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

What we use: $f=\frac{R}{2}$ (focus sits halfway to the centre of curvature). Why: The focus–radius relation is pure geometry — nothing to compute but a halving. $f=\frac{R}{2}=\frac{+30}{2}=+15\text{ cm}$ Positive, because a convex mirror's focus lies behind the surface (with the outgoing light). ✔

A plane mirror is the flat limit of a spherical one ($R\to \infty$, so $f\to \infty$, and $\frac{1}{f}\to 0$). $\frac{1}{v}+\frac{1}{u}=0\Rightarrow v=-u$ With $u=-12\text{ cm}$, we get $v=+12\text{ cm}$ — image $12\text{ cm}$ behind the mirror (virtual). $m=-\frac{v}{u}=-\frac{+12}{-12}=+1$ Erect, same size. This is the "equal distance behind, same size" rule falling straight out of the equation. ✔

Rearrange first so the unknown is alone: $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{-20}-\frac{1}{-30}$ Common denominator $60$: $-\frac{3}{60}+\frac{2}{60}=-\frac{1}{60}$. $v=-60\text{ cm}$ $v<0$ ⇒ image in front ⇒ real. $m=-\frac{v}{u}=-\frac{-60}{-30}=-2$ $m<0$ ⇒ inverted; $|m|=2$ ⇒ magnified. So: real, inverted, twice the size, farther than the object. ✔

$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{15}-\frac{1}{-10}=\frac{1}{15}+\frac{1}{10}$ Common denominator $30$: $\frac{2}{30}+\frac{3}{30}=\frac{5}{30}=\frac{1}{6}$. $v=+6\text{ cm}$ $v>0$ ⇒ virtual, behind the mirror. $m=-\frac{v}{u}=-\frac{+6}{-10}=+0.6$ Erect, diminished — the standard convex-mirror image. ✔

Read the words as signs first. Erect ⇒ $m>0$; magnified 3× ⇒ $m=+3$. From $m=-\frac{v}{u}$: $+3=-\frac{v}{u}\Rightarrow v=-3u$ Substitute into the mirror equation: $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{-3u}+\frac{1}{u}=\frac{1}{-12}$ $\frac{-1+3}{3u}=\frac{1}{-12}\Rightarrow \frac{2}{3u}=\frac{-1}{12}$ $3u=2\times (-12)=-24\Rightarrow u=-8\text{ cm}$ Then $v=-3u=+24\text{ cm}$ (behind, virtual — consistent with erect & magnified). The object is 8 cm in front, which is inside the focus ($|u|=8<|f|=12$) — exactly the shaving-mirror regime. ✔

First find $v$: $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{20}-\frac{1}{-30}=\frac{1}{20}+\frac{1}{30}=\frac{3+2}{60}=\frac{5}{60}=\frac{1}{12}$ $v=+12\text{ cm}$ Magnification: $m=-\frac{v}{u}=-\frac{+12}{-30}=+0.4$ Image height: $h^{\prime }=mh=0.4\times 4=+1.6\text{ cm}$ Positive height ⇒ erect; smaller ⇒ diminished. ✔

Same size + real ⇒ $m=-1$ (real image of a single mirror is inverted, so the negative sign is forced). $m=-\frac{v}{u}=-1\Rightarrow v=u$ With $u=-40\text{ cm}$, $v=-40\text{ cm}$ (both in front — real). $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}=\frac{1}{-40}+\frac{1}{-40}=\frac{-2}{40}=\frac{-1}{20}$ $f=-20\text{ cm},R=2f=-40\text{ cm}$ The object sits at $|u|=40=|R|$ — it is at the centre of curvature C. (Object at C ⇒ image at C, real, inverted, same size.) ✔

$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{-10}-\frac{1}{-6}=-\frac{1}{10}+\frac{1}{6}$ Common denominator $30$: $-\frac{3}{30}+\frac{5}{30}=\frac{2}{30}=\frac{1}{15}$. $v=+15\text{ cm}$ $v>0$ ⇒ image is behind the mirror ⇒ virtual ⇒ it cannot fall on a screen. The student is wrong. $m=-\frac{v}{u}=-\frac{+15}{-6}=+2.5$ Erect, magnified — a virtual image you can only see, not project. ✔

Concave ⇒ $f=-15\text{ cm}$. We need $|m|=2$, so two cases.

Case A — real image (inverted), $m=-2$: $v=-2u\cdot (-1)$… let's be careful. $m=-v/u=-2\Rightarrow v=2u$. $\frac{1}{2u}+\frac{1}{u}=\frac{1}{-15}\Rightarrow \frac{1+2}{2u}=\frac{-1}{15}\Rightarrow \frac{3}{2u}=\frac{-1}{15}$ $2u=-45\Rightarrow u=-22.5\text{ cm},v=2u=-45\text{ cm (real, in front).}$ Check: $m=-v/u=-(-45)/(-22.5)=-2$ ✔ (real, inverted, magnified — object between F and C).

Case B — virtual image (erect), $m=+2$: $-v/u=+2\Rightarrow v=-2u$. $\frac{1}{-2u}+\frac{1}{u}=\frac{1}{-15}\Rightarrow \frac{-1+2}{2u}=\frac{-1}{15}\Rightarrow \frac{1}{2u}=\frac{-1}{15}$ $2u=-15\Rightarrow u=-7.5\text{ cm},v=-2u=+15\text{ cm (virtual, behind).}$ Check: $m=-v/u=-(15)/(-7.5)=+2$ ✔ (virtual, erect — object inside F).

Answer: $u=-22.5\text{ cm}$ (real image) or $u=-7.5\text{ cm}$ (virtual image). Same $|m|$, two totally different physical situations. ✔

(a) $f=R/2=+1.0\text{ m}=+100\text{ cm}$. Object $u=-10\text{ m}=-1000\text{ cm}$. $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{100}-\frac{1}{-1000}=\frac{1}{100}+\frac{1}{1000}=\frac{10+1}{1000}=\frac{11}{1000}$ $v=\frac{1000}{11}\approx +90.9\text{ cm}\approx +0.91\text{ m (virtual, behind).}$ (b) $m=-\frac{v}{u}=-\frac{+90.9}{-1000}\approx +0.091$ The image is only about 9% of true size, erect. Because it is shrunk toward the tiny mirror focus, your brain reads the small image as a distant, small car — but the car is actually much closer. That mismatch is exactly why the safety warning is stamped on every convex mirror. ✔

1. Object at C of a concave mirror → image? → at C, real, inverted, same size.
2. Convex mirror, any real object → image? → always virtual, erect, diminished.
3. $|m|=2$, concave: how many object positions? → two — one real ($m=-2$), one virtual ($m=+2$).
4. Sign that tells real vs virtual? → sign of $v$: $v<0$ real, $v>0$ virtual.