Total internal reflection — critical angle derivation: Worked examples

This page is the drill floor for the critical-angle idea. The parent note built the one formula we need:

Before we drill, let us list every kind of question this one formula can be dressed up as. If we solve one example per row, nothing on an exam can surprise us.

The scenario matrix

Every critical-angle problem is really a rearrangement of $sin θ_{c} = n_{2} / n_{1}$ . There are only three "unknowns" you can be asked for, plus a set of edge cases where the formula behaves strangely. Here is the full map:

#	Case class	What is given	What is asked	Which cell
1	Standard forward	$n_{1}, n_{2}$ (dense→rare)	$θ_{c}$	forward, two real media
2	Against air	$n_{1}$ , air	$θ_{c}$	forward, $n_{2} = 1$
3	Inverse	$θ_{c}$ , air	$n_{1}$	solve for index
4	Two-media inverse	$θ_{c}$ , one index	other index	solve for index
5	Degenerate: equal indices	$n_{1} = n_{2}$	$θ_{c}$	limiting value $θ_{c} = 90°$
6	Impossible: rarer→denser	$n_{1} < n_{2}$	$θ_{c}$ ?	no solution case
7	Beyond critical	$n_{1}, n_{2}, θ_{1} > θ_{c}$	what happens?	sign/regime test
8	Word problem	fibre / underwater	$θ_{c}$ + meaning	real-world
9	Exam twist	prism / $45°$ geometry	does TIR occur?	compare-to- $45°$

We now hit every cell.

Worked examples

Worked example Example 1 — Cell 1: standard forward, two real media

Light travels from a dense plastic ( $n_{1} = 1.60$ ) into oil ( $n_{2} = 1.40$ ). Find $θ_{c}$ .

Forecast: the two indices are close, so the ray barely bends — guess: is $θ_{c}$ closer to $90°$ or to $30°$ ?

Check direction: $n_{1} = 1.60 > n_{2} = 1.40$ , dense→rare. ✅ TIR is possible. Why this step? If the light went the other way, no critical angle exists at all (Cell 6). Always verify direction first.
Apply the formula: $sin θ_{c} = \frac{n _{2}}{n _{1}} = \frac{1.40}{1.60} = 0.875$ . Why this step? This is the definition — set the refracted angle to $90°$ and solve, as in Snell's Law.
Invert the sine: $θ_{c} = sin^{- 1} (0.875) = 61.0°$ . Why this step? We know the sine of the angle; $sin^{- 1}$ is the operation that undoes "sine of" and hands back the angle.

Verify: the indices are close (ratio near 1), so $θ_{c}$ should be near $90°$ — and $61°$ is indeed steep. Forecast confirmed. Plug back: $sin 61.0° = 0.875 = 1.40/1.60$ . ✅

Figure — Total internal reflection — critical angle derivation

Worked example Example 6 — Cell 6: the impossible direction (rarer → denser)

Light travels from air ( $n_{1} = 1.00$ ) into glass ( $n_{2} = 1.50$ ). Someone asks for the critical angle. Find it — or explain why you cannot.

Forecast: the light is going into the denser medium. Does TIR even make sense here?

Blindly apply the formula: $sin θ_{c} = \frac{n _{2}}{n _{1}} = \frac{1.50}{1.00} = 1.50$ . Why this step? To expose the contradiction, we let the formula speak.
Ask: can $sin θ_{c} = 1.50$ ? No. Sine of any real angle lives in $[- 1, 1]$ . Why this step? $sin θ_{c} = 1.5$ has no real solution — geometrically, no such angle exists.
Conclude: there is no critical angle going rarer→denser. TIR is impossible in this direction. Why this step? Refraction into the denser medium always succeeds (the ray bends toward the normal), so the light never gets "trapped."

Verify: this matches the parent note's first condition — TIR needs $n_{1} > n_{2}$ . Here $n_{1} < n_{2}$ , so the machinery correctly refuses. ✅

Worked example Example 7 — Cell 7: beyond the critical angle

Inside water ( $n_{1} = 1.33$ , air $n_{2} = 1$ ) a ray strikes the surface at $θ_{1} = 60°$ . First find $θ_{c}$ , then decide what happens to this ray.

Forecast: is $60°$ above or below the critical angle for water? What does the ray do?

