Geometric optics — rectilinear propagation, reflection, refraction: Worked examples

This page builds on the parent topic and leans on Fermat's principle, Total internal reflection, and Optical fibres.

The scenario matrix

Every problem in ray optics is one of a handful of case classes. Each row below is a distinct "shape" of question, defined by which quantity is unknown and what special/degenerate feature it has. The rightmost column tells you which worked example covers that cell.

#	Case class	Special/degenerate feature	Covered by
A	Rarer → denser refraction ( $n_{1} < n_{2}$ )	ray bends toward normal, $θ_{2} < θ_{1}$	Ex 1
B	Denser → rarer refraction ( $n_{1} > n_{2}$ )	ray bends away, $θ_{2} > θ_{1}$	Ex 2
C	Normal incidence	$θ_{1} = 0$ (degenerate: no bending)	Ex 3
D	Grazing / critical angle	$θ_{2} = 90°$ limit, then beyond it (TIR)	Ex 4
E	Slab with parallel faces	ray emerges parallel, only shifted	Ex 5
F	Wavelength & speed inside medium	$f$ fixed, $λ$ & $v$ change	Ex 6
G	Reflection geometry	$θ_{i} = θ_{r}$ , rotating mirror	Ex 7
H	Rectilinear / pinhole (real-world)	similar triangles, inversion	Ex 8
I	Exam twist (multi-step + limiting check)	TIR inside a fibre bent at entry face	Ex 9

We now solve one example per row.

Ex 1 — Cell A: rarer → denser (bend toward normal)

Worked example Light from air into water

Light travels from air ( $n_{1} = 1.00$ ) into water ( $n_{2} = 1.33$ ), striking the surface at $θ_{1} = 40°$ from the normal. Find the refraction angle $θ_{2}$ .

Forecast: guess first — will $θ_{2}$ be bigger or smaller than $40°$ ?

Write Snell's law. $n_{1} sin θ_{1} = n_{2} sin θ_{2}$ . Why this step? This is the one equation that links angles across a boundary; every refraction problem starts here.
Solve for the unknown sine. $sin θ_{2} = \frac{n _{1} sin θ _{1}}{n _{2}} = \frac{1.00 \times sin 40°}{1.33} = \frac{0.643}{1.33} = 0.483$ . Why this step? We isolate the thing we don't know; sine is the natural variable Snell hands us.
Undo the sine. $θ_{2} = arcsin (0.483) = 28.9°$ . Why this step? $arcsin$ answers "which angle has this sine?" — it converts the ratio back into a geometric angle.

Verify: $28.9° < 40°$ ✔ — entering a denser medium bends the ray toward the normal, exactly as forecast. Also $n_{1} sin 40° = 0.643 = n_{2} sin 28.9°$ balances. ✔

The figure below draws exactly this case. Look at how the orange incident ray leans $40°$ off the dashed grey normal in the (lighter) air region, and how the green refracted ray below the boundary hugs closer to the normal at $28.9°$ — the shorter green angle arc is the visual signature of bending toward the normal.

Figure — Geometric optics — rectilinear propagation, reflection, refraction

Ex 2 — Cell B: denser → rarer (bend away from normal)

Worked example Light from glass into air

A ray inside glass ( $n_{1} = 1.50$ ) meets the glass–air surface ( $n_{2} = 1.00$ ) at $θ_{1} = 25°$ . Find $θ_{2}$ .

Forecast: bigger or smaller than $25°$ this time?

Snell's law. $1.50 sin 25° = 1.00 sin θ_{2}$ . Why this step? Same master equation — the direction of travel does not change which law applies.
Isolate. $sin θ_{2} = 1.50 \times 0.4226 = 0.634$ . Why this step? Now the multiplying factor $n_{1} / n_{2} = 1.5 > 1$ , so $sin θ_{2} > sin θ_{1}$ — the angle must grow.
Undo the sine. $θ_{2} = arcsin (0.634) = 39.4°$ .

Verify: $39.4° > 25°$ ✔ — leaving a dense medium bends the ray away from the normal. Since $sin θ_{2} = 0.634 < 1$ , a real refracted ray still exists (we haven't hit the critical angle yet — see Ex 4). ✔

(This is just the mirror-image of the Ex 1 figure with the media swapped — the green ray fans further from the normal instead of hugging it, so no separate diagram is needed.)

