Refraction of Light - Bending Light Paths
The Movie Hook 🎬
Ever noticed how Interstellar’s black hole bends light creating gravitational lensing? Or why the pool in The Graduate looks shallower than it is? Or how Cast Away could use water droplets as magnifying glasses? That’s all refraction - light bending as it enters different materials!
The Big Picture
Refraction is the bending of light when it passes from one medium to another due to change in speed. This is the foundation for:
- Lenses (glasses, cameras, microscopes)
- Optical fibers (internet cables!)
- Atmospheric phenomena (mirages, rainbows)
- Total Internal Reflection (diamonds sparkle!)
Connection Alert: Links to Thin Lens, Prism, and Wave Optics.
Core Concepts
1. Refractive Index
Definition: Ratio of speed of light in vacuum to speed in medium
$$\boxed{n = \frac{c}{v}}$$Where:
- n = refractive index (dimensionless)
- c = speed of light in vacuum = 3 × 10⁸ m/s
- v = speed of light in medium
Key Values to Remember:
| Medium | Refractive Index |
|---|---|
| Vacuum | 1.0 (exactly) |
| Air | ~1.0003 ≈ 1 |
| Water | 1.33 |
| Glass | 1.5 |
| Diamond | 2.42 |
Memory Trick: “Very Wet Girls Dance” (Vacuum < Water < Glass < Diamond)
2. Snell’s Law
When light passes from medium 1 to medium 2:
$$\boxed{n_1 \sin i = n_2 \sin r}$$Or equivalently:
$$\boxed{\frac{n_2}{n_1} = \frac{\sin i}{\sin r} = \frac{v_1}{v_2} = \frac{\lambda_1}{\lambda_2}}$$Critical Insight: Frequency remains constant during refraction!
Memory Trick: “Silly Nice Index Students Rule” (sn × i = sn × r)
3. Behavior at Interface
Case 1: Rarer to Denser (n₂ > n₁)
- Light bends towards normal
- r < i (angle of refraction < angle of incidence)
- Speed decreases
- Wavelength decreases
Case 2: Denser to Rarer (n₂ < n₁)
- Light bends away from normal
- r > i
- Speed increases
- Wavelength increases
4. Apparent Depth & Real Depth
When viewing object in water from air:
$$\boxed{\text{Apparent Depth} = \frac{\text{Real Depth}}{n}}$$Or more generally:
$$\boxed{n = \frac{\text{Real Depth}}{\text{Apparent Depth}}}$$For small angles:
$$\boxed{\text{Shift} = t\left(1 - \frac{1}{n}\right)}$$Where t = thickness of medium
JEE Trick: Object appears closer when viewing from denser to rarer medium!
5. Total Internal Reflection (TIR)
Conditions for TIR:
- Light travels from denser to rarer medium (n₁ > n₂)
- Angle of incidence > Critical angle (i > iₓ)
Critical Angle
$$\boxed{\sin i_c = \frac{n_2}{n_1}}$$For air-water:
$$i_c = \sin^{-1}\left(\frac{1}{1.33}\right) \approx 48.6°$$For air-glass:
$$i_c = \sin^{-1}\left(\frac{1}{1.5}\right) \approx 41.8°$$For air-diamond:
$$i_c = \sin^{-1}\left(\frac{1}{2.42}\right) \approx 24.4°$$Memory Trick: “TIR requires Denser to Rarer and Critical” (DRC)
6. Applications of TIR
Optical Fibers: Light trapped by repeated TIR
- Core has higher n than cladding
- Used in telecommunications, internet
Diamonds: Small critical angle → multiple TIR → sparkle
Mirage: Hot air (rarer) below, cooler air (denser) above → TIR
Prisms: 45-45-90 prism used in binoculars
7. Refraction Through Glass Slab
Key Result: Emergent ray is parallel to incident ray but laterally displaced
$$\boxed{\text{Lateral Displacement} = t \frac{\sin(i - r)}{\cos r}}$$Where:
- t = thickness of slab
- i = angle of incidence
- r = angle of refraction
For small angles:
$$\boxed{d \approx t(i - r) = t \cdot i\left(1 - \frac{1}{n}\right)}$$The Formula Cheat Sheet
REFRACTION ESSENTIALS:
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
1. Refractive index: n = c/v
2. Snell's law: n₁sin i = n₂sin r
3. Apparent depth: d_app = d_real / n
4. Critical angle: sin iₓ = n₂/n₁
5. TIR condition: i > iₓ (denser→rarer)
6. Slab shift: s = t(1 - 1/n)
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
Common Traps & Mistakes
Trap #1: Confusing Rarer and Denser
❌ Wrong: Assuming higher density → higher refractive index ✅ Right: Optically denser means higher n (not physical density!)
Trap #2: TIR Direction
❌ Wrong: Expecting TIR when going rarer → denser ✅ Right: TIR only when denser → rarer
Trap #3: Critical Angle Formula
❌ Wrong: sin iₓ = n₁/n₂ ✅ Right: sin iₓ = n₂/n₁ (smaller n in numerator!)
