Exercises — Total internal reflection — critical angle derivation

WHAT: TIR needs both conditions from the checklist above. WHY: Condition 1 is denser→rarer, i.e. $n_1>n_2$. Here the ray starts in air ($n_1=1.00$) and enters glass ($n_2=1.50$), so $n_1<n_2$ — it goes rarer→denser. Condition 1 already fails, so we needn't even check the angle. Result: $\boxed{\text{No.}}$ Going into a denser medium, $\sin {\theta }_2=\frac{n_1}{n_2}\sin {\theta }_1<1$ always, so a refracted ray always exists. TIR is impossible in this direction.

WHAT: Compare ${\theta }_c$ for two indices without a calculator. WHY: $\sin {\theta }_c=1/n$. A bigger $n$ makes the fraction $1/n$ smaller, and a smaller sine means a smaller angle (both angles live in $[0°,90°]$, where sine strictly increases). Diamond has the larger $n$, so it has the smaller ${\theta }_c$. Therefore glass's ${\theta }_c$ is larger. Result: $\boxed{\text{True.}}$ (Numerically $41.8°>24.4°$ — verified below.)

Result: $\boxed{\text{(b) }\sin {\theta }_c=n_2/n_1.}$ WHY: Set ${\theta }_2=90°$ in Snell's law: $n_1\sin {\theta }_c=n_2\sin 90°=n_2$. Divide by $n_1$. Since the ray goes denser→rarer, $n_2<n_1$, so the fraction is $<1$ — exactly what a sine must be. Option (a) would give a value $>1$, which no sine can equal: that impossibility is the built-in "you flipped it" alarm.

WHAT: direct plug-in. Rarer medium is air ($n_2=1$), denser is water ($n_1=1.33$). $\sin {\theta }_c=\frac{n_2}{n_1}=\frac{1}{1.33}=0.7519$ WHY invert the sine: we know the sine of the angle; the angle itself is ${\sin }^{-1}$ of that number (principal value in $[0°,90°]$). ${\theta }_c={\sin }^{-1}(0.7519)=\boxed{48.8°}$

WHAT: unknown is the denser index; air is $n_2=1$. $\sin {\theta }_c=\frac{1}{n}\Rightarrow n=\frac{1}{\sin {\theta }_c}=\frac{1}{\sin 40.0°}=\frac{1}{0.6428}$ $n=\boxed{1.556}$ Check: $n>1$ as any real medium must be, and ${\theta }_c<90°$ gives $n>1$ automatically. ✓

WHAT: neither medium is air — use both indices. $\sin {\theta }_c=\frac{n_2}{n_1}=\frac{1.33}{1.62}=0.8210$ ${\theta }_c={\sin }^{-1}(0.8210)=\boxed{55.2°}$ WHY it's larger than the glass–air case: the two indices are closer together, so the fraction $n_2/n_1$ is closer to $1$, pushing ${\theta }_c$ closer to $90°$.

$\sin {\theta }_c=\frac{1}{2.42}=0.4132\Rightarrow {\theta }_c={\sin }^{-1}(0.4132)=\boxed{24.4°}$ Meaning: almost any internal ray ($>24.4°$) is trapped, so light bounces many times before escaping — the source of a diamond's sparkle.

Compare each incidence angle to ${\theta }_c=41.8°$:

• $30°$ (mint ray): $30°<{\theta }_c$, so a real refracted ray exists — light refracts out into the air, bending away from the normal (see the mint arrow escaping upward in the figure).
• $41.8°$ (butter ray): exactly ${\theta }_c$, so ${\theta }_2=90°$ — the refracted ray grazes flat along the surface (butter arrow lying along the boundary).
• $55°$ (coral ray): $55°>{\theta }_c$, so Snell would need $\sin {\theta }_2=1.5\sin 55°=1.229>1$, impossible — the ray is totally internally reflected back into the glass (coral arrow bouncing back down). $\boxed{30°\to \text{refracts},41.8°\to \text{grazes},55°\to \text{TIR}}$

