Question bank — Mirrors — plane, concave, convex; mirror equation 1 - v + 1 - u = 1 - f

True — a plane mirror has $R=\infty$ so $f=\infty$; the mirror equation gives $v=-u$, meaning $m=+1$, so image is erect, same size, and as far behind as the object is in front.

False — only when the object is beyond $F$. If the object sits inside the focus ($|u|<|f|$) the concave mirror gives a virtual, erect, magnified image (the shaving-mirror mode).

False — with $f>0$ and any real object $u<0$, the equation forces $v>0$ always, so the image is invariably virtual, erect, and diminished.

False — sign comes from geometry, not behaviour. The focus lies in front of the mirror (to the left, against incident light), so $f<0$ for concave regardless of how strongly it converges.

True — $m>0$ means erect, and a single mirror can only make an erect image on the far (virtual) side, so the image is virtual and diminished (this is a convex-mirror signature).

False — the magnitude $|f|=|R|/2$ matches, but a concave one gives $f<0$ and a convex one $f=+|R|/2$; the sign differs, and that sign controls all image behaviour.

False — plane-mirror magnification is always exactly $+1$; the image only moves closer (still equally far behind), it never changes size.

True in the limiting sense — as the object approaches $F$, $v\to \infty$, so $|m|\to \infty$; the "image" runs off to infinity and is effectively unboundedly large just before it flips from real to virtual.

Error: they dropped the sign. A real object in front means the pole-to-object distance is measured against the incident light, so $u=-30\text{ cm}$; using $+30$ misplaces the image and can predict a nonexistent virtual image.

Error: $f=R/2$ is a paraxial approximation. Rays far from the axis focus slightly closer to the mirror, causing spherical aberration; the clean halfway result only holds near the axis.

Error: the normal at any surface point $M$ runs along the radius $R=PC$ to the centre of curvature $C$; only at the pole $P$ does that radius coincide with the principal axis.

Error: $v>0$ means the image is behind the mirror, so it is virtual, not real. For mirrors, real images have $v<0$ (in front) — this student swapped the convention used for lenses.

Error: you must keep the signed values inside. E.g. $v=-15,u=-30$ gives $m=-(-15)/(-30)=-0.5$; using bare magnitudes would wrongly report the orientation.

Error: at $C$ the image also forms at $C$ with $m=-1$ — same size, inverted, real. There is no magnification; the object and image just coincide in size.

Partly wrong: the equation is symmetric, so swapping object and image positions is geometrically valid (reversibility of light), but their heights $h,h^{\prime }$ and the magnification are not symmetric — $m=-v/u$ inverts to $-u/v$ when swapped.

The ray from the object top (height $h$) to the pole $P$ reflects with equal angles about the axis (the axis is the normal at $P$), forming two similar right triangles whose side ratios give $|h^{\prime }/h|=|v/u|$; the sign turns negative because the image (height $h^{\prime }$) drops below the axis.

It always forms a diminished, erect virtual image, so it squeezes a wide scene into a small mirror — a large field of view, at the cost of things looking farther than they are.

It was derived by comparing signed similar-triangle lengths; without consistent signs the same formula would give contradictory answers for objects and images on different sides of the pole.

A concave surface bends parallel rays to actually cross in front, so $F$ is a real point ($f<0$); a convex surface spreads them so they only appear to diverge from a point behind, making $F$ virtual ($f>0$).

The isosceles-triangle argument assumes the strike point $M$ is so close to $P$ that $FM\approx FP$; for wide rays that approximation breaks, and the exact focus creeps toward the mirror — the origin of spherical aberration.

With $f=\infty$ the equation gives $v=-u$, always positive-behind for a front object, so the image is forever virtual — reflected rays only diverge as if from a point behind the glass, never converge.

The mirror flips the axis pointing into it (front-back), which our brains read as left-right swap; vertical and horizontal directions in the mirror plane are preserved, so the image stays erect.

The equation gives $\frac{1}{v}=\frac{1}{f}-\frac{1}{f}=0$, so $v\to \infty$ — the reflected rays emerge parallel and the image forms "at infinity," which is why a bulb at $F$ makes a searchlight beam.

As $u\to 0^{-}$, $\frac{1}{v}\to -\frac{1}{u}\to +\infty$ so $v\to 0^{-}$ (a hair in front of the pole), and $m=-v/u\to +1$: a tiny object touching the mirror sees an image right at the surface, same size and erect, behaving locally like a flat mirror.

$\frac{1}{v}=\frac{1}{f}-0=\frac{1}{f}$, so $v=f$; since a concave mirror has $f<0$, this $v$ is negative — a real image in front of the mirror at the focus, tiny and inverted (how a concave mirror forms a sharp Sun image).

Since $v=-u$, we get $m=-v/u=-(-u)/u=+1$ for every object distance — a formal proof that plane mirrors never enlarge or shrink.

$v=f>0$, so the image sits at the (virtual) focus behind the mirror, a point-sized erect virtual image — the smallest, farthest-back image a convex mirror can make.

Just outside $F$ the image is real and huge in front; exactly at $F$ it vanishes to infinity; just inside $F$ it reappears virtual, erect and magnified behind the mirror — the sign of $v$ flips as the object passes the focus.