Ex 3 — Cell C: normal incidence (the degenerate case)

Worked example Straight-in ray

Light hits a water surface ( $n_{2} = 1.33$ ) from air along the normal itself, i.e. $θ_{1} = 0°$ . Find $θ_{2}$ .

Forecast: does a straight-in ray bend at all?

Snell's law. $1.00 sin 0° = 1.33 sin θ_{2}$ . Why this step? We must not "reason from intuition" — plug $θ_{1} = 0$ into the actual law and let it decide.
Evaluate. $sin 0° = 0$ , so $1.33 sin θ_{2} = 0 \Rightarrow sin θ_{2} = 0$ . Why this step? A product is zero only if a factor is zero; $n_{2} \neq = 0$ , so $sin θ_{2} = 0$ .
Undo. $θ_{2} = arcsin (0) = 0°$ .

Verify: the ray passes straight through, unbent — the only refraction case with no bending. This is why looking straight down into a pool doesn't distort the bottom directly beneath you. ✔ (The speed still changes; only the direction stays put.)

(Picture the Ex 1 figure with both rays collapsed onto the dashed normal line — a single straight arrow — which is why no figure is drawn for this degenerate case.)

Ex 4 — Cell D: critical angle and beyond (limiting behaviour)

Worked example Approaching and passing the critical angle

For a glass–air surface ( $n_{1} = 1.50$ , $n_{2} = 1.00$ ): (a) find the critical angle $θ_{c}$ ; (b) what happens at $θ_{1} = 50°$ ?

Forecast: at what incidence does the refracted ray disappear?

Set the refracted ray to graze. At the critical angle the ray bends to $θ_{2} = 90°$ (the largest refraction possible). $1.50 sin θ_{c} = 1.00 sin 90° = 1$ . Why this step? $θ_{2} = 90°$ is the boundary between "a refracted ray exists" and "it can't" — the exact edge we're hunting.
Solve. $sin θ_{c} = \frac{1.00}{1.50} = 0.667 \Rightarrow θ_{c} = arcsin (0.667) = 41.8°$ . Why this step? $arcsin$ turns the index ratio into the threshold angle.
Test $θ_{1} = 50°$ . $sin θ_{2} = 1.50 \times sin 50° = 1.50 \times 0.766 = 1.149 > 1$ . Why this step? Since $50° > 41.8°$ , Snell demands a sine above 1 — impossible for any real angle.
Interpret. No refracted ray can exist → total internal reflection: all the light bounces back with $θ_{r} = 50°$ (here $θ_{r}$ is the reflection angle).

Verify: $41.8°$ matches the parent note's critical-angle example ✔. The demand $sin θ_{2} = 1.149 > 1$ is the mathematical fingerprint of TIR — see Total internal reflection and Optical fibres. ✔

The figure shows the "beyond critical" case at $θ_{1} = 50°$ . Notice there is no green refracted arrow crossing into the air — instead the red arrow bounces back into the glass at the same $50°$ . The caption text on the plot spells out "no refracted ray" to make the impossibility visible.

Ex 5 — Cell E: parallel-sided slab (emerges parallel, shifted)

Worked example Ray through a glass slab

A ray in air ( $n = 1.00$ ) hits a glass slab ( $n = 1.50$ , parallel faces) at $θ_{1} = 45°$ . Find the exit angle after it leaves the far face back into air.

Forecast: will the exit ray be steeper, shallower, or the same slant as the entry ray?

Refract in at the top face. $1.00 sin 45° = 1.50 sin θ_{2} \Rightarrow sin θ_{2} = \frac{0.7071}{1.5} = 0.4714$ , so $θ_{2} = 28.1°$ . Why this step? Air→glass is a rarer→denser refraction (cell A); we find the angle inside the slab.
The inside angle hits the bottom face. Because the faces are parallel, the normal is the same direction, so the angle of incidence at the bottom face equals $θ_{2} = 28.1°$ . Why this step? Parallel faces share a normal; the internal ray keeps its slant relative to that shared normal.
Refract out at the bottom face. $1.50 sin 28.1° = 1.00 sin θ_{3} \Rightarrow sin θ_{3} = 1.50 \times 0.4714 = 0.7071$ , so $θ_{3} = 45°$ . Why this step? This is glass→air (cell B); we solve for the final exit angle.

