Physics · Geometric optics

Critical angle for total internal reflection

θc is incidence where the limiting propagating refracted direction becomes tangent. Above it, all normal power is reflected in the ideal two-medium model.

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Formula

Calculate

sin θ_{c} = \frac{n _{2}}{n _{1}} (n_{1} > n_{2})

n₁,n₂ dimensionless; θc in degrees or radians. Differential angle uncertainty uses radians.

Variables & units

Variables & units
Symbol	Meaning	Unit
$n1$	Incident phase index, larger than n₂	dimensionless
$n2$	Second medium phase index at the same frequency	dimensionless
$θc$	Critical angle from the local normal	rad or °

LaTeX

\sin\theta_c=\frac{n_2}{n_1}\quad(n_1>n_2)

QUANTA STUDYquanta-study.de

Three separate ray diagrams with angles from the normal: 35° refracts to 59.36°; at 41.81° the limiting direction is tangent; at 50° reflection and a decaying evanescent field remain. — The same model interface from n₁ = 1.50 to n₂ = 1.00: below, at and above the critical angle 41.81°. The decay profile is schematic. Above it an evanescent field remains, but no propagating refracted ray.

Critical angle calculator

Find the critical angle for total internal reflection from two refractive indices and check whether a ray is totally reflected at its angle of incidence.

Calculation model

Refractive index n₁ (incoming side)

Refractive index of the medium the light travels in, larger than n₂, e.g. 1.50 for glass.

Refractive index n₂ (far side)

Refractive index of the medium behind the boundary, e.g. 1.00 for air.

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for θc, substitute, evaluate

1. Rearrange for θc
$\theta_c=\arcsin\frac{n_2}{n_1}$
2. Substitute values (in SI units)
$\arcsin\frac{1}{1.5}$
3. Result
θc = 41.81 °

Result: Critical angle from the normal θc

θc = 41.81 °

With n₁ = 1.5 and n₂ = 1, θc = 41.81 °.

Quick tips

Total internal reflection only happens from the optically denser into the less dense medium, so n₁ > n₂. From air into glass there is no critical angle.
The larger n₁ compared with n₂, the smaller the critical angle: water to air 48.8°, glass to air 41.8°, diamond to air 24.4°.

Assumptions and limits

Stationary smooth planar boundary of ordinary isotropic lossless media with n₁>n₂>0 at the same frequency. Angles from the local normal, not surface or fibre axis. Equality is the grazing threshold; strict total internal reflection requires a larger angle. No loss, thin-gap or guided-mode solver.

Derivation & background

Set θ₂=90° in Snell’s law. Incident index must be larger. Above threshold, the second medium’s normal wavevector becomes imaginary; its field amplitude decays without carrying normal net power into the ideal semi-infinite medium. A nearby third medium can couple to it. Fiber entry links sidewall threshold to end-face refraction; axial acceptance angle and normal critical angle are different quantities.

Exam blueprint

Validity range

A stationary smooth planar boundary of ordinary isotropic lossless media with $n_1>n_2$ and indices at the same frequency. Total reflection for $\theta_1>\theta_c$ ; equality is the threshold. Absorption, roughness, bends and thin-gap coupling can produce real losses or transmission.

Derivation steps

Insert the limiting propagating direction into Snell’s law.

1Assign n₁ to the incident medium and n₂ to the other medium; total reflection requires $n_1>n_2$ .
2Measure both angles from the local normal and use $n_1\sin\theta_1=n_2\sin\theta_2$ .
3As θ₁ increases, θ₂ reaches the limiting 90° tangent direction.
4Set $\sin\theta_2=1$ , giving $n_1\sin\theta_c=n_2$ .
5Divide by n₁ and take inverse sine: $\theta_c=\arcsin\!\left(\frac{n_2}{n_1}\right)$ .
6For $\theta_1>\theta_c$ , the required refraction sine exceeds one. No propagating transmitted ray exists, but an evanescent field remains in medium 2.

Rearrangements

Critical angle

\theta_c=\arcsin\left(\frac{n_2}{n_1}\right)

Requires $n_1>n_2$ ; angle is measured from the normal.

Task variant

Light travels from model glass n₁ = 1.520 into air n₂ = 1.000. Is incidence 45° ± 1° from the normal above the ideal total-reflection threshold throughout? Also compare the critical angle for n₁ = 1.50. A straight meridional fiber has core index 1.46 and cladding index 1.44: find its acceptance half-angle for entry from air.

Approach: $\theta_c=\arcsin\frac{n_2}{n_1}$ .
For 1.520 toward 1.000 the threshold is 41.1395°; even 44° exceeds it by 2.8605°.
For 1.50 toward 1.00 the threshold is 41.8103°.
The straight meridional fiber has sidewall critical angle 80.5055°.
At entry from air, $\sin\alpha_{\max}=\sqrt{1.46^2-1.44^2}$ gives acceptance half-angle 13.9356°.
Sidewall and entrance angles use different normals.

Common mistakes

Ignoring units or assumptions stated for the formula: n₁,n₂ dimensionless; θc in degrees or radians. Differential angle uncertainty uses radians.

