Refraction of Light — Class 10 Physical Science
"Light doesn't just BOUNCE. It BENDS. When it passes from air to water, from water to glass — it changes SPEED, and therefore DIRECTION. This is REFRACTION."
1. What Is Refraction?
The BENDING of light when it passes OBLIQUELY from one TRANSPARENT medium to another. WHY does it bend? Because LIGHT CHANGES SPEED. 'Light travels FASTEST in vacuum (c = 3×10⁸ m/s). It slows down in any medium. When it enters at an ANGLE: one side of the wavefront slows down BEFORE the other → the light BENDS.'
Key Observations
- Light ray, normal, and refracted ray all lie in the SAME PLANE.
- Going from RARER to DENSER (air → water): light bends TOWARD the NORMAL. Angle of refraction is SMALLER.
- Going from DENSER to RARER (water → air): light bends AWAY from the NORMAL. Angle of refraction is LARGER.
- If light strikes PERPENDICULARLY (i = 0°): NO BENDING. r = 0°. 'Only when light strikes at an ANGLE does it refract. Normal incidence = straight through.'
2. Snell's Law: n₁ sin i = n₂ sin r
n₁ = refractive index of first medium. n₂ = refractive index of second medium.
3. Refractive Index (n): n = c/v
c = speed of light in vacuum (3×10⁸ m/s). v = speed of light in the medium. 'n is ALWAYS ≥ 1. Higher n = light travels SLOWER in that medium = MORE bending.' n for water = 1.33. n for glass = 1.5. n for diamond = 2.42 (very high — diamond sparkles because light bends so much inside it).
4. Total Internal Reflection (TIR)
Conditions
- Light must travel from DENSER to RARER medium (e.g., water → air, glass → air).
- Angle of incidence MUST be GREATER than the CRITICAL ANGLE (i > i_c).
Critical Angle: i_c = sin⁻¹(n₂/n₁). For water → air: i_c = sin⁻¹(1/1.33) ≈ 49°. For glass → air: i_c ≈ 42°.
Applications
Optical Fibres: Thin strands of glass. Light enters at one end. Undergoes REPEATED TIR down the fibre. Used for: INTERNET (fibre optic cables carry data as light pulses). MEDICAL ENDOSCOPY (seeing inside the body without surgery).
Mirages: On a HOT ROAD: air near the road is HOTTER (less dense, lower n) than air above. Light from the sky undergoes TIR at the boundary → reaches your eye → the road looks WET (reflecting the sky).
Diamond Sparkle: Diamond has a VERY high refractive index (2.42) → LOW critical angle (24°). Light entering a diamond undergoes MANY TIRs before emerging → BRILLIANT sparkle.
5. Common Mistakes
- 'Refraction always bends light' — When i = 0° (perpendicular), there is NO bending. The ray goes straight through.
- 'Light speeds up in denser medium' — Light SLOWS DOWN in denser medium. It's FASTEST in vacuum.
- 'Total internal reflection = reflection from a mirror' — TIR occurs at a BOUNDARY between media. Mirror reflection uses a COATED surface.
6. AP SSC Exam Focus
| Topic | Marks |
|---|---|
| Snell's Law problems | 3-4 |
| Refractive index definition | 2-3 |
| TIR and applications | 3-4 |
| Critical angle | 2-3 |
7. Worked Numerical Problems — Snell's Law and Refractive Index
Example 1: Light enters from air to water at an angle of incidence of 30°. If the refractive index of water is 1.33, find the angle of refraction. Solution: n₁ sin i = n₂ sin r. n_air = 1. 1 × sin 30° = 1.33 × sin r. 0.5 = 1.33 × sin r. sin r = 0.5/1.33 = 0.376. r = sin⁻¹(0.376) ≈ 22°. 'Light BENDS TOWARD the normal as it enters water.'
Example 2: The speed of light in a medium is 2 × 10⁸ m/s. Find the refractive index of the medium. (c = 3 × 10⁸ m/s) Solution: n = c/v = (3 × 10⁸)/(2 × 10⁸) = 1.5. The medium is likely GLASS (n ≈ 1.5).
Example 3: If the refractive index of diamond is 2.42, find the speed of light in diamond. Solution: n = c/v → v = c/n = (3 × 10⁸)/2.42 = 1.24 × 10⁸ m/s. 'Light travels at HALF the speed in diamond compared to vacuum — that's why diamond has such HIGH dispersive power.'
Example 4: Light passes from glass (n = 1.5) into water (n = 1.33). The angle of incidence in glass is 25°. Find the angle of refraction in water. Solution: n₁ sin i = n₂ sin r. 1.5 × sin 25° = 1.33 × sin r. 1.5 × 0.423 = 1.33 × sin r. 0.634 = 1.33 × sin r. sin r = 0.634/1.33 = 0.477. r = sin⁻¹(0.477) ≈ 28.5°. 'Since r > i, light bends AWAY from the normal — going from denser to rarer.'
8. Refraction Through a Glass Slab
Lateral Displacement: When light passes through a PARALLEL-SIDED glass slab: the EMERGENT ray is PARALLEL to the INCIDENT ray — but SHIFTED sideways. This shift is called LATERAL DISPLACEMENT (d).
Factors affecting Lateral Displacement: (1) Thickness of slab (t) — more thickness = more displacement. (2) Refractive index of glass (n) — higher n = more bending = more displacement. (3) Angle of incidence (i) — displacement changes with angle.
Verification of Snell's Law: 'A glass slab experiment is used to VERIFY Snell's law. Measure i and r at the FIRST surface. Measure e (emergence angle) and r' at the SECOND surface. You should find: i = e (for parallel sides) and r = r'.'