Verify: $θ_{3} = 45° = θ_{1}$ ✔ — the exit ray is parallel to the entry ray; the slab only shifts it sideways, it never changes its direction. The two $n$ -factors cancel because we cross the same pair of media in reverse. ✔

Trace the three arrows in the figure: the orange entry ray at $45°$ , the blue ray inside the slab tilted closer to the normal at $28.1°$ , and the green exit ray that comes out again at $45°$ . The two dashed grey lines mark the (parallel) normals at the top and bottom faces — the green ray is parallel to the orange one but pushed sideways, which is the lateral shift.

Ex 6 — Cell F: speed, wavelength and frequency inside a medium

Worked example What changes inside glass?

Green light of vacuum wavelength $λ_{0} = 500 nm$ enters glass of $n = 1.50$ . Find its speed $v$ , wavelength $λ_{m e d}$ , and frequency $f$ inside the glass. Take $c = 3.00 \times 1 0^{8} m/s$ .

Forecast: which of the three ( $v$ , $λ$ , $f$ ) stays the same?

Speed inside. $v = \frac{c}{n} = \frac{3.00 \times 1 0 ^{8}}{1.50} = 2.00 \times 1 0^{8} m/s$ . Why this step? $n = c / v$ is the definition of index; light slows in a denser medium.
Wavelength inside. $λ_{m e d} = \frac{λ _{0}}{n} = \frac{500}{1.50} = 333 nm$ . Why this step? The wave gets "squeezed" because it slows but its frequency can't change.
Frequency (vacuum). $f = \frac{c}{λ _{0}} = \frac{3.00 \times 1 0 ^{8}}{500 \times 1 0 ^{- 9}} = 6.00 \times 1 0^{14} Hz$ . Why this step? Frequency is fixed by the source; we compute it once and it stays put.
Frequency (in glass) cross-check. $\frac{v}{λ _{m e d}} = \frac{2.00 \times 1 0 ^{8}}{333 \times 1 0 ^{- 9}} = 6.00 \times 1 0^{14} Hz$ . Why this step? $v = f λ$ must hold inside glass too — this proves $f$ is unchanged.

Verify: frequency is identical in vacuum and glass ( $6.00 \times 1 0^{14} Hz$ ) ✔; only $v$ and $λ$ dropped by the factor $n$ . Units: $m/s \div m = 1/s = Hz$ ✔. This is why colour (tied to $f$ ) does not change — see Dispersion. ✔

Ex 7 — Cell G: reflection geometry (rotating mirror)

Worked example Rotate the mirror, watch the ray

A ray strikes a mirror at $θ_{i} = 20°$ from the normal. (a) State the reflected angle. (b) The mirror is now rotated by $10°$ (incidence becomes $30°$ ). By how much does the reflected ray turn?

Forecast: does the reflected ray turn by $10°$ , $20°$ , or something else?

Apply the law of reflection. $θ_{r} = θ_{i} = 20°$ . Why this step? Reflection always mirrors the angle across the normal — no exceptions.
Angle between incident and reflected ray (first case). They sit symmetrically about the normal, so the angle between them is $θ_{i} + θ_{r} = 20° + 20° = 40°$ . Why this step? Each ray leans $θ$ off the normal on opposite sides; add them.
New case. Incidence $30° \Rightarrow$ reflected $30° \Rightarrow$ angle between rays $= 60°$ . Why this step? Same rule, larger angle.
Change in reflected-ray direction. The between-rays angle changed by $60° - 40° = 20°$ , i.e. the reflected ray turned twice the mirror's $10°$ rotation. Why this step? Compare the two configurations for the same incident ray.

Verify: reflected ray swings by $20° = 2 \times 10°$ ✔ — the classic "reflected ray turns by double the mirror rotation." ✔

Ex 8 — Cell H: rectilinear propagation (pinhole word problem)

Worked example Pinhole camera

A candle of height $H = 20 cm$ stands $u = 50 cm$ in front of a pinhole. A screen is $v = 15 cm$ behind it. Find the image height $h$ and its orientation.

