Physics · Optics

Total Internal Reflection

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Wrap-up

What you learned

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Read next: dispersion-refractive-index-colorUse index changes with color

Key takeaway

Check the travel direction first: TIR only appears when light starts in the higher-index medium.
Use sin(theta_c) = n_2 / n_1 to find the boundary threshold for that media pair.
Read theta_1 - theta_c as the live margin between ordinary refraction and total internal reflection.
Explain why changing n_2 can move the same ray back across the threshold.

Common misconception

Do not treat TIR as any steep ray bouncing off a surface. Without higher-to-lower travel, there is no critical-angle cutoff in this model.

Angle alone is not enough. The light must be traveling from higher $n$ to lower $n$ so that a real critical angle exists.

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Equation snapshotCritical-angle threshold · Critical angleShow 3 equations

Start with Snell's law, check that light is leaving the higher-index side, then compare theta_1 with theta_c.

Critical-angle threshold
$n_{1} sin (θ_{1}) = n_{2} sin (θ_{2})$
This is the same refraction law used below threshold, and it is the law that stops giving a real $θ_{2}$ above the critical angle.
Critical angle
$sin (θ_{c}) = \frac{n _{2}}{n _{1}} (n_{1} > n_{2})$
This is the largest incident angle that still allows a real transmitted ray when light goes from higher $n$ to lower $n$ .
Reflected angle
$θ_{r} = θ_{1}$
In total internal reflection, the ray remains in medium 1 and the reflected angle still equals the incident angle.

Worked examples

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Frozen walkthrough

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Frozen walkthrough

Use the same indices and incident angle shown on the stage. The numbers you substitute are the same ones controlling the ray diagram and the threshold graphs.

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Example 1 of 2

Frozen valuesUsing frozen parameters

For the current media pair, does a critical angle exist, and if it does, what is its value?

n_{1}

Incident-medium index

1.52

n_{2}

Transmitted-medium index

θ_{c}

Critical angle

41.14 °

1. Check whether the ray starts in the higher-index medium

A critical angle only exists when light starts in the higher-index medium, so first compare

n_{1} = 1.52

with

n_{2} = 1

2. Use the critical-angle relation if it exists

When

n_{1} > n_{2}

, use

sin (θ_{c}) = \frac{n _{2}}{n _{1}}

3. Evaluate the threshold

θ_{c} = 41.1 4^{\circ}

Critical-angle result

θ_{c} = 41.1 4^{\circ}

The current incident angle is 8.86^\circ above the critical angle, so the refracted branch has already ended.

Mini challenge

Keep the incident angle fixed above the glass-to-air critical angle, then raise

n_{2}

closer to

n_{1}

. Before you test it, what should happen to the critical angle and to the boundary outcome?

Make a prediction before you reveal the next step.

Decide whether theta_c rises or falls, and whether the same ray is more or less likely to keep reflecting internally.

Check your reasoning against the live bench.

The critical angle rises, so the same incident angle becomes less likely to stay in total internal reflection.

Raising

n_{2}

toward

n_{1}

weakens the index contrast. Since

n_{2} / n_{1}

becomes larger,

θ_{c}

increases, and a setup that used to be above threshold can return to ordinary refraction.

Common misconception

Use this only when you want to pressure-test a mistaken intuition.

Any large incident angle can produce total internal reflection.

Angle alone is not enough. The light must be traveling from higher $n$ to lower $n$ so that a real critical angle exists.

If $n_{1} \leq n_{2}$ , this model still gives a real transmitted angle even at very large incident angles, so there is no TIR threshold to cross.

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