Sign convention for mirrors and lenses: Visual walkthrough

2.5.3 · D2 · Physics › Optics › Mirrors aur lenses ke liye sign convention

Step 1 — Axis banao aur choose karo ki "positive" direction kaun si hai

KYA. Kisi bhi physics se pehle, hum ek single horizontal line draw karte hain aur ek end par mirror lagate hain. Light baayi taraf se aayegi. Hum declare karte hain ki left→right measurement ka positive direction hai.

KYUN. Ek distance tab tak meaningless hai jab tak tum yeh na bolo ki "kahan se measure kiya, kis direction mein." Optics mein bahut saari distances hain ( $u$ , $v$ , $f$ , $R$ ) aur yeh tab hi sahi se add hoti hain jab sabka ek hi ruler ho. Woh shared ruler principal axis hai, aur uska positive end jis taraf light travel kar rahi hai us taraf point karta hai. Yeh ek choice hi hai jo ek equation ko dozen special cases ki jagah kaam karne deti hai — parent ke "why bother" argument mein dekho.

PICTURE. Safed horizontal line principal axis hai. Jahan yeh mirror se milti hai woh point pole $P$ hai — hamara origin, ruler ka " $0$ ". Amber arrow incident light ko left→right jaate dikhaata hai, yaani $+$ direction.

Figure — Sign convention for mirrors and lenses

Step 2 — Object rakho aur uski distance ko $u$ naam do

KYA. Hum mirror ke baayi taraf axis par ek seedha arrow (object, height $h$ ) khadte hain. Pole $P$ se object ke foot tak ki distance ko hum $u$ kehte hain.

KYUN. $u$ "kitna door hai" woh raw length nahi hai — yeh hamare ruler par ek coordinate hai. Object $P$ ke baayi taraf hai, jo Step 1 mein choose ki gayi incident-light direction ke against hai. "Against light $=$ negative" rule se, ek real object hamesha negative coordinate deta hai. Hum likhte hain $u = - a$ , jahan $a = ∣ u ∣$ woh plain positive length hai jo tum tape measure se padhoge.

PICTURE. Height $h$ ka cyan arrow object hai. $P$ se baayi taraf safed bracket length $a$ hai. Label yaad dilaata hai: coordinate $u = - a$ hai, length $a > 0$ hai.

Step 3 — Do rays bhejo aur unhe image build karne do

KYA. Hum object ki tip se do special rays draw karte hain:

ek ray axis ke parallel, jo reflect hone ke baad focus $F$ se guzarni chahiye;
ek ray centre $C$ se guzarti hui, jo mirror ko sidha hit karti hai aur seedha wapas bounce hoti hai. Jahan do reflected rays cross hoti hain woh image ki tip hai, height $h^{'}$ ka ek arrow $P$ se $v$ distance par.

KYUN. Ek point locate karne ke liye do rays kaafi hain: ek point ke liye do lines ka cross karna zaroori hai. Hum yeh do isliye choose karte hain kyunki unka behavior jaana-maana hai — parallel-through-focus focus ki definition hai, aur through- $C$ sphere ko radius ke saath hit karta hai (surface ke perpendicular) isliye woh khud ko retrace karta hai. Abhi tak koi calculus nahi; bas "do jaani-maani lines kahan milti hain."

PICTURE. Do amber rays object tip se nikalti hain, reflect hoti hain, aur baayi taraf cross karti hain, height $h^{'}$ ka inverted cyan image banaati hain. Uska foot $P$ se $b = ∣ v ∣$ length par hai; kyunki image bhi baayi taraf (against light) hai, uska coordinate $v = - b$ hai.

Step 4 — Do similar triangles dhundho (iska dil yehi hai)

KYA. Dekho kahan har parallel incident ray pole ke paas mirror ko touch karti hai. Woh reflected ray image ki tip tak jaati hai aur focus $F$ se guzarti hai. Isse do triangles bante hain jo same angles share karte hain — yeh similar hain — isliye unke side-ratios equal hain.

KYUN similar triangles aur koi fancy cheez nahi? Humhe $a$ , $b$ , aur focal length $F$ ke beech ek relation chahiye. Similar triangles sabse simple tool hai jo "yeh angles equal hain" ko "yeh ratios equal hain" mein convert karta hai. Hum paraxial approximation mein kaam kar rahe hain: saari rays axis ke paas rehti hain, saari angles chhoti hain, isliye curved mirror $P$ se guzarti ek flat vertical line ki tarah behave karta hai aur chhoti heights aur lengths saaf line up hoti hain.

PICTURE. Do shaded triangles $F$ par vertex share karte hain. Uncha wala object height $h$ rakhta hai length $(a - F)$ ke upar; chhota wala image height $h^{'}$ rakhta hai length $(F - b)$ ke upar... lekin saaf pair woh hai jo neeche hai. Do matching angle-ticks dekho.

Do similar triangles (object-side vs image-side, dono $F$ par milte hain) plain positive lengths mein dete hain:

\frac{h}{h ^{'}} = \frac{a - F}{F} aur \frac{h}{h ^{'}} = \frac{a}{b} .

Step 5 — Ratios ko combine karke ek magnitude equation banao

KYA. $h / h^{'}$ ke liye dono expressions ko equal set karo aur simplify karo.

KYUN. Dono same cheez ( $h / h^{'}$ ) ke equal hain, isliye woh ek doosre ke equal hain. Isse heights eliminate ho jaati hain aur sirf distances $a$ , $b$ , $F$ bachi rehti hain — exactly woh relation jise hum dhundh rahe hain.

\frac{a - F}{F} = \frac{a}{b}

Cross-multiply karo ( $b (a - F) = a F$ ), expand karo ( $ab - b F = a F$ ), aur har term ko $ab F$ se divide karo:

\frac{ab}{ab F} - \frac{b F}{ab F} = \frac{a F}{ab F} ⟹ \frac{1}{a} + \frac{1}{b} = \frac{1}{F}

PICTURE. Algebra visualise ki gayi: teen reciprocal "slices" $\frac{1}{a}$ , $\frac{1}{b}$ , $\frac{1}{F}$ bars ke roop mein dikhaye gaye jinki lengths sum satisfy karti hain. Yeh magnitude mirror formula hai — true numbers, abhi tak koi signs nahi.

Step 6 — Lengths ko signed coordinates mein promote karo

KYA. Har plain length ko Steps 2–3 ke signed coordinate se replace karo. Is concave-mirror-with-real-image picture ke liye: object left ⇒ $u = - a$ ; image left ⇒ $v = - b$ ; concave focus left ⇒ $f = - F$ .

KYUN. Magnitude formula sirf isi ek picture ke liye kaam karta hai. Hum ek aisi single equation chahte hain jo tab bhi survive kare jab image mirror ke peeche flip ho jaaye, ya mirror convex ho jaaye. Trick yeh hai: sab kuch un coordinates mein express karo jo already sign carry karte hain, taaki geometry ko kabhi redraw na karna pade.

Magnitude relation se shuru karo aur $- 1$ se multiply karo:

- \frac{1}{a} - \frac{1}{b} = - \frac{1}{F} .

Ab substitute karo $- a = u$ , $- b = v$ , $- F = f$ , yaani $\frac{1}{- a} = \frac{1}{u}$ etc.:

\frac{1}{u} + \frac{1}{v} = \frac{1}{f} ⟹ \frac{1}{v} + \frac{1}{u} = \frac{1}{f}