Trap #4: Frequency vs Wavelength
❌ Wrong: Thinking frequency changes during refraction ✅ Right: Only speed and wavelength change; frequency constant!
Practice Problems
Level 1: JEE Main Warmup
Problem 1.1: Light travels from air (n=1) into water (n=1.33) at 60° to normal. Find angle of refraction.
Solution
Given: n₁ = 1, n₂ = 1.33, i = 60°
Using Snell’s law:
$$n_1 \sin i = n_2 \sin r$$ $$1 \times \sin 60° = 1.33 \times \sin r$$ $$\sin r = \frac{\sqrt{3}/2}{1.33} = \frac{0.866}{1.33} = 0.651$$ $$r = \sin^{-1}(0.651) \approx 40.6°$$Answer: 40.6° (bends towards normal as expected)
Problem 1.2: A coin lies at the bottom of a pool 3m deep (n=1.33). What is its apparent depth when viewed from above?
Solution
Answer: 2.26 m (appears closer!)
Problem 1.3: Find the critical angle for water-air interface (n_water = 1.33).
Solution
For water → air: n₁ = 1.33, n₂ = 1
$$\sin i_c = \frac{n_2}{n_1} = \frac{1}{1.33} = 0.752$$ $$i_c = \sin^{-1}(0.752) \approx 48.8°$$Answer: 48.8°
Level 2: JEE Main/Advanced
Problem 2.1: A ray of light passes through a glass slab (n=1.5) of thickness 6 cm at an angle of incidence 60°. Find the lateral displacement.
Solution
Given: n = 1.5, t = 6 cm, i = 60°
First find r using Snell’s law:
$$\sin r = \frac{n_1 \sin i}{n_2} = \frac{1 \times \sin 60°}{1.5} = \frac{\sqrt{3}/2}{1.5} = 0.577$$ $$r = \sin^{-1}(0.577) = 35.3°$$Lateral displacement:
$$d = t \frac{\sin(i-r)}{\cos r}$$ $$d = 6 \times \frac{\sin(60° - 35.3°)}{\cos 35.3°}$$ $$d = 6 \times \frac{\sin 24.7°}{\cos 35.3°} = 6 \times \frac{0.418}{0.816} = 3.07 \text{ cm}$$Answer: 3.07 cm
Problem 2.2: A light ray is incident on a water surface at polarizing angle (Brewster’s angle). If refractive index of water is 4/3, find the angle of refraction.
Solution
At Brewster’s angle:
$$\tan i_p = n = \frac{4}{3}$$This gives:
$$i_p = \tan^{-1}(4/3) = 53.1°$$Special property: At Brewster’s angle:
$$i_p + r = 90°$$Therefore:
$$r = 90° - 53.1° = 36.9°$$We can verify using Snell’s law:
$$1 \times \sin 53.1° = \frac{4}{3} \times \sin r$$ $$\sin r = \frac{3 \times 0.8}{4} = 0.6$$ $$r = 36.9°$$✓
Answer: 36.9° (perpendicular to incident ray!)
Problem 2.3: What should be the minimum refractive index of a material for which critical angle is 30°?
Solution
For critical angle with air (n₂ = 1):
$$\sin i_c = \frac{n_2}{n_1} = \frac{1}{n_1}$$Given iₓ = 30°:
$$\sin 30° = \frac{1}{n_1}$$ $$\frac{1}{2} = \frac{1}{n_1}$$ $$n_1 = 2$$Answer: Refractive index = 2
Level 3: JEE Advanced
Problem 3.1: A point source is placed at the bottom of a tank filled with water (n=4/3) to height h. An opaque disc of radius r is placed on water surface with its center vertically above the source. For total darkness above the disc, find minimum r in terms of h.
Solution
For total darkness, all light from source must undergo TIR at water surface.
Critical angle:
$$\sin i_c = \frac{n_{\text{air}}}{n_{\text{water}}} = \frac{1}{4/3} = \frac{3}{4}$$At the edge of disc, ray makes critical angle:
$$\tan i_c = \frac{r}{h}$$From sin iₓ = 3/4, we can find cos iₓ:
$$\cos i_c = \sqrt{1 - \sin^2 i_c} = \sqrt{1 - \frac{9}{16}} = \frac{\sqrt{7}}{4}$$ $$\tan i_c = \frac{\sin i_c}{\cos i_c} = \frac{3/4}{\sqrt{7}/4} = \frac{3}{\sqrt{7}}$$Therefore:
$$r = h \tan i_c = \frac{3h}{\sqrt{7}}$$Answer:
$$r_{\min} = \frac{3h}{\sqrt{7}} = \frac{3h\sqrt{7}}{7}$$Problem 3.2: A ray of light is incident at an angle of 60° on one face of a rectangular glass slab of thickness 0.1 m and refractive index √3. Calculate the lateral shift and time taken by light to pass through the slab.
