minimal loss.

Structure:
Core: Central glass strand with high refractive index (nₑᵓᵅᵃ ≈ 1.48).
Cladding: Outer glass layer with lower refractive index (nₒᵈᵃₒₒᵄᵗᵊ ≈ 1.46).
Buffer coating: Protective plastic outer layer.

Working principle: Light enters the core at one end. At every core-cladding interface, the angle of incidence (measured from the normal to the curved surface) exceeds the critical angle C. Therefore, TIR occurs at each bounce, and the light zigzags through the core from one end to the other with virtually no loss — even through bends and curves.

Applications: (1) High-speed internet (broadband fibre cables). (2) International submarine cables carrying internet traffic between continents. (3) Medical endoscopes (imaging inside body). (4) Industrial inspection cameras. (5) Decorative optical fibre lamps.

Q5. Why does a pool of water appear shallower than it is? Write the formula for apparent depth.

When an observer in air looks at an object submerged in water, light rays from the object refract at the water-air boundary, bending away from the normal (going from denser water to rarer air). The observer’s eye receives the diverging refracted rays and, extending them backward in straight lines (the eye/brain does not account for refraction), appears to see the object at a point closer to the surface than its actual position.

Formula:
n (of water) = Real Depth / Apparent Depth
⇒ Apparent Depth = Real Depth / n
For water (n = 1.33): A pool 2 m deep appears only 2/1.33 = 1.5 m deep.

Apparent shift (how much closer it appears):
Shift = Real Depth − Apparent Depth = d(1 − 1/n)
For 2 m pool: Shift = 2(1 − 1/1.33) = 2 × 0.25 = 0.5 m (appears 0.5 m closer than it is).

Caution in swimming pools: This is why pools always appear shallower than they are — a potentially dangerous illusion for non-swimmers who underestimate the depth.

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