Lateral Displacement Formula: d = t × sin(i−r) / cos r. 'You don't need to MEMORISE the formula — but understand that d ∝ t and d ∝ sin(i−r).'
9. Lenses — Convex and Concave
| Convex Lens (Converging) | Concave Lens (Diverging) | |
|---|---|---|
| Shape | THICKER in middle | THINNER in middle |
| Effect on parallel rays | CONVERGES to focus | DIVERGES (appears to come from focus) |
| Focal length sign | POSITIVE (+) | NEGATIVE (−) |
| Image types | Real or virtual | ALWAYS virtual, erect, diminished |
Lens Formula: 1/f = 1/v − 1/u
Magnification: m = v/u = h'/h
'Note the DIFFERENCE from mirror formula: mirror has 1/f = 1/u + 1/v. Lens has 1/f = 1/v − 1/u. The sign of u is NEGATIVE (same as mirrors — object in front). f is POSITIVE for convex, NEGATIVE for concave.'
Power of a Lens: P = 1/f (in metres)
Unit: DIOPTRE (D). 1 D = 1 m⁻¹. Convex lens → P is POSITIVE. Concave lens → P is NEGATIVE. 'Power tells you HOW STRONG the lens is — higher |P| means more bending. A lens with f = 20 cm has P = 1/0.2 = 5 D. A lens with f = −50 cm has P = −2 D.'
Ray Diagrams for Convex Lens (6 Cases)
| Object Position | Image Position | Size | Nature |
|---|---|---|---|
| At infinity | At F | Point-sized | REAL, inverted |
| Beyond 2F | Between F and 2F | Diminished | REAL, inverted |
| At 2F | At 2F | Same size | REAL, inverted |
| Between F and 2F | Beyond 2F | Magnified | REAL, inverted |
| At F | At infinity | Highly magnified | REAL, inverted |
| Between F and O | On SAME side as object | Magnified | VIRTUAL, erect |
Ray Diagrams for Concave Lens (Always the Same)
Image is ALWAYS: VIRTUAL, ERECT, DIMINISHED, between F and O on the same side as the object.
10. Worked Numerical Problems — Lenses
Example 5: An object is placed 30 cm from a convex lens of focal length 20 cm. Find the image position and magnification. Solution: u = −30 cm, f = +20 cm. 1/f = 1/v − 1/u → 1/v = 1/f + 1/u = 1/20 + 1/(−30) = 1/20 − 1/30 = (3−2)/60 = 1/60. v = +60 cm. m = v/u = 60/(−30) = −2. 'Image is REAL (v positive for a lens means real), INVERTED (m negative), MAGNIFIED (2×), at 60 cm on the other side of the lens.'
Example 6: A concave lens of focal length 15 cm has an object placed 30 cm from it. Find the image position. Solution: u = −30 cm, f = −15 cm. 1/f = 1/v − 1/u → 1/v = 1/f + 1/u = 1/(−15) + 1/(−30) = −1/15 − 1/30 = (−2−1)/30 = −3/30 = −1/10. v = −10 cm. m = v/u = (−10)/(−30) = +1/3. 'Image is VIRTUAL (v negative), ERECT (m positive), DIMINISHED (|m| = 1/3), on the SAME side as the object — exactly as expected for a concave lens.'
Example 7: A convex lens produces a real image at 40 cm when the object is at 20 cm. Find the focal length. Solution: u = −20 cm, v = +40 cm (real → v positive). 1/f = 1/v − 1/u = 1/40 − 1/(−20) = 1/40 + 1/20 = 1/40 + 2/40 = 3/40. f = 40/3 ≈ 13.33 cm.
11. Self-Test
Q1: Light enters from air to glass (n = 1.5) at an angle of incidence of 45°. Calculate the angle of refraction. A1: n₁ sin i = n₂ sin r. 1 × sin 45° = 1.5 × sin r. 0.707 = 1.5 × sin r. sin r = 0.707/1.5 = 0.471. r = sin⁻¹(0.471) ≈ 28.1°.
Q2: What is the difference between the mirror formula and the lens formula? A2: Mirror formula: 1/f = 1/u + 1/v. Lens formula: 1/f = 1/v − 1/u. In both, u is NEGATIVE. In the mirror formula, f is negative for concave and positive for convex. In the lens formula, f is positive for convex and negative for concave.
Q3: An object is placed at 2F of a convex lens. Where is the image formed? What are its characteristics? A3: Image is formed at 2F on the OTHER side. Characteristics: REAL, INVERTED, SAME SIZE (m = −1).
Q4: Calculate the power of a convex lens of focal length 25 cm. A4: f = 25 cm = 0.25 m. P = 1/f = 1/0.25 = +4 D. Positive because it's a convex (converging) lens.
Q5: A ray of light passes through a glass slab. Why is the emergent ray parallel to the incident ray? A5: The light bends TOWARD the normal when entering glass (rarer→denser) and AWAY from the normal when exiting (denser→rarer). Since the two surfaces are PARALLEL, the NET bending is ZERO — the emergent ray is parallel but LATERALLY DISPLACED.
Q6: Why do diamond and glass sparkle? A6: Diamond has a very HIGH refractive index (2.42) → LOW critical angle (~24°) → light entering the diamond undergoes TOTAL INTERNAL REFLECTION multiple times before emerging → intense SPARKLE. Glass has lower n (~1.5) → higher critical angle (~42°) → less TIR → less sparkle.
Q7: What is the critical angle for a glass-air interface if n_glass = 1.5? A7: i_c = sin⁻¹(n₂/n₁) = sin⁻¹(1/1.5) = sin⁻¹(0.667) ≈ 41.8°. Total internal reflection occurs when i > 41.8°.