Forecast: taller or shorter than the candle, upright or upside-down?

Use similar triangles. Straight rays through one pinhole vertex give $\frac{h}{H} = \frac{v}{u}$ . Why this step? Rectilinear propagation makes the rays perfectly straight, so the two triangles (object side, image side) are similar.
Solve for $h$ . $h = H \times \frac{v}{u} = 20 \times \frac{15}{50} = 6 cm$ . Why this step? Plug the known lengths into the proportion.
Orientation. Top rays cross to the bottom through the hole → the image is inverted. Why this step? Each straight ray passes through the single hole, swapping top and bottom.

Verify: magnification $m = v / u = 15/50 = 0.30$ , and $h = 0.30 \times 20 = 6 cm$ ✔; since $v < u$ the image is smaller, and inverted ✔. Units: $cm \times (cm / cm) = cm$ ✔.

Follow the two blue rays in the figure: both leave the tip and base of the orange candle on the left, pass through the single grey pinhole dot, and cross — so the top ray lands below the axis, producing the shorter green inverted image on the right. The crossing at the hole is exactly why the image flips.

Ex 9 — Cell I: exam twist (fibre entry face + TIR check)

Worked example Will light stay trapped in the fibre?

Light enters the flat end of an optical fibre from air ( $n_{0} = 1.00$ ) at $θ_{in} = 20°$ . The fibre core has $n_{1} = 1.50$ , surrounded by air cladding ( $n_{2} = 1.00$ ). (a) Find the angle the ray makes with the fibre axis after entering. (b) Find the angle at which it hits the side wall, and decide if it is totally internally reflected. Critical angle at the wall is $θ_{c} = 41.8°$ (from Ex 4).

Here we write $θ_{f}$ for the refracted angle inside the fibre (measured from the entry-face normal, which points along the fibre axis). We deliberately avoid the letter $r$ here, because in this book $θ_{r}$ already means the reflection angle — a different thing.

Forecast: does this ray escape the side, or bounce and travel down the fibre?

Refract at the flat entry face. $1.00 sin 20° = 1.50 sin θ_{f} \Rightarrow sin θ_{f} = \frac{0.3420}{1.50} = 0.2280$ , so $θ_{f} = 13.2°$ . Why this step? The entry face is a normal refraction; $θ_{f}$ is measured from the face's normal, which points along the fibre axis.
Angle at the side wall. The side wall's normal is perpendicular to the axis, so the wall incidence angle is $90° - θ_{f} = 90° - 13.2° = 76.8°$ . Why this step? Two normals at right angles means the two angles are complementary — a ray shallow to the axis is steep to the wall.
Compare with critical angle. $76.8° > 41.8° = θ_{c}$ . Why this step? TIR happens whenever the wall incidence exceeds the critical angle.
Conclude. Since $76.8° > θ_{c}$ , the light is totally internally reflected at the wall and stays trapped, zig-zagging down the fibre.

Verify: $θ_{f} = 13.2°$ from Snell ✔; wall angle $76.8°$ complements it ✔; $76.8° > 41.8°$ guarantees TIR ✔. This is the principle behind Optical fibres and Total internal reflection. ✔

Active recall

Recall Which cell is this? (cover the answers)

Air → water at $30°$ , find $θ_{2}$ ::: Cell A (rarer→denser); $sin θ_{2} = sin 30°/1.33$ . A ray goes straight into a surface, $θ_{1} = 0$ ::: Cell C; degenerate, $θ_{2} = 0$ , no bending. Glass→air at $50°$ with $n = 1.5$ ::: Cell D; $sin θ_{2} > 1$ ⇒ total internal reflection. Ray through a parallel slab ::: Cell E; emerges parallel, only laterally shifted. Which of $v$ , $λ$ , $f$ is unchanged in a medium? ::: Frequency $f$ ; $v$ and $λ$ shrink by factor $n$ . Mirror rotates $10°$ , reflected ray turns...? ::: $20°$ — double the mirror rotation.

Connections

Fermat's principle — every angle law used above descends from least time.
Total internal reflection, Optical fibres — Ex 4 and Ex 9.
Dispersion — why frequency-fixed refraction (Ex 6) splits colours.
Mirrors and Lenses — reflection geometry of Ex 7 extended to curved surfaces.