Convert units before substitution and verify the validity range; round only at the end.

Exam context

Analyze fiber guides, reflection prisms, endoscopy and evanescent sensing geometries; account separately for real losses and angular margin.

Formula cluster

Geometric optics

Related relations: brechungsgesetz-snellius, linsengleichung, malus-gesetz-polarisation.

n₁·sin(α₁) = n₂·sin(α₂)

Snell's Law of Refraction

1/f = 1/g + 1/b

Thin Lens Equation

I = I₀ cos²θ

Malus’s law

Worked example

Model glass 1.50 toward air 1.00 gives θc≈41.81°. For 1.520, θc=41.1395°;45°±1° remains above it in the specified angle model. A 1.46 core with 1.44 cladding gives θc=80.5055° and a meridional acceptance half-angle in air of 13.9356°.

Applications

Analyze fiber guides, reflection prisms, endoscopy and evanescent sensing geometries; account separately for real losses and angular margin.

How do you calculate with Critical angle for total internal reflection?

Here is how to work through a typical Critical angle for total internal reflection (sin θc = n₂/n₁) task step by step:

1
Task
Light travels from model glass n₁ = 1.520 into air n₂ = 1.000. Is incidence 45° ± 1° from the normal above the ideal total-reflection threshold throughout? Also compare the critical angle for n₁ = 1.50. A straight meridional fiber has core index 1.46 and cladding index 1.44: find its acceptance half-angle for entry from air.
Solution path
1. Approach: $\theta_c=\arcsin\frac{n_2}{n_1}$ .
2. For 1.520 toward 1.000 the threshold is 41.1395°; even 44° exceeds it by 2.8605°.
3. For 1.50 toward 1.00 the threshold is 41.8103°.
4. The straight meridional fiber has sidewall critical angle 80.5055°.
5. At entry from air, $\sin\alpha_{\max}=\sqrt{1.46^2-1.44^2}$ gives acceptance half-angle 13.9356°.
6. Sidewall and entrance angles use different normals.

Understand and apply the equation

Inside a prism, light can return from a face without a mirror coating. Propagation direction and incidence angle decide whether this happens. Critical angle separates propagating transmission from total internal reflection. First check the threshold, then a guide’s angular margin and finally the field that remains beyond the reflecting interface.

Before calculating

Recall that Snell’s law measures angles from the normal. Total internal reflection requires incidence from higher toward lower refractive index. A large angle alone is insufficient. Before calculating, identify the incident medium and the normal to the actual interface.[1]

The basic model uses two lossless isotropic media and a planar boundary; initially the second medium is sufficiently extended. Within this model, “total” means unit reflected normal power fraction. A real fiber with bends, roughness or absorption additionally needs a loss model.[3][4]

What the threshold separates

The wave must approach from larger index n₁ toward smaller index n₂. Increasing its angle from the normal increases the refracted angle faster. At critical incidence, the limiting refracted direction reaches 90° from the normal, tangent to the surface. Ray-model total internal reflection occurs for $\theta_1>\theta_c$ . Equality is the threshold case and should be identified separately.[1]

\begin{aligned}&\theta_c=\arcsin\frac{n_2}{n_1}\\&n_1>n_2\end{aligned}

Use indices at the same frequency; measure angles from the local normal.[1]

When $n_1<n_2$ , the ratio exceeds one and no critical angle of this kind exists. Equal indices give the algebraic limit 90°, but no index step and no total-reflection interval. An inadmissible inverse sine is a physical result, not something to repair by swapping indices.[1]

The refracted-angle branch ends at threshold

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The curve ends at 90° refraction. Beyond its last incidence angle the model has no propagating refracted ray; missing points do not mean zero degrees.[1]

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Derive the boundary from Snell and the wavevector

Insert θ₂=90° into $n_1\sin\theta_1=n_2\sin\theta_2$ . Since $\sin\theta_2=1$ , $n_1\sin\theta_c=n_2$ . With $n_1>n_2$ the ratio lies between zero and one, giving a unique critical angle between 0° and 90°. It follows from refraction for ordinary positive indices rather than defining a new empirical material constant.[2]

\begin{aligned}n_1\sin\theta_c&=n_2\sin90^\circ\\\sin\theta_c&=\frac{n_2}{n_1}\\\theta_c&=\arcsin\frac{n_2}{n_1}\qquad(n_1>n_2)\end{aligned}

Insert the grazing limiting direction before solving for incidence.[1][2]

k_{2z}^2=\left(\frac{2\pi}{\lambda_0}\right)^2\!\left(n_2^2-n_1^2\sin^2\theta_1\right)

z is the normal coordinate in medium 2. Above threshold the normal wavevector component becomes imaginary.[3]

The tangential component remains real. Thus the field can have a phase along the boundary while decaying perpendicular to it. Saying no light enters medium 2 may describe the absence of a propagating transmitted ray, but it is false as a claim about the entire electromagnetic field.[3]

Snell’s law supplies the underlying angle relation; the critical-angle calculation inserts grazing refraction into that relation. Keep the same propagation order and local normals rather than swapping indices merely to obtain an admissible inverse sine.[1][2]

An angle alone does not guarantee a reflection margin

Model glass against air

Given: n₁=1.520; n₂=1.000; incidence 45.0° with possible±1.0° deviation.
Find: Check critical angle and worst-case incidence.