WHAT: study ${\theta }_c={\sin }^{-1}(n_2/n_1)$ as its input grows. WHY: ${\sin }^{-1}$ is an increasing function on $[0,1]$ (returning the principal angle in $[0°,90°]$) — a bigger input gives a bigger angle. So as $n_2/n_1$ rises, ${\theta }_c$ increases. Limit: when $n_2\to n_1$, $n_2/n_1\to 1$ and ${\theta }_c\to {\sin }^{-1}(1)=90°$. Physical meaning: if the two media are nearly identical, light barely bends, so you must hit almost parallel to the surface ($\approx 90°$) before TIR sets in. $\boxed{{\theta }_c\text{increases, reaching }90°\text{ as }{n}_2\to n_1.}$

The algebra is correct: with $n_2>n_1$, $\sin {\theta }_c=n_2/n_1>1$ has no real solution. The interpretation is the point: "no critical angle" means there is no incidence angle at which refraction ceases — going rarer→denser, a refracted ray always exists for every ${\theta }_1$ up to $90°$. So TIR simply cannot occur in that direction. $\boxed{\text{Reasoning valid; it correctly signals TIR is impossible rarer}\to \text{denser.}}$

WHAT: TIR at the hypotenuse needs the incidence angle ($45°$) to exceed the critical angle: $45°>{\theta }_c\iff \sin 45°>\sin {\theta }_c=\frac{1}{n}$ WHY the direction of the inequality holds: sine is increasing on $[0°,90°]$, so $45°>{\theta }_c$ is the same as $\sin 45°>\sin {\theta }_c$. Solve for $n$: $\frac{1}{n}<\sin 45°=0.7071\Rightarrow n>\frac{1}{0.7071}=1.414$ $\boxed{n>1.414}$ Meaning: ordinary glass ($n=1.5$) easily satisfies this, which is why right-angle prisms in binoculars and periscopes reflect without any silvering. See Prisms and Total Internal Reflection.

WHAT: the boundary is core (denser) → cladding (rarer), so $n_1=1.50$, $n_2=1.45$. $\sin {\theta }_c=\frac{n_2}{n_1}=\frac{1.45}{1.50}=0.9667$ ${\theta }_c={\sin }^{-1}(0.9667)=\boxed{75.2°}$ Meaning: any ray striking the wall at more than $75.2°$ (from the normal) is trapped and guided down the fibre. Because ${\theta }_c$ is so close to $90°$, only rays travelling nearly parallel to the axis stay guided — this is the design idea behind Optical Fibres.

(a) Water→oil goes $n=1.33\to 1.46$, i.e. rarer→denser. TIR needs denser→rarer, so $\boxed{\text{no TIR at the water–oil boundary.}}$ (b) Oil→air is denser→rarer with $n_1=1.46$, $n_2=1$: $\sin {\theta }_c=\frac{1}{1.46}=0.6849\Rightarrow {\theta }_c={\sin }^{-1}(0.6849)=\boxed{43.2°}$

Step 1 — critical angle. WHAT/WHY: only rays inside the cone of half-angle ${\theta }_c$ escape, so first find ${\theta }_c$ for water against air: $\sin {\theta }_c=\frac{1}{1.33}=0.7519,{\theta }_c={\sin }^{-1}(0.7519)=48.75°$ Step 2 — why tangent, and where it comes from. Look at the shaded right triangle in the figure. Its vertical leg is the depth $h$ (this is the side adjacent to the angle ${\theta }_c$, because ${\theta }_c$ is measured from the vertical normal). Its horizontal leg is the radius $R$ (the side opposite to ${\theta }_c$). We want $R$ from $h$ and the angle, i.e. we want opposite-from-adjacent — and the trig ratio that connects an angle to opposite-over-adjacent is exactly the tangent: $\tan {\theta }_c=\frac{\text{opposite}}{\text{adjacent}}=\frac{R}{h}.$ That is why tangent and not sine or cosine: sine needs the hypotenuse (the ray length, which we don't have and don't want), whereas tangent uses only the two legs we actually know. Step 3 — solve for $R$. Multiply both sides by $h$: $R=h\tan {\theta }_c=2.0\times \tan (48.75°)=2.0\times 1.1391=\boxed{2.28\text{m}}$ So the lit circle on the surface has radius about $2.28$ m; everything outside it looks mirror-like from below.

