Total Internal Reflection

A Steep Angle Traps the Light

Angle of incidence θ20°

👀 See It

①Light leaves a denser medium (higher index) for a rarer one

②As the incidence angle grows, the refracted ray lies down toward the boundary

③Beyond the critical angle, all light bounces back — total reflection

Refraction and the Critical Angle

Snell's law

n₁ sinθ₁ = n₂ sinθ₂

Index × sine is equal on both sides of the boundary

Critical angle

sinθ_c = n₂n₁ (n₁ > n₂)

The incidence angle that makes the refraction angle 90° — beyond it, total reflection

Conditions and Optical Fiber

🔦 Two Conditions Needed

①denser → rarer medium (n₁ > n₂)

②incidence angle > critical angle

③optical fiber has a core denser than the cladding, trapping light by total reflection

Compute It Directly

Example 1

Light leaves a medium of index 2 for air (index 1). Find the critical angle.

Substitute into sinθ_c = n₂/n₁.

sinθ_c = 12

Find the angle whose sine is 1/2.

θ_c = 30°

▸ 30°

Note the critical angle uses sinθc = n₂/n₁, not the indices reversed.

Example 2

Light leaves a medium of index √2 for air. Find the critical angle.

sinθ_c = 1/√2.

Find the angle whose sine is 1/√2.

θ_c = 45°

▸ 45°

A smaller index gives a larger critical angle (total reflection is harder).

Wrap-up

Key result

sinθ_c = n₂n₁, incidence > θ_c ⇒ total reflection

From denser to rarer, exceeding the critical angle gives total reflection — the fiber principle

2021 CSAT Science (Physics Ⅰ) type, adapted

When light leaves a medium for air, the critical angle is 30°. What is the medium’s index?

①1.5

②√2

③√3

④2

⑤2.5

▸ ④ 2

From sinθ_c = 1/n, we get n = 1/sinθ_c.

n = 1sin30° = 11/2

Compute.

n = 2

🎯 Exam Points

①Snell's law n₁sinθ₁=n₂sinθ₂

②critical angle sinθc=n₂/n₁ (n₁>n₂)

③total reflection needs denser→rarer AND angle>critical

④larger index, smaller critical angle

⑤optical fiber = core (dense) + cladding (rare)

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