Approach: The critical limit has grazing transmission. Check the smallest possible incidence angle when including tolerance.
$\begin{aligned}\theta_c&=\arcsin\frac{n_2}{n_1}\\\theta_{1,\min}&=\theta_{1,\mathrm{nom}}-\Delta\theta\\\theta_{1,\min}&>\theta_c\end{aligned}$
Form the ratio and find threshold.
$\theta_c=\arcsin\frac{1}{1.520}=41.1395^\circ$
The smallest possible angle from the normal is 44°.
$\theta_{1,\min}-\theta_c=44.0^\circ-41.1395^\circ=2.8605^\circ$

Result: Even the worst-case angle leaves about 2.86° above threshold.

Check: Check: at 44°, $\frac{n_1\sin\theta_1}{n_2}\approx1.056>1$ , excluding a propagating refracted ray. This margin covers the specified angular uncertainty, not coatings or changes in index.

From internal threshold to external acceptance angle

In a straight step-index fiber, a meridional ray is guided when incidence at the core-cladding boundary exceeds the threshold. If β is the core ray angle to the fiber axis, sidewall incidence is $\theta_1=90^\circ-\beta$ . Entry at a flat end face and reflection at the wall therefore use different normals.[1]

A weakly guiding model fiber

Given: ncore=1.460; ncladding=1.440; external n₀=1.000; flat end face normal to the axis.
Find: Find critical angle, maximum β and external acceptance half-angle α.

Approach: For a meridional ray, combine sidewall reflection with entry refraction; axial and normal angles are complementary.
$\begin{aligned}\theta_c&=\arcsin\frac{n_{\mathrm{clad}}}{n_{\mathrm{core}}}\\\beta_{\max}&=90^\circ-\theta_c\\n_0\sin\alpha_{\max}&=n_{\mathrm{core}}\sin\beta_{\max}\end{aligned}$
First calculate internal sidewall threshold.
$\begin{aligned}&\theta_c=\arcsin\frac{1.440}{1.460}=80.5055^\circ\\&\beta_{\max}=9.4945^\circ\end{aligned}$
Combine end-face refraction with the internal limiting condition.
$n_0\sin\alpha_{\max}=n_{\mathrm{core}}\sin\beta_{\max}=\sqrt{n_{\mathrm{core}}^2-n_{\mathrm{clad}}^2}=0.240832$
Convert numerical aperture into the external angle.
$\alpha_{\max}=13.9356^\circ$

Result: The acceptance half-angle is about 13.94°, giving a full geometric cone of about 27.87°. The edge is the threshold; reliable guidance requires margin.

Check: Sinβmax≈0.16495; multiplying by 1.460 gives 0.240832. A large internal normal angle 80.51° is consistent with a small external axial angle 13.94°. This ray model does not determine the number of guided modes.

Explore further: evanescent field and decay length

For lossless semi-infinite media with $\theta_1>\theta_c$ , field amplitude in medium 2 decays as $\exp(-\kappa z)$ . The length $\frac1\kappa$ is its amplitude decay length. Squared amplitude decays twice as fast, with 1/e length $\frac1{2\kappa}$ . Always state which definition of penetration depth is being used.[4]

\begin{aligned}&\kappa=\frac{2\pi}{\lambda_0}\sqrt{n_1^2\sin^2\theta_1-n_2^2}\\&d_A=\kappa^{-1}\end{aligned}

For n₁=1.50, n₂=1.00, θ₁=50° and λ₀=550 nm, dA≈154.7 nm and squared-amplitude decay length≈77.3 nm.[4]

A nearby suitable third medium can couple to the evanescent field and receive energy: frustrated total internal reflection. A thin air gap cannot therefore be treated as an infinitely thick surrounding medium. The critical-angle formula identifies the initial threshold, not transmission through a finite gap.[4]

How far the evanescent field extends

Given: n₁=1.50; n₂=1.00; θ₁=45°; λ₀=600 nm; lossless semi-infinite media.
Find: Find amplitude decay length and squared field at distance dA.

Approach: check threshold first, then use the imaginary normal wavevector component for decay length.
$\begin{aligned}\theta_c&=\arcsin\frac{n_2}{n_1}\\d_A&=\frac{\lambda_0}{2\pi\sqrt{n_1^2\sin^2\theta_1-n_2^2}}\\\frac{|E(z)|^2}{|E(0)|^2}&=\exp\!\left(-\frac{2z}{d_A}\right)\end{aligned}$
45° exceeds 41.81°, and the square-root argument is positive.
$\begin{aligned}n_1^2\sin^245^\circ-n_2^2&=2.25\cdot\frac12-1=0.125\\d_A&=\frac{600\,\mathrm{nm}}{2\pi\sqrt{0.125}}\approx270.095\,\mathrm{nm}\end{aligned}$
The squared field has half the decay length.
$\begin{aligned}d_{|E|^2}&=\frac{d_A}{2}\approx135.047\,\mathrm{nm}\\\frac{|E(d_A)|^2}{|E(0)|^2}&=e^{-2}\approx0.135335\end{aligned}$

Result: Over its 270 nm decay length amplitude falls to e⁻¹; the squared field has already fallen to about 13.53%.

