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

Hum kya use karenge: $f=\frac{R}{2}$ (focus, centre of curvature ke aadhe raaste pe hota hai). Kyun: Focus–radius relation purely geometry hai — sirf half karna hai, kuch compute nahi. $f=\frac{R}{2}=\frac{+30}{2}=+15\text{ cm}$ Positive, kyunki convex mirror ka focus surface ke peeche hota hai (outgoing light ke saath). ✔

Plane mirror ek spherical mirror ki flat limit hai ($R\to \infty$, toh $f\to \infty$, aur $\frac{1}{f}\to 0$). $\frac{1}{v}+\frac{1}{u}=0\Rightarrow v=-u$ $u=-12\text{ cm}$ ke saath, hume milta hai $v=+12\text{ cm}$ — image mirror ke $12\text{ cm}$ peeche (virtual). $m=-\frac{v}{u}=-\frac{+12}{-12}=+1$ Erect, same size. Yeh "equal distance peeche, same size" rule seedha equation se nikal aata hai. ✔

Pehle rearrange karo taaki unknown akela ho: $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{-20}-\frac{1}{-30}$ Common denominator $60$: $-\frac{3}{60}+\frac{2}{60}=-\frac{1}{60}$. $v=-60\text{ cm}$ $v<0$ ⇒ image saamne ⇒ real. $m=-\frac{v}{u}=-\frac{-60}{-30}=-2$ $m<0$ ⇒ inverted; $|m|=2$ ⇒ magnified. Toh: real, inverted, double size, object se door. ✔

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

Pehle words ko signs mein padhо. Erect ⇒ $m>0$; magnified 3× ⇒ $m=+3$. $m=-\frac{v}{u}$ se: $+3=-\frac{v}{u}\Rightarrow v=-3u$ Mirror equation mein substitute karo: $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{-3u}+\frac{1}{u}=\frac{1}{-12}$ $\frac{-1+3}{3u}=\frac{1}{-12}\Rightarrow \frac{2}{3u}=\frac{-1}{12}$ $3u=2\times (-12)=-24\Rightarrow u=-8\text{ cm}$ Phir $v=-3u=+24\text{ cm}$ (peeche, virtual — erect & magnified ke saath consistent). Object 8 cm saamne hai, jo focus ke andar hai ($|u|=8<|f|=12$) — bilkul shaving-mirror regime. ✔

Pehle $v$ nikalo: $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{20}-\frac{1}{-30}=\frac{1}{20}+\frac{1}{30}=\frac{3+2}{60}=\frac{5}{60}=\frac{1}{12}$ $v=+12\text{ cm}$ 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; chhota ⇒ diminished. ✔

Same size + real ⇒ $m=-1$ (single mirror ki real image inverted hoti hai, toh negative sign forced hai). $m=-\frac{v}{u}=-1\Rightarrow v=u$ $u=-40\text{ cm}$ ke saath, $v=-40\text{ cm}$ (dono saamne — real). $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}=\frac{1}{-40}+\frac{1}{-40}=\frac{-2}{40}=\frac{-1}{20}$ $f=-20\text{ cm},R=2f=-40\text{ cm}$ Object $|u|=40=|R|$ pe hai — yeh centre of curvature C pe hai. (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 mirror ke peeche hai ⇒ virtual ⇒ yeh screen pe nahi aaegi. Student galat hai. $m=-\frac{v}{u}=-\frac{+15}{-6}=+2.5$ Erect, magnified — ek virtual image jo tum sirf dekh sakte ho, project nahi kar sakte. ✔

Concave ⇒ $f=-15\text{ cm}$. Hume chahiye $|m|=2$, toh do cases.

Case A — real image (inverted), $m=-2$: $v=-2u\cdot (-1)$… dhyan se karte hain. $m=-v/u=-2\Rightarrow v=2u$. $\frac{1}{2u}+\frac{1}{u}=\frac{1}{-15}\Rightarrow \frac{1+2}{2u}=\frac{-1}{15}\Rightarrow \frac{3}{2u}=\frac{-1}{15}$ $2u=-45\Rightarrow u=-22.5\text{ cm},v=2u=-45\text{ cm (real, saamne).}$ Check: $m=-v/u=-(-45)/(-22.5)=-2$ ✔ (real, inverted, magnified — object F aur C ke beech).

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

Answer: $u=-22.5\text{ cm}$ (real image) ya $u=-7.5\text{ cm}$ (virtual image). Same $|m|$, do bilkul alag physical situations. ✔

(a) $f=R/2=+1.0\text{ m}=+100\text{ cm}$. Object $u=-10\text{ m}=-1000\text{ cm}$. $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{100}-\frac{1}{-1000}=\frac{1}{100}+\frac{1}{1000}=\frac{10+1}{1000}=\frac{11}{1000}$ $v=\frac{1000}{11}\approx +90.9\text{ cm}\approx +0.91\text{ m (virtual, peeche).}$ (b) $m=-\frac{v}{u}=-\frac{+90.9}{-1000}\approx +0.091$ Image sirf asli size ka 9% hai, erect. Kyunki yeh tiny mirror focus ki taraf shrink ho jaati hai, tumhara brain chhoti image ko ek door, chhoti car padhta hai — lekin car actually bahut paas hai. Yahi mismatch exactly hai jis wajah se safety warning har convex mirror pe stamp hoti hai. ✔

1. Concave mirror ke C pe object → image? → C pe, real, inverted, same size.
2. Convex mirror, koi bhi real object → image? → hamesha virtual, erect, diminished.
3. $|m|=2$, concave: kitne object positions? → do — ek real ($m=-2$), ek virtual ($m=+2$).
4. Sign jo real vs virtual batata hai? → $v$ ka sign: $v<0$ real, $v>0$ virtual.