Question bank — Snell's law — derivation from Fermat's principle

False. In a single uniform medium it does, but across media light minimizes time, which equals minimizing OPL $=\sum n_i{d}_i$, not raw length — so the path kinks at the boundary.

False. That is the law of reflection. Refraction gives $n_1\sin {\theta }_1=n_2\sin {\theta }_2$; the angles are generally unequal unless $n_1=n_2$.

Mostly true but imprecise: it demands a stationary time ($dT/dx=0$). For refraction at a flat boundary this stationary point is a minimum, but the exact statement is "stationary," which also covers maxima and saddle paths (e.g. mirrors). See Fermat's Principle.

False. Denser means larger $n$; since $n\sin \theta$ is conserved, larger $n_2$ forces smaller $\sin {\theta }_2$ — the ray bends toward the normal.

False. Frequency is fixed by the source and is continuous across the boundary; it is the speed and hence wavelength that change ($\lambda =v/f$, and $v=c/n$).

True. Since $n=c/v$, the law becomes $\frac{\sin {\theta }_1}{v_1}=\frac{\sin {\theta }_2}{v_2}$; the $c$ cancels. See Refractive Index.

True. At ${\theta }_1=0$, $\sin {\theta }_1=0$, so $\sin {\theta }_2=0$ and ${\theta }_2=0$ for any indices — the ray passes straight through (its speed still changes, only its direction doesn't).

False. TIR needs $n_1>n_2$ (dense to rare). Going air→glass is rare→dense, so $\sin {\theta }_2=(n_1/n_2)\sin {\theta }_1<\sin {\theta }_1$ always has a real solution — no TIR. See Total Internal Reflection.

The word "more" is misapplied. The conserved quantity is $n\sin \theta$; a bigger $n$ must be paired with a smaller $\sin \theta$ to keep the product constant. Denser ⇒ smaller angle.

Snell's law uses the angle from the normal, not the surface. The true incidence angle is ${\theta }_1=90^{\circ }-60^{\circ }=30^{\circ }$.

The chain-rule derivative of $\sqrt{a^2+x^2}$ with respect to $x$ is $\frac{x}{\sqrt{a^2+x^2}}$, not $\frac{a}{\sqrt{a^2+x^2}}$ — you differentiate with respect to the free variable $x$, and $a$ is a fixed constant. See Calculus — minimization and stationary points.

The premise (spend less path in the slow medium) is right, but that is achieved by bending, not by going straight. A straight line would not minimize time; the kink that trades slow-length for fast-length does.

No — it's inverted. Going dense→rare with ${\theta }_2=90^{\circ }$: $n_1\sin {\theta }_c=n_2\sin 90^{\circ }$, so $\sin {\theta }_c=n_2/n_1$ (the smaller index on top). See Critical Angle.

No. Solving $dT/dx=0$ finds a stationary point; for a flat refracting boundary the second derivative is positive, so it is a minimum time. The wording "maximum" is wrong for this case.

Because time is what matters: $t=d/v=nd/c$. The speed slows by a factor $n$, so a distance in a dense medium "costs" $n$ times more time — encoding this as $n\cdot d$ lets us minimize time by minimizing a length-like quantity.

The endpoints $A$ and $B$ are fixed, so their heights $a,b$ and separation $w$ are locked. The single choice light makes is where on the boundary to cross — every path is parameterized by that one number $x$.

The derivative $\frac{x}{\sqrt{a^2+x^2}}$ is exactly (horizontal run)/(hypotenuse) of the incidence right triangle, and in that triangle the run is the side opposite the angle-from-normal — so the ratio is $\sin \theta$ by definition.

Snell's law says $n_1\sin {\theta }_1=n_2\sin {\theta }_2$: this combination has the same value on both sides of the interface. Physically it is the horizontal component of the wave direction (tangential wave-vector), which the flat boundary cannot change.

The lifeguard is fast on sand, slow in water — like light in a rare then dense medium. Minimizing total time makes them run farther on the fast sand and cut short the slow swim, producing a bent path with a kink at the shoreline exactly as light bends at a boundary.

Snell's law is a purely geometric (kinematic) statement about directions of the transmitted ray. The amount of energy split between reflection and refraction is a separate dynamical question answered by the Fresnel equations, not by Fermat's least-time condition.

Both encode the same physical fact — that wavefronts stay continuous and light slows by $n$ in a medium. Huygens matches wavefront geometry at the boundary, Fermat minimizes time; they are two views of one truth. See Huygens' Principle and Snell's Law.

Then $\sin {\theta }_1=\sin {\theta }_2$ so ${\theta }_2={\theta }_1$ — no bending. There is effectively no boundary, and the path is a straight line, consistent with least-time in a uniform medium.

It would require $\sin {\theta }_2=(n_1/n_2)\sin {\theta }_1>1$, which has no real angle. There is no transmitted ray: all light reflects internally — total internal reflection.

To ${\theta }_2=90^{\circ }$: the refracted ray skims along the surface (grazing), the limiting orientation before transmission vanishes entirely.

This is grazing incidence; going rare→dense it gives the largest possible ${\theta }_2$, and by symmetry that limiting ${\theta }_2$ equals the critical angle of the reverse (dense→rare) trip.

Then $n_1>n_2$ again in effect: the ray bends away from the normal, and a critical angle for TIR can exist for the vacuum→medium direction — the "denser" role is played by whichever side has the larger $n$, regardless of everyday intuition about materials.

Yes, locally. At each point you use the local normal to the curved surface; Snell's law is a point condition. The overall path just uses a different normal direction at each crossing point.