Check: At half that distance the squared field retains e⁻¹. This is spatial field decay, not normally transmitted ray power. Both indices and vacuum wavelength are used consistently.

Units, index changes and proximity to threshold

The index ratio is dimensionless; θc may be reported in degrees or radians. Differential angle changes must use radians. As r= $\frac{n_2}{n_1}$ approaches one, $\frac{d\theta_c}{dr}=\frac1{\sqrt{1-r^2}}$ becomes large. A small change in index contrast can then produce a substantial angle change. Rounding margins matter particularly for weak contrast.[1]

Indices depend on frequency and material conditions. Temperature, wavelength, cladding and a wet contacting surface can change an earlier margin. For model water 1.33 against air 1.00, threshold is 48.7535°, rather than exactly the often quoted 49°. For broadband illumination, check the relevant frequencies separately.[1]

What is ideally total and what is lost in reality

Power reflectance equals one above threshold in the ideal planar lossless two-medium problem. This does not make every real fiber lossless. Absorption, roughness, bends, scattering and finite cladding thickness can remove or couple out power. Reflection phase also depends on angle and polarization, so repeated total reflection can alter polarization.[3]

Ray geometry is useful when core dimensions are large compared with wavelength. Small guides require an electromagnetic modal solution to determine guidance and propagation. Showing $n_1>n_2$ alone is insufficient; a particular ray must meet the wall at the appropriate local angle.[4]

The Brewster angle describes a different condition: p reflection vanishes at a suitable single transparent interface. The critical-angle threshold instead separates propagating transmission from its absence. A p-reflection minimum must therefore not be interpreted as evidence of total internal reflection.[5][1]

Five decisions with complete solutions

Calculate · Foundation

A ray from n₁=1.60 toward n₂=1.30 hits at 55°. Test total reflection.

Show hint

Calculate threshold before comparing.

Learning goal: Compare incidence with a calculated critical angle.

Show solution and calculation

Approach: incidence is from larger to smaller index.
Calculate the threshold from their ratio and compare it with the specified normal angle. θc≈54.3409° is below 55°, giving total reflection in the ideal lossless model.
Angular margin is only 0.6591°.

Check: $1.60\sin\theta_c=1.30$ ; do not swap the indices.

\begin{aligned}\theta_c&=\arcsin\frac{1.30}{1.60}\approx54.3409^\circ\\55^\circ-\theta_c&\approx0.6591^\circ\end{aligned}

Interpret · Foundation

From n=1.00 into n=1.50 at 80°: can total reflection occur?

Show hint

Inspect propagation direction and index ordering.

Learning goal: Check the required direction of refractive-index change.

Show solution and calculation

Approach: check the required propagation direction before calculating an angle.
The incident index is smaller, so there is no critical angle for total internal reflection.
The formal ratio $\frac{1.50}{1.00}$ exceeds one.
This does not indicate calculator failure.
Partial reflection can still be large at 80°, but it is not total internal reflection.

n_1<n_2

Transfer · Application

A meridional core ray at β=8° to the axis meets cladding with ncore=1.46, ncladding=1.44. Is it guided?

Show hint

Convert to the sidewall normal angle.

Learning goal: Convert fibre-axis angle to wall incidence and test guidance.

Show solution and calculation

Approach: axial angle β and sidewall-normal angle add to 90°.
At β=8°, θ₁=82°.
Critical angle is θc≈80.5055°, leaving 1.4945° margin.
This ray is guided in the straight ideal meridional model.
Check at β=12°: θ₁=78° lies below threshold and does not meet the same guidance condition.

\theta_1=90^\circ-\beta=82^\circ

Interpret · Foundation

Light travels from a lossless medium with $n_1=1.50$ into a second with $n_2=1.00$ . A sketch shows the limiting refracted ray exactly along the surface. Find incidence measured from the normal and classify it as below, at, or above the critical angle. Is incidence 1° smaller already totally internally reflected?

Show hint

A ray along the surface has refraction angle 90° from the normal. Insert this into Snell’s law.

Learning goal: Distinguish the critical limit from strict total internal reflection.

Show solution and calculation

Approach: set the sine of the tangential limiting refraction angle equal to one.
This gives incidence 41.81°.
The sketch represents the threshold itself, not an angle above it.
At 40.81°, a propagating refracted ray still exists.