WHAT: we turn $\tan {\theta }_c$ into an expression in $n$ so no calculator angle is needed. WHY / the triangle behind it: from $\sin {\theta }_c=1/n$ build a right triangle whose hypotenuse is $n$ and whose side opposite ${\theta }_c$ is $1$ (that pair realises $\sin {\theta }_c=1/n$). Pythagoras then fixes the adjacent side: $\text{adjacent}=\sqrt{{\text{hyp}}^2-{\text{opp}}^2}=\sqrt{n^2-1}.$ Therefore $\tan {\theta }_c=\frac{\text{opposite}}{\text{adjacent}}=\frac{1}{\sqrt{n^2-1}}.$ Substitute into $R=h\tan {\theta }_c$: $R=\frac{h}{\sqrt{n^2-1}}=\frac{2.0}{\sqrt{1.33^2-1}}=\frac{2.0}{\sqrt{0.7689}}=\frac{2.0}{0.8769}=\boxed{2.28\text{m}}✓$ The two methods agree, confirming the clean formula.

Step 1 — refraction at the top face (air→glass, so the ray bends toward the normal). WHAT/WHY: apply Snell's law at the top interface, air $n=1$ outside, glass $n=1.5$ inside: $1\cdot \sin \alpha =1.5\sin r\Rightarrow \sin r=\frac{\sin 40°}{1.5}=\frac{0.6428}{1.5}=0.4285,r={\sin }^{-1}(0.4285)=25.37°$ Here $r$ is the refraction angle inside the glass, measured from the top face's normal (the vertical). Step 2 — geometry from the top normal to the side normal. WHAT: the top face is horizontal, so its normal is vertical; the side face is vertical, so its normal is horizontal. These two normals are perpendicular. WHY the incidence on the side is $90°-r$: the refracted ray leans $r$ away from the vertical. Since the side-face normal points horizontally (rotated $90°$ from the vertical), the same ray makes an angle $90°-r$ with it — the two angles a single straight line makes with two perpendicular reference lines always add to $90°$. So ${\theta }_{\text{side}}=90°-r=90°-25.37°=64.63°.$ What it looks like: picture the ray as a slightly-tilted spear inside the block; it is nearly vertical ($25°$ from the top normal), hence nearly perpendicular to the horizontal side normal, i.e. a large ${\theta }_{\text{side}}$. Step 3 — compare with ${\theta }_c$. For glass, ${\theta }_c={\sin }^{-1}(1/1.5)=41.8°$. Since ${\theta }_{\text{side}}=64.63°>41.8°={\theta }_c$, the ray undergoes TIR at the side face. $\boxed{{\theta }_{\text{side}}=64.6°>41.8°\Rightarrow \text{TIR occurs.}}$

Recognition — Which direction allows TIR? ::: Denser to rarer only ($n_1>n_2$). Application — ${\theta }_c$ for $n=1.33$ vs air? ::: ${\sin }^{-1}(1/1.33)\approx 48.8°$. Analysis — What happens at exactly ${\theta }_c$? ::: Refracted ray grazes at $90°$; TIR needs strictly ${\theta }_1>{\theta }_c$. Synthesis — Why do $45°$ prisms reflect? ::: $45°>{\theta }_c$ for glass (${\theta }_c\approx 41.8°$), so TIR without silvering. Mastery — Escape-window radius formula? ::: $R=h/\sqrt{n^2-1}$.