Exercises — Geometric optics — rectilinear propagation, reflection, refraction

WHAT: Inside a single uniform medium the ray travels in a straight line. WHY: In a homogeneous medium the speed $v$ is the same at every point. The least-time path between two points at constant speed is the shortest path, which is a straight line. Any bend would add length, hence time. This is rectilinear propagation.

Trap-aware step: the $37^{\circ }$ was given from the surface, but optics uses the normal. Convert: ${\theta }_i=90^{\circ }-37^{\circ }=53^{\circ }(\text{from the normal}).$ Law of reflection: ${\theta }_r={\theta }_i=53^{\circ }$.

WHY Snell: crossing a boundary changes the speed, so the ray bends; the quantity $n\sin \theta$ is conserved across the surface. $n_1\sin {\theta }_1=n_2\sin {\theta }_2\Rightarrow \sin {\theta }_2=\frac{1.00\cdot \sin 40^{\circ }}{1.33}.$ $\sin 40^{\circ }=0.6428$, so $\sin {\theta }_2=\frac{0.6428}{1.33}=0.4833$. ${\theta }_2=\arcsin (0.4833)=28.9^{\circ }.$ Sanity check: water is denser than air, so the ray bends toward the normal: ${\theta }_2<{\theta }_1$. ✔

$n=\frac{c}{v}=\frac{3.00\times 10^8}{2.00\times 10^8}=\mathrm{1.50.}$ Read it out: light is $1.5\times$ slower in this plastic than in vacuum.

Frequency is set by the source and does not change on refraction: $f=\frac{c}{{\lambda }_0}=\frac{3.00\times 10^8}{500\times 10^{-9}}=6.00\times 10^{14}\text{Hz}.$ Wavelength shrinks by the factor $n$ (speed dropped, frequency fixed, so $v=f\lambda$ forces $\lambda$ down): ${\lambda }_{\text{med}}=\frac{{\lambda }_0}{n}=\frac{500}{1.50}=333\text{nm}.$

WHY: As ${\theta }_1$ grows, ${\theta }_2$ grows faster (going to the rarer, faster medium bends the ray away from the normal). At the critical angle the refracted ray grazes exactly along the surface: ${\theta }_2=90^{\circ }$ — this is the flat yellow ray in figure s01. $n_1\sin {\theta }_c=n_2\sin 90^{\circ }=n_2\Rightarrow \sin {\theta }_c=\frac{n_2}{n_1}=\frac{1.00}{1.50}=\mathrm{0.6667.}$ ${\theta }_c=\arcsin (0.6667)=41.8^{\circ }.$ Beyond $41.8^{\circ }$, Snell would demand $\sin {\theta }_2>1$ (impossible) — so all the light reflects internally. See Total internal reflection and Optical fibres.

Let air index $n_a$, glass index $n_g$. At the top face: $n_a\sin {\theta }_1=n_g\sin {\theta }_2.$ The two faces are parallel, so the normal at the bottom face is parallel to the normal at the top; the internal ray meets the bottom face at the same angle ${\theta }_2$. At the bottom face: $n_g\sin {\theta }_2=n_a\sin {\theta }_3.$ Chain them: $n_a\sin {\theta }_1=n_g\sin {\theta }_2=n_a\sin {\theta }_3$, so $\sin {\theta }_1=\sin {\theta }_3$, giving ${\theta }_3={\theta }_1$. The exit ray leaves at the same angle it entered → parallel (merely shifted sideways). The glass "steals" $n_g$ inside but hands it back on the way out.

Step 1 — is $60^{\circ }$ past the critical angle? From water to air: $\sin {\theta }_c=\frac{n_2}{n_1}=\frac{1.00}{1.33}=0.7519\Rightarrow {\theta }_c=48.8^{\circ }.$ Since $60^{\circ }>48.8^{\circ }$, we are beyond critical. Step 2 — check Snell directly: $\sin {\theta }_2=\frac{1.33\sin 60^{\circ }}{1.00}=1.33\times 0.8660=1.152>1$: impossible. Conclusion: No light escapes; it is totally internally reflected back into the water. This is why the underside of a calm water surface acts like a mirror when you look up at a steep angle.