Critical angle: $sin θ_{c} = \frac{1}{1.33} = 0.7519 \Rightarrow θ_{c} = 48.8°$ . Why this step? We need the threshold before we can compare.
Compare: $θ_{1} = 60° > 48.8° = θ_{c}$ . Why this step? The regime is decided by whether we are above or below $θ_{c}$ .
Since $θ_{1} > θ_{c}$ , Snell would need $sin θ_{2} = \frac{n _{1}}{n _{2}} sin θ_{1} = 1.33 \times sin 60° = 1.33 \times 0.8660 = 1.152 > 1$ . Why this step? Checking Snell's demand shows it asks the impossible.
No refracted ray exists → total internal reflection: 100% of the light bounces back into the water. Why this step? Energy must go somewhere; with refraction forbidden, reflection takes all of it.

Verify: $sin θ_{2} = 1.152 > 1$ confirms impossibility, and $60° > 48.8°$ confirms we're past threshold. Both tests agree. ✅

Worked example Example 8 — Cell 8: real-world word problem (optical fibre)

An optical fibre core has $n_{1} = 1.50$ , surrounded by cladding $n_{2} = 1.45$ . Light bouncing along the core hits the core–cladding wall. Find $θ_{c}$ , and state the range of angles that stay trapped.

Forecast: the two indices are very close. Will the trapping angle be near $90°$ (a narrow trap) or near $0°$ ?

Direction check: $1.50 > 1.45$ , dense→rare. ✅ Why this step? Fibres deliberately put the denser glass in the core so light can be trapped.
$sin θ_{c} = \frac{1.45}{1.50} = 0.9667$ . Why this step? Rarer (cladding) over denser (core).
$θ_{c} = sin^{- 1} (0.9667) = 75.2°$ . Why this step? Undo the sine.
Trapped range: any wall-strike with $θ_{1}$ between $75.2°$ and $90°$ undergoes TIR. Why this step? All incidence angles above $θ_{c}$ (up to grazing) are totally reflected.

Verify: the close indices give a large $θ_{c}$ near $90°$ , meaning light must travel nearly parallel to the fibre axis to stay in — exactly why fibres are engineered with tiny index differences and light is launched at shallow angles. ✅

Worked example Example 9 — Cell 9: exam twist (45° prism)

A right-angled glass prism ( $n = 1.50$ , air outside) is used as a mirror: light hits the slanted face at exactly $45°$ . Does it undergo TIR? (This is the periscope prism trick.)

Forecast: the incidence is fixed at $45°$ . We must compare it to $θ_{c}$ — will $45°$ win?

Find $θ_{c}$ for glass–air: $sin θ_{c} = \frac{1}{1.50} = 0.6667 \Rightarrow θ_{c} = 41.8°$ . Why this step? The threshold is what $45°$ must beat.
Compare: $45° > 41.8° = θ_{c}$ . Why this step? TIR happens precisely when incidence exceeds the critical angle.
Conclusion: yes — the light is totally internally reflected, so the prism acts as a perfect mirror with no silvering needed. Why this step? Since $45°$ clears the threshold, all energy reflects.

Verify: $45° - 41.8° = 3.2°$ of margin — TIR occurs, but only just. If the glass were low-index ( $n < 1/ sin 45° = 1.414$ ), $θ_{c}$ would exceed $45°$ and the trick would fail. Our $n = 1.50 > 1.414$ , so we are safely above. ✅

Recall Rapid-fire self-test (cover the answers)

Glass $n = 1.5$ into air, $θ_{c}$ ? ::: $sin^{- 1} (1/1.5) \approx 41.8°$ . Diamond $n = 2.42$ into air, $θ_{c}$ ? ::: $sin^{- 1} (1/2.42) \approx 24.4°$ . Liquid with $θ_{c} = 45°$ against air, its $n$ ? ::: $1/ sin 45° = 1.414$ . Equal indices $n_{1} = n_{2}$ , $θ_{c}$ ? ::: $90°$ (no real trapping). Air into glass, $θ_{c}$ ? ::: none exists — wrong direction, $sin θ_{c} = 1.5 > 1$ . Fibre core $1.50$ , cladding $1.45$ , $θ_{c}$ ? ::: $sin^{- 1} (1.45/1.50) \approx 75.2°$ .

Connections

Total internal reflection — critical angle derivation — the parent note these drills serve.
Snell's Law — every example is a rearrangement of it.
Refractive Index — the numbers we plug in.
Refraction of Light — TIR as the limiting case.
Optical Fibres — Example 8 in action.
Prisms and Total Internal Reflection — Example 9 in action.
Brewster's Angle — a different named angle; do not confuse.
Mirage and Atmospheric Refraction — TIR from gradual air layers.