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Critical Angle (Total Internal Reflection)

Calculates the minimum angle of incidence at which light traveling from a denser to a less dense medium undergoes total internal reflection.

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Critical Angle CalculatorTotal Internal Reflection

θc = arcsin( n2 / n1 )
θc = critical angle (°)  ·  n1 = index of incident medium  ·  n2 = index of refractive medium (n2 < n1)
⟹ Solveθc, n1, n2
°
Please fix the errors above.
Solve for:
Presets:
Critical Angle
θc: n1: n2:
Condition: n₁ > n₂ (light travels from denser to rarer medium)
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Critical Angle
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θc = arcsin(n2/n1)  ·  Total internal reflection occurs when angle of incidence > θc.

Interpretation

Critical angle: θ_c = arcsin(n₂/n₁), where n₁ > n₂. It is the angle of incidence above which total internal reflection occurs. Example: n₁=1.5 (glass), n₂=1.0 (air) → θ_c = arcsin(0.667) ≈ 41.8°.

theta_c = arcsin(n2 / n1)
Critical Angle (Total Internal Reflection)

Variables

SymbolQuantityUnit
theta_cCritical angledegrees
n1Index of refraction of the denser (incident) medium
n2Index of refraction of the less dense (transmitted) medium

What it means

The critical angle is the angle of incidence in a denser medium (higher n) for which the refracted ray in the rarer medium grazes the interface (angle of refraction = 90°). For any incidence angle greater than θ_c, total internal reflection occurs, meaning all light is reflected back into the denser medium. This phenomenon is the basis for optical fibres, where light is guided along the core by repeated total internal reflection. It is also used in prisms, endoscopes, and in diamond cutting to exploit its brilliance. The formula θ_c = arcsin(n₂/n₁) is derived from Snell’s law. Understanding the critical angle is essential for designing fibre optics and for studying light propagation in media.

Worked example

Critical Angle – Two Examples

Real‑World
Scenario: Light travels from glass (n₁ = 1.5) to air (n₂ = 1.0). Find the critical angle.
ParameterValue
n₁1.5
n₂1.0
1θ_c = arcsin(n₂/n₁) = arcsin(1/1.5) = arcsin(0.667) = 41.8°
Result 41.8° ✓ Total internal reflection
Scenario: Water (n₁ = 1.33) to air (n₂ = 1.0). Find critical angle.
ParameterValue
n₁1.33
n₂1.0
1θ_c = arcsin(1/1.33) = arcsin(0.752) = 48.8°
Result 48.8° ✓ Larger
Key insight: Critical angle = arcsin(n₂/n₁) – beyond this, light undergoes total internal reflection.

Common mistakes

  • Critical angle: Only defined when n₁ > n₂ (light going from denser to rarer medium).
  • Arcsine: θ_c = arcsin(n₂/n₁) – ensure the ratio ≤ 1; if >1, no total internal reflection.
  • Units: Result in degrees or radians – be consistent.
  • Total internal reflection: Occurs when angle of incidence > θ_c.
  • Media: n₁ is the incident medium, n₂ is the refracting medium.

Applications

The critical angle for total internal reflection, θ_c = arcsin(n₂/n₁), determines the angle above which light is totally reflected within a medium. This principle is the basis for optical fibres, which rely on total internal reflection to guide light with minimal loss. Engineers use it to design fibre optic cables, endoscopes, and telecommunications systems. It is also applied in prism spectrometers, in refractive index measurement, and in light‑guiding displays. By understanding the critical angle, professionals can ensure efficient light transmission and create innovative optical devices that exploit this effect.

  • Design of optical fibre communication networks
  • Endoscopes and medical imaging devices
  • Prism spectrometers and refractometers
  • Optical waveguides and couplers
  • Light‑guiding displays and signage

Frequently Asked Questions

Q01What is the critical angle for total internal reflection?
A01

The critical angle θ_c is the angle of incidence in the denser medium (n₁) for which the refracted ray in the less dense medium (n₂) travels along the interface. It is given by θ_c = arcsin(n₂/n₁), where n₁ > n₂.

Q02What is the common mistake when applying this formula?
A02

Applying it when light travels from a less dense to a denser medium (n₂ > n₁). In that case, total internal reflection is not possible because the sine of the critical angle would be > 1.

Q03What happens when the angle of incidence is greater than the critical angle?
A03

All light is reflected back into the denser medium; no light transmits into the second medium. This is the principle behind optical fibres.

Q04What is the critical angle for water‑air?
A04

For water (n₁=1.333) to air (n₂=1.000), θ_c = arcsin(1/1.333) ≈ 48.6°.

Q05What is the critical angle for glass‑air?
A05

For typical glass (n₁=1.5) to air, θ_c ≈ 41.8°. This is why glass prisms can reflect light internally.

Q06How is total internal reflection used in optical fibres?
A06

Light is confined within the fibre core by repeated total internal reflection at the core‑cladding interface. The cladding has a lower refractive index than the core.

Q07What is the condition for the maximum acceptance angle in a fibre?
A07

The acceptance angle is determined by the numerical aperture: NA = n₀·sinθ_a = √(n₁² – n₂²).

Q08What is the significance of the critical angle in geology?
A08

In seismology, the critical angle is used to analyse the refraction of seismic waves at layer boundaries, helping to determine Earth's internal structure.