WHY inverted: every point of the tree sends a straight ray through the single pinhole; a ray from the top travels down-and-through, landing low on the screen, while a bottom ray lands high — top and bottom swap → inverted (the crossed chalk rays in figure s02). Similar triangles share the pinhole vertex, so corresponding sides are proportional. In the standard pinhole sign convention we attach a minus to mark inversion: $m=\frac{h}{H}=-\frac{v}{u}\Rightarrow h=-H\cdot \frac{v}{u}=-6.0\times \frac{0.30}{2.0}=-0.90\text{m}.$ The magnitude is $0.90\text{m}$; the negative sign is the algebra's way of saying "inverted." Image: $0.90\text{m}$ tall, inverted.

WHY least length: speed is constant off a mirror, so least time = least path length $L(x)=\sqrt{x^2+3^2}+\sqrt{(8-x{)}^2+1^2}.$ The image trick (fastest route to the answer): reflect $B=(8,1)$ across the mirror to $B^{\prime }=(8,-1)$. Any path $A\to P\to B$ has the same length as $A\to P\to B^{\prime }$ (the second leg is mirrored), and the shortest $A\to B^{\prime }$ is the straight line — the yellow dashed line in figure s03. So $P$ is where line $A{B}^{\prime }$ crosses the $x$-axis. Line from $A=(0,3)$ to $B^{\prime }=(8,-1)$: slope $=\frac{-1-3}{8-0}=-\frac{1}{2}$, so $y=3-\frac{1}{2}x$. Set $y=0$: $0=3-\frac{1}{2}x\Rightarrow x=6.0\text{m},P=(6,0).$ Verify equal angles. Measure both from the vertical normal at $P$.

• Incident side ($A$ to $P$): horizontal run $=6-0=6$, vertical drop $=3$. $\tan {\theta }_i=\frac{6}{3}=2$, so ${\theta }_i=\arctan 2=63.43^{\circ }$.
• Reflected side ($P$ to $B$): horizontal run $=8-6=2$, vertical rise $=1$. $\tan {\theta }_r=\frac{2}{1}=2$, so ${\theta }_r=63.43^{\circ }$. ${\theta }_i={\theta }_r$ ✔ — Fermat's minimum reproduces the law of reflection exactly.

WHY not least-length now: the two speeds differ, so time $\ne$ distance. Minimise time: $T(x)=\frac{\sqrt{x^2+9}}{v_1}+\frac{\sqrt{(6-x{)}^2+9}}{v_2}.$ Setting $\frac{dT}{dx}=0$ gives $\frac{x}{v_1\sqrt{x^2+9}}=\frac{6-x}{v_2\sqrt{(6-x{)}^2+9}},$ which is exactly $\frac{\sin {\theta }_1}{v_1}=\frac{\sin {\theta }_2}{v_2}$, i.e. Snell's law, because each fraction is (horizontal offset)/(hypotenuse) = sine measured from the vertical normal. Solve numerically. Cross-multiply the balance condition. With $v_1=3$ and $v_2=2$ (in units of $10^8\text{m/s}$), $v_1=1.5v_2$, so the $v$-factors give $v_{2}x\sqrt{(6-x{)}^2+9}=v_1(6-x)\sqrt{x^2+9}\Rightarrow x\sqrt{(6-x{)}^2+9}=1.5(6-x)\sqrt{x^2+9}.$ Solving this in $0<x<6$ gives $x\approx 4.09\text{m}$ (the physical root). Confirm Snell. With $x=4.09$:

• $\sin {\theta }_1=\frac{x}{\sqrt{x^2+9}}=\frac{4.09}{\sqrt{4.09^2+9}}=0.8062$, so $n_1\sin {\theta }_1=1\times 0.8062=0.8062$.
• $\sin {\theta }_2=\frac{6-x}{\sqrt{(6-x{)}^2+9}}=\frac{1.91}{\sqrt{1.91^2+9}}=0.5375$, so $n_2\sin {\theta }_2=1.5\times 0.5375=0.8062$. Both sides equal $0.8062$ ✔. The ray bends toward the normal entering the slower medium: ${\theta }_2<{\theta }_1$.

• Angles are always from the normal.
• Denser medium ($\uparrow n$) → slower light → bends toward the normal.
• Critical angle exists only going to the rarer medium: $\sin {\theta }_c=n_2/n_1$.
• Parallel-faced slab → exit ray parallel to entry.
• Pinhole: $|m|=v/u$, image inverted (sign of $m$ is negative).
• Fermat: reflection minimises path length; refraction minimises time (weight legs by $1/v$).