Check: strict total internal reflection requires travel toward smaller refractive index and incidence greater than the threshold.

\begin{aligned}n_1\sin\theta_c&=n_2\sin90^\circ\\\theta_c&=\arcsin\frac{1.00}{1.50}\approx41.81^\circ\\\theta_1&=40.81^\circ<\theta_c\end{aligned}

[1]

Error analysis · Application

In the ideal lossless two-medium model above the critical angle, the evanescent amplitude is $E(z)=E(0)\exp\!\left(-\frac z{d_A}\right)$ , with $d_A=180\,\mathrm{nm}$ . Someone also calls 180 nm the $e^{-1}$ length of the normalized squared field amplitude. Check this claim and find the normalized field square at $z=180\,\mathrm{nm}$ . This is not normally transmitted ray power.

Show hint

Square the amplitude function. The exponent doubles; an amplitude length and a squared-field length are different quantities.

Learning goal: Distinguish amplitude and intensity decay lengths of an evanescent field.

Show solution and calculation

Approach: form the normalized field square from the amplitude and compare its exponent with a separate $e^{-1}$ length.
That length is 90 nm.
At 180 nm, about 0.1353 of the field square remains, not $e^{-1}$ .

Check: at 90 nm the fraction is indeed $e^{-1}$ . An evanescent field therefore does not mean that the second medium has no field.

\begin{aligned}\frac{|E(z)|^2}{|E(0)|^2}&=\exp\!\left(-\frac{2z}{d_A}\right)\\d_{|E|^2}&=\frac{d_A}{2}=90\,\mathrm{nm}\\\frac{|E(180\,\mathrm{nm})|^2}{|E(0)|^2}&=e^{-2}\approx0.1353\end{aligned}

[3][4]

What to take away

Check index ordering first and local incidence second. Above threshold the propagating refracted ray is absent, not the entire field. Angular margin covers only the tolerances considered. In a fiber, axial angle differs from sidewall-normal angle. State whether a field decay length describes amplitude or its square.[1][3]

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

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This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

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An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 2, 2026Added relevant connections; preserved existing derivations, calculations, exercises, figures and model limits.
October 1, 2026Explained threshold, margin and fiber angles step by step; added a decay-length example and threshold curve; marked the field extension as optional.
October 1, 2026Added prerequisites and model assumptions; clarified derivation steps and arithmetic checks; included an additional original example, diagram and summary.

Cite this reference page

Quanta. Critical angle for total internal reflection. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/grenzwinkel-totalreflexion

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Frequently asked questions about Critical angle for total internal reflection

What does the critical angle for total internal reflection mean?+

Critical angle separates transmission with a propagating refracted ray from ideal total internal reflection. It belongs to propagation from larger toward smaller index. At threshold the limiting transmitted direction is tangent to the surface,90° from the normal. Above it no real refraction angle can satisfy Snell’s law. The ideal lossless two-medium problem then reflects all normal incident power. An evanescent field nevertheless remains in the other medium. For model glass 1.50 toward air 1.00, threshold is 41.8103°. Saying no light exists in medium 2 can only describe the absent propagating transmitted ray, not the disappearance of all electromagnetic fields.

How do I check critical angle and incidence angle?+

First identify the medium containing the incoming ray. Only when its index n₁ exceeds n₂ does their ratio n₂ divided by n₁ define this threshold. For 1.520 toward 1.000, inverse sine gives 41.1395°. Back-substitution must give $n_1\sin\theta_c=n_2$ . Then compare actual incidence from the normal with that value.45° is above,40° below, and equality is the limiting case. Retain unrounded intermediate values near threshold. With 45°±1°, even minimum incidence 44° stays 2.8605° above it. That margin covers the specified angular deviation, not every possible material change or surface imperfection.

Under what assumptions does the total-reflection threshold apply?+

The simple calculation assumes a stationary smooth boundary of ordinary isotropic transparent media with real indices and larger incident index. Complete power reflection additionally belongs to the ideal lossless two-medium model. Real absorption, roughness, bends or finite cladding can remove or couple out power. A thin gap next to another medium permits evanescent coupling and frustrated total internal reflection. Small guides also require electromagnetic modal solutions rather than only ray sketches. Critical angle remains a useful initial geometric check, but it determines neither all propagation losses nor a universal guided-mode count nor polarization after repeated reflections.

How do I check angle units and scale?+

Both indices and their ratio are dimensionless. Inverse sine returns an angle in the chosen calculator mode; label it explicitly as degrees or radians. For $n_1>n_2$ , θc lies between 0° and 90°. As the ratio approaches one, threshold approaches 90° and becomes sensitive to index changes. A 1.46 core with 1.44 cladding gives 80.5055°, not the ray’s axial angle. Small angle deviations in differential uncertainty calculations use radians. Equal indices give no genuine total-reflection interval; smaller incident index makes inverse sine inadmissible because the necessary physical condition is absent. Never swap the indices simply to obtain a numerical answer.

How is critical angle related to a fiber’s acceptance angle?+

For a fiber, combine sidewall reflection threshold with Snell’s law at the entrance face. A meridional core ray at axial angleβ strikes the wall at $90^\circ-\beta$ from its local normal. With core 1.46 and cladding 1.44, limitingβ is 9.4945°. Entry from air gives an external acceptance half-angle 13.9356°. Those angles reference different normals and must not be confused. Describing a thin gap or evanescent measurement instead requires field decay and multilayer coupling. Critical angle supplies the initial threshold but does not yet calculate power transmitted across that gap. Guidance also needs the actual local ray geometry.