Solution
Part A: Lateral Shift
Given: i = 60°, n = √3, t = 0.1 m
Find angle of refraction:
$$\sin r = \frac{\sin i}{n} = \frac{\sin 60°}{\sqrt{3}} = \frac{\sqrt{3}/2}{\sqrt{3}} = \frac{1}{2}$$ $$r = 30°$$Lateral displacement:
$$d = t \frac{\sin(i-r)}{\cos r} = 0.1 \times \frac{\sin 30°}{\cos 30°} = 0.1 \times \frac{1/2}{\sqrt{3}/2} = \frac{0.1}{\sqrt{3}} = 0.0577 \text{ m}$$Part B: Time Taken
Distance traveled in glass:
$$l = \frac{t}{\cos r} = \frac{0.1}{\cos 30°} = \frac{0.1}{\sqrt{3}/2} = \frac{0.2}{\sqrt{3}}$$Speed in glass:
$$v = \frac{c}{n} = \frac{3 \times 10^8}{\sqrt{3}}$$Time:
$$t = \frac{l}{v} = \frac{0.2/\sqrt{3}}{3 \times 10^8/\sqrt{3}} = \frac{0.2}{3 \times 10^8} = 6.67 \times 10^{-10} \text{ s}$$Answers:
- Lateral shift = 5.77 cm
- Time = 6.67 × 10⁻¹⁰ s = 0.667 ns
Problem 3.3: A container is filled with water (n=4/3) up to height 20 cm. A point source is placed at the bottom. If we pour oil (n=1.2) on water to form a 5 cm layer, find the apparent depth of source as seen from above oil.
Solution
This is a two-layer problem. Work backwards from top.
Step 1: Apparent depth of source as seen from oil-air interface
The source appears at some depth d₁ when viewed from water’s top surface:
$$d_1 = \frac{20}{4/3} = 15 \text{ cm below water surface}$$Step 2: Now this “apparent source” is at 15 cm in water, but below 5 cm of oil
Total depth from oil surface = 5 + 15 = 20 cm (but 5 cm is in oil, 15 cm is apparent position in water)
Actually, we need to use the refraction through each layer:
Correct approach: When viewing from air through oil then water:
Shift due to water:
$$s_w = h_w\left(1 - \frac{1}{n_w}\right) = 20\left(1 - \frac{3}{4}\right) = 5 \text{ cm}$$Object appears at: 20 - 5 = 15 cm from water’s bottom
Now this appears through oil of thickness 5 cm:
$$s_o = 5\left(1 - \frac{1}{1.2}\right) = 5 \times 0.167 = 0.833 \text{ cm}$$Total apparent depth from top surface = 15 + 5 - 0.833 = 19.17 cm
Answer: 19.17 cm from top surface
Better formula: For composite medium:
$$d_{\text{app}} = \frac{t_1}{n_1} + \frac{t_2}{n_2} = \frac{5}{1.2} + \frac{20}{1.33} = 4.17 + 15.04 = 19.21 \text{ cm}$$Answer: Approximately 19.2 cm
Advanced Concepts
1. Relative Refractive Index
$$\boxed{_1n_2 = \frac{n_2}{n_1} = \frac{v_1}{v_2}}$$Reads as “refractive index of medium 2 with respect to medium 1”
2. Cauchy’s Formula (Dispersion)
$$\boxed{n = A + \frac{B}{\lambda^2}}$$Shows that n depends on wavelength (leads to dispersion in prisms!)
3. Vector Form of Snell’s Law
For advanced ray tracing problems - rarely asked in JEE but good to know.
Quick Revision Cards
╔══════════════════════════════════════╗
║ REFRACTION QUICK FACTS ║
╚══════════════════════════════════════╝
🎯 n = c/v (always ≥ 1)
🎯 n₁sin i = n₂sin r
🎯 Rarer→Denser: bends toward normal
🎯 Denser→Rarer: bends away from normal
🎯 TIR: denser→rarer AND i > iₓ
🎯 sin iₓ = n_rarer/n_denser
🎯 Apparent depth = Real depth / n
🎯 Frequency constant in refraction
Cross-Topic Connections
- Reflection: Combined with refraction in lenses
- Thin Lens: Uses refraction at curved surfaces
- Prism: Double refraction + dispersion
- Wave Optics: Huygens’ principle explains refraction
- EM Waves: Speed of light in different media
Exam Strategy
Time Allocation:
- Snell’s law application: 2 min
- TIR problems: 3 min
- Multi-layer problems: 4-5 min
Common JEE Patterns:
- Apparent depth + slab combinations
- Critical angle calculations
- TIR in optical fibers
- Fish looking at bird problems (reverse apparent depth)
- Swimming pool problems
Red Flags:
- Watch for “fish observing bird” - reverse the formula!
- Multiple media - apply Snell’s law at each interface
- TIR questions - check both conditions
Final Tips
- Draw ray diagrams - shows bending direction instantly
- Check if i > iₓ before concluding TIR
- Remember: Denser medium has larger n, smaller v
- For quick checks: If going into water, angle decreases; if leaving water, angle increases
Pro Tip: In JEE, if problem mentions “critical angle” or “just undergoes TIR,” angle is exactly iₓ!
Next up: Thin Lens Formula - where we combine refraction at two surfaces!