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Grade 10-11 (age 15-17)

high school physics Total Internal Reflection

When light passes from a denser medium with a higher refractive index into a less dense one with a lower index, it bends away from the normal. As the angle of incidence grows, the angle of refraction reaches 90° first; the angle of incidence at that moment is the critical angle θc, found from sinθc = n₂/n₁. If the angle of incidence is larger than the critical angle, the light cannot get out through the boundary and is all reflected; this is total internal reflection, the principle that lets optical fiber carry light over long distances. The picture shows light going from a medium of index 2 into one of index 1, so the critical angle is 30°, and raising the angle of incidence past 30° makes the refracted ray disappear.

At Large Angles the Light Cannot Get Out

20°
👀 Reading the Picture
①Orange is the incoming light, green is the refracted ray escaping upward, and red is the reflected ray coming back down
②Because the light goes from denser to less dense, the angle of refraction θ' is larger than the angle of incidence θ
③Some light is reflected even below the critical angle, but past 30° the refracted ray vanishes and all the light is reflected

Refraction and the Critical Angle

Law of Refraction (Snell's Law)
n1 sinθ1 = n2 sinθ2
The refractive index times the sine of the angle is the same on both sides of the boundary
Critical Angle
sinθc = n2n1 (n1 > n2)
The angle of incidence at which the angle of refraction becomes 90°; beyond it, total internal reflection occurs

Conditions for Total Internal Reflection and Optical Communication

🔦 Both Conditions Are Needed
①The light must go from a denser medium into a less dense one (n₁ > n₂)
②The angle of incidence must be larger than the critical angle
③In optical fiber, a core with a higher index is wrapped in cladding with a lower index, so the light keeps undergoing total internal reflection inside the core and travels a long way

Examples

Example 1
Light leaves a medium of index 2 for air (index 1). Find the critical angle.
1
Put n₁ = 2 and n₂ = 1 into sinθc = n₂/n₁.
sinθc = 12
2
Find the angle whose sine is 1/2.
θc = 30°
▸ 30°
The less dense medium's index n₂ goes on top and the denser medium's index n₁ on the bottom. Swapping them gives sinθc = 2, which has no angle.
Example 2
Light leaves a medium of index √2 for air. Find the critical angle.
1
The index of air is 1, so sinθc = 1/√2.
2
Find the angle whose sine is 1/√2.
θc = 45°
▸ 45°
When the index drops from 2 to √2, the critical angle grows from 30° to 45°. The smaller the difference in index between the two media, the harder total internal reflection becomes.

Summary

Key Result
sinθc = n2n1, incidence > θc ⇒ total reflection
Light going from a denser to a less dense medium is all reflected once the angle of incidence passes the critical angle, and light in an optical fiber is kept inside the core the same way
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
1
The light leaves into air, so sinθc = 1/n and n = 1/sinθc.
n = 1sin30° = 11/2
2
Substitute sin 30° = 1/2 and compute.
n = 2
🎯 Exam Points
①Snell's law n₁sinθ₁ = n₂sinθ₂
②The critical angle is sinθc = n₂/n₁ (n₁ > n₂)
③Total internal reflection needs both conditions: going from a denser to a less dense medium, and an angle of incidence larger than the critical angle
④With the same outer medium, the higher the index of the inner medium, the smaller the critical angle
⑤Optical fiber is made of a high-index core and low-index cladding and carries light by total internal reflection
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