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\tan\theta_B=\frac{n_2}{n_1}

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\theta_{\min}\approx1.22\frac{\lambda}{D}

Rayleigh criterion for a circular aperture

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Critical angle for total internal reflection

θc is incidence where the limiting propagating refracted direction becomes tangent. Above it, all normal power is reflected in the ideal two-medium model.

AdvancedExam-relevantUpper secondaryUniversity/collegeBachelor foundations

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Symbol

Meaning

Unit

n1

Incident phase index, larger than n₂

dimensionless

n2

Second medium phase index at the same frequency

dimensionless

θc

Critical angle from the local normal

rad or °

Derivation & background

Exam blueprint

Validity range

Derivation steps

Insert the limiting propagating direction into Snell’s law.

1Assign n₁ to the incident medium and n₂ to the other medium; total reflection requires $n_1>n_2$ .
2Measure both angles from the local normal and use $n_1\sin\theta_1=n_2\sin\theta_2$ .
3As θ₁ increases, θ₂ reaches the limiting 90° tangent direction.
4Set $\sin\theta_2=1$ , giving $n_1\sin\theta_c=n_2$ .
5Divide by n₁ and take inverse sine: $\theta_c=\arcsin\!\left(\frac{n_2}{n_1}\right)$ .
6For $\theta_1>\theta_c$ , the required refraction sine exceeds one. No propagating transmitted ray exists, but an evanescent field remains in medium 2.

How do you calculate with Critical angle for total internal reflection?

Here is how to work through a typical Critical angle for total internal reflection (sin θc = n₂/n₁) task step by step:

Task

Solution path

Approach: $\theta_c=\arcsin\frac{n_2}{n_1}$ .
For 1.520 toward 1.000 the threshold is 41.1395°; even 44° exceeds it by 2.8605°.
For 1.50 toward 1.00 the threshold is 41.8103°.
The straight meridional fiber has sidewall critical angle 80.5055°.
At entry from air, $\sin\alpha_{\max}=\sqrt{1.46^2-1.44^2}$ gives acceptance half-angle 13.9356°.
Sidewall and entrance angles use different normals.

Understand and apply the equation

Before calculating

What the threshold separates

\begin{aligned}&\theta_c=\arcsin\frac{n_2}{n_1}\\&n_1>n_2\end{aligned}

Use indices at the same frequency; measure angles from the local normal.[1]

The refracted-angle branch ends at threshold

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The curve ends at 90° refraction. Beyond its last incidence angle the model has no propagating refracted ray; missing points do not mean zero degrees.[1]

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Derive the boundary from Snell and the wavevector

\begin{aligned}n_1\sin\theta_c&=n_2\sin90^\circ\\\sin\theta_c&=\frac{n_2}{n_1}\\\theta_c&=\arcsin\frac{n_2}{n_1}\qquad(n_1>n_2)\end{aligned}

Insert the grazing limiting direction before solving for incidence.[1][2]

k_{2z}^2=\left(\frac{2\pi}{\lambda_0}\right)^2\!\left(n_2^2-n_1^2\sin^2\theta_1\right)

z is the normal coordinate in medium 2. Above threshold the normal wavevector component becomes imaginary.[3]

An angle alone does not guarantee a reflection margin

Model glass against air

Given: n₁=1.520; n₂=1.000; incidence 45.0° with possible±1.0° deviation.
Find: Check critical angle and worst-case incidence.

Approach: The critical limit has grazing transmission. Check the smallest possible incidence angle when including tolerance.
$\begin{aligned}\theta_c&=\arcsin\frac{n_2}{n_1}\\\theta_{1,\min}&=\theta_{1,\mathrm{nom}}-\Delta\theta\\\theta_{1,\min}&>\theta_c\end{aligned}$
Form the ratio and find threshold.
$\theta_c=\arcsin\frac{1}{1.520}=41.1395^\circ$
The smallest possible angle from the normal is 44°.
$\theta_{1,\min}-\theta_c=44.0^\circ-41.1395^\circ=2.8605^\circ$

Result: Even the worst-case angle leaves about 2.86° above threshold.

Check: Check: at 44°, $\frac{n_1\sin\theta_1}{n_2}\approx1.056>1$ , excluding a propagating refracted ray. This margin covers the specified angular uncertainty, not coatings or changes in index.

From internal threshold to external acceptance angle

A weakly guiding model fiber

Given: ncore=1.460; ncladding=1.440; external n₀=1.000; flat end face normal to the axis.
Find: Find critical angle, maximum β and external acceptance half-angle α.

