About the Critical Angle Calculator
This critical angle calculator finds the angle of incidence above which light travelling inside a denser material can no longer escape and is totally internally reflected. Pick the material the light is travelling in (the higher-index side) and the material outside it from a list — water, glass, acrylic, diamond, air and others — or enter custom refractive indices, and the critical angle is returned in degrees.
Physics students use it for refraction and total internal reflection problems; it is also handy for understanding optical fibres, prisms, light pipes, gemstone brilliance and why a swimming pool surface looks like a mirror from underwater. Treating the inner material as a fibre core and the outer one as cladding, the calculator also reports the numerical aperture and the maximum acceptance angle in air. Enter an incidence angle to check whether a particular ray will be totally reflected.
Total internal reflection can only happen when light goes from a higher refractive index to a lower one. Angles are measured from the normal to the surface. Refractive indices vary slightly with wavelength; the listed values are typical figures for yellow light around 589 nm.
With the default inputs, the critical angle (θc) is 41.825 °. Change any value above to recalculate instantly.
How to use the critical angle calculator
- 1Select the material the light is travelling in (higher refractive index).
- 2Select the material on the other side of the boundary.
- 3Use custom indices for a specific glass, fibre core or cladding.
- 4Optionally enter an incidence angle to see whether the ray escapes.
- 5Read the critical angle, numerical aperture and acceptance angle.
Formula and method
The critical angle follows from Snell's law, n₁ sin θ₁ = n₂ sin θ₂. As the incidence angle grows, the refracted ray bends further from the normal until it skims along the surface at θ₂ = 90°. Setting sin θ₂ = 1 gives sin θc = n₂/n₁. For any incidence angle larger than θc there is no refracted ray and all of the light is reflected — total internal reflection. This requires n₁ > n₂.
In a step-index optical fibre, light must hit the core–cladding boundary beyond the critical angle to stay trapped. Working back through the entrance face gives the numerical aperture NA = √(n₁² − n₂²) and the acceptance half-angle in air θa = arcsin(NA). An NA of 1 or more means every ray from air is guided.
- θc
- Critical angle, measured from the normal
- n₁
- Refractive index of the material the light is in (higher)
- n₂
- Refractive index of the material outside (lower)
- NA
- Numerical aperture of a fibre with core n₁ and cladding n₂
- θa
- Acceptance half-angle in air
Worked examples
Glass (n = 1.50) to air
θc = arcsin(1.000293 / 1.5) = 41.83°. A ray hitting the inside of the glass at 45° is beyond this, so it is totally internally reflected — the principle behind 45° reflecting prisms in binoculars.
Water to air (looking up from a pool)
θc = arcsin(1.000293 / 1.333) = 48.63°. A ray at 30° is below that and leaves the water at 41.78° from the normal. Beyond 48.6° the surface acts as a mirror, which is why divers see a bright circle of sky overhead (Snell’s window).
Optical fibre: core 1.47, cladding 1.46
θc = arcsin(1.46 / 1.47) = 83.31°, so light must travel nearly parallel to the fibre axis. The numerical aperture √(1.47² − 1.46²) = 0.171 gives an acceptance half-angle of 9.86° from air.
Frequently asked questions
What is the critical angle?+
It is the angle of incidence, measured from the normal inside the denser material, at which the refracted ray travels along the boundary (90°). At larger angles light cannot escape and is totally internally reflected.
What is the critical angle for glass and water?+
For typical glass (n ≈ 1.50) into air it is about 41.8°, for water (n ≈ 1.333) into air about 48.6°, and for diamond (n ≈ 2.417) into air only about 24.4°, which is why diamonds trap and sparkle light.
Why is there no critical angle from air into glass?+
Light entering a higher-index material bends toward the normal, so a refracted ray always exists. Total internal reflection is only possible when going from a higher index to a lower index.
How do optical fibres use the critical angle?+
The core has a slightly higher index than the cladding. Light striking the boundary beyond the critical angle is reflected without loss over and over, carrying signals for kilometres.
How is the critical angle related to Snell’s law?+
Setting the refracted angle to 90° in n₁ sin θ₁ = n₂ sin θ₂ gives sin θc = n₂ / n₁. It is simply the limiting case of Snell’s law.