Approach: For a meridional ray, combine sidewall reflection with entry refraction; axial and normal angles are complementary.
$\begin{aligned}\theta_c&=\arcsin\frac{n_{\mathrm{clad}}}{n_{\mathrm{core}}}\\\beta_{\max}&=90^\circ-\theta_c\\n_0\sin\alpha_{\max}&=n_{\mathrm{core}}\sin\beta_{\max}\end{aligned}$
First calculate internal sidewall threshold.
$\begin{aligned}&\theta_c=\arcsin\frac{1.440}{1.460}=80.5055^\circ\\&\beta_{\max}=9.4945^\circ\end{aligned}$
Combine end-face refraction with the internal limiting condition.
$n_0\sin\alpha_{\max}=n_{\mathrm{core}}\sin\beta_{\max}=\sqrt{n_{\mathrm{core}}^2-n_{\mathrm{clad}}^2}=0.240832$
Convert numerical aperture into the external angle.
$\alpha_{\max}=13.9356^\circ$

Result: The acceptance half-angle is about 13.94°, giving a full geometric cone of about 27.87°. The edge is the threshold; reliable guidance requires margin.

Explore further: evanescent field and decay length

\begin{aligned}&\kappa=\frac{2\pi}{\lambda_0}\sqrt{n_1^2\sin^2\theta_1-n_2^2}\\&d_A=\kappa^{-1}\end{aligned}

For n₁=1.50, n₂=1.00, θ₁=50° and λ₀=550 nm, dA≈154.7 nm and squared-amplitude decay length≈77.3 nm.[4]

How far the evanescent field extends

Given: n₁=1.50; n₂=1.00; θ₁=45°; λ₀=600 nm; lossless semi-infinite media.
Find: Find amplitude decay length and squared field at distance dA.

Approach: check threshold first, then use the imaginary normal wavevector component for decay length.
$\begin{aligned}\theta_c&=\arcsin\frac{n_2}{n_1}\\d_A&=\frac{\lambda_0}{2\pi\sqrt{n_1^2\sin^2\theta_1-n_2^2}}\\\frac{|E(z)|^2}{|E(0)|^2}&=\exp\!\left(-\frac{2z}{d_A}\right)\end{aligned}$
45° exceeds 41.81°, and the square-root argument is positive.
$\begin{aligned}n_1^2\sin^245^\circ-n_2^2&=2.25\cdot\frac12-1=0.125\\d_A&=\frac{600\,\mathrm{nm}}{2\pi\sqrt{0.125}}\approx270.095\,\mathrm{nm}\end{aligned}$
The squared field has half the decay length.
$\begin{aligned}d_{|E|^2}&=\frac{d_A}{2}\approx135.047\,\mathrm{nm}\\\frac{|E(d_A)|^2}{|E(0)|^2}&=e^{-2}\approx0.135335\end{aligned}$

Result: Over its 270 nm decay length amplitude falls to e⁻¹; the squared field has already fallen to about 13.53%.

Check: At half that distance the squared field retains e⁻¹. This is spatial field decay, not normally transmitted ray power. Both indices and vacuum wavelength are used consistently.

Units, index changes and proximity to threshold

What is ideally total and what is lost in reality

Five decisions with complete solutions

Calculate · Foundation

A ray from n₁=1.60 toward n₂=1.30 hits at 55°. Test total reflection.

Show hint

Calculate threshold before comparing.

Learning goal: Compare incidence with a calculated critical angle.

Show solution and calculation

Approach: incidence is from larger to smaller index.
Calculate the threshold from their ratio and compare it with the specified normal angle. θc≈54.3409° is below 55°, giving total reflection in the ideal lossless model.
Angular margin is only 0.6591°.

Check: $1.60\sin\theta_c=1.30$ ; do not swap the indices.

\begin{aligned}\theta_c&=\arcsin\frac{1.30}{1.60}\approx54.3409^\circ\\55^\circ-\theta_c&\approx0.6591^\circ\end{aligned}

Interpret · Foundation

From n=1.00 into n=1.50 at 80°: can total reflection occur?

Show hint

Inspect propagation direction and index ordering.

Learning goal: Check the required direction of refractive-index change.

Show solution and calculation

Approach: check the required propagation direction before calculating an angle.
The incident index is smaller, so there is no critical angle for total internal reflection.
The formal ratio $\frac{1.50}{1.00}$ exceeds one.
This does not indicate calculator failure.
Partial reflection can still be large at 80°, but it is not total internal reflection.

n_1<n_2

Transfer · Application

A meridional core ray at β=8° to the axis meets cladding with ncore=1.46, ncladding=1.44. Is it guided?

Show hint

Convert to the sidewall normal angle.

Learning goal: Convert fibre-axis angle to wall incidence and test guidance.

Show solution and calculation

Approach: axial angle β and sidewall-normal angle add to 90°.
At β=8°, θ₁=82°.
Critical angle is θc≈80.5055°, leaving 1.4945° margin.
This ray is guided in the straight ideal meridional model.
Check at β=12°: θ₁=78° lies below threshold and does not meet the same guidance condition.

\theta_1=90^\circ-\beta=82^\circ

Interpret · Foundation

Light travels from a lossless medium with $n_1=1.50$ into a second with $n_2=1.00$ . A sketch shows the limiting refracted ray exactly along the surface. Find incidence measured from the normal and classify it as below, at, or above the critical angle. Is incidence 1° smaller already totally internally reflected?

Show hint

A ray along the surface has refraction angle 90° from the normal. Insert this into Snell’s law.

Learning goal: Distinguish the critical limit from strict total internal reflection.

Show solution and calculation

Approach: set the sine of the tangential limiting refraction angle equal to one.
This gives incidence 41.81°.
The sketch represents the threshold itself, not an angle above it.
At 40.81°, a propagating refracted ray still exists.

Check: strict total internal reflection requires travel toward smaller refractive index and incidence greater than the threshold.

\begin{aligned}n_1\sin\theta_c&=n_2\sin90^\circ\\\theta_c&=\arcsin\frac{1.00}{1.50}\approx41.81^\circ\\\theta_1&=40.81^\circ<\theta_c\end{aligned}

[1]

Error analysis · Application

In the ideal lossless two-medium model above the critical angle, the evanescent amplitude is $E(z)=E(0)\exp\!\left(-\frac z{d_A}\right)$ , with $d_A=180\,\mathrm{nm}$ . Someone also calls 180 nm the $e^{-1}$ length of the normalized squared field amplitude. Check this claim and find the normalized field square at $z=180\,\mathrm{nm}$ . This is not normally transmitted ray power.

Show hint

Square the amplitude function. The exponent doubles; an amplitude length and a squared-field length are different quantities.

Learning goal: Distinguish amplitude and intensity decay lengths of an evanescent field.

Show solution and calculation

Approach: form the normalized field square from the amplitude and compare its exponent with a separate $e^{-1}$ length.
That length is 90 nm.
At 180 nm, about 0.1353 of the field square remains, not $e^{-1}$ .

Check: at 90 nm the fraction is indeed $e^{-1}$ . An evanescent field therefore does not mean that the second medium has no field.

\begin{aligned}\frac{|E(z)|^2}{|E(0)|^2}&=\exp\!\left(-\frac{2z}{d_A}\right)\\d_{|E|^2}&=\frac{d_A}{2}=90\,\mathrm{nm}\\\frac{|E(180\,\mathrm{nm})|^2}{|E(0)|^2}&=e^{-2}\approx0.1353\end{aligned}

[3][4]

What to take away

Sources for this explanation

Quanta · Updated October 2, 2026

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An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 2, 2026Added relevant connections; preserved existing derivations, calculations, exercises, figures and model limits.
October 1, 2026Explained threshold, margin and fiber angles step by step; added a decay-length example and threshold curve; marked the field extension as optional.
October 1, 2026Added prerequisites and model assumptions; clarified derivation steps and arithmetic checks; included an additional original example, diagram and summary.

Cite this reference page

Quanta. Critical angle for total internal reflection. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/grenzwinkel-totalreflexion

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Critical angle for total internal reflection

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How do you calculate with Critical angle for total internal reflection?

Before calculating

What the threshold separates

The refracted-angle branch ends at threshold

Derive the boundary from Snell and the wavevector

An angle alone does not guarantee a reflection margin

Model glass against air

From internal threshold to external acceptance angle

A weakly guiding model fiber

Explore further: evanescent field and decay length

How far the evanescent field extends

Units, index changes and proximity to threshold

What is ideally total and what is lost in reality

Five decisions with complete solutions

A ray from n1=1.60 toward n2=1.30 hits at 55°. Test total reflection.

From n=1.00 into n=1.50 at 80°: can total reflection occur?

A meridional core ray at β=8° to the axis meets cladding with ncore=1.46, ncladding=1.44. Is it guided?

What to take away

Sources for this explanation

Frequently asked questions about Critical angle for total internal reflection

Subject references and further reading

More formulas: Optics

Retain Critical angle for total internal reflection for exams

Critical angle for total internal reflection

Variables & units

Critical angle calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Geometric optics

Worked example

Applications

How do you calculate with Critical angle for total internal reflection?

Before calculating

What the threshold separates

The refracted-angle branch ends at threshold

Derive the boundary from Snell and the wavevector

An angle alone does not guarantee a reflection margin

Model glass against air

From internal threshold to external acceptance angle

A weakly guiding model fiber

Explore further: evanescent field and decay length

How far the evanescent field extends

Units, index changes and proximity to threshold

What is ideally total and what is lost in reality

Five decisions with complete solutions

A ray from n1=1.60 toward n2=1.30 hits at 55°. Test total reflection.

From n=1.00 into n=1.50 at 80°: can total reflection occur?

A meridional core ray at β=8° to the axis meets cladding with ncore=1.46, ncladding=1.44. Is it guided?

What to take away

Sources for this explanation

Frequently asked questions about Critical angle for total internal reflection

Subject references and further reading

More formulas: Optics

Retain Critical angle for total internal reflection for exams

A ray from n₁=1.60 toward n₂=1.30 hits at 55°. Test total reflection.

A ray from n₁=1.60 toward n₂=1.30 hits at 55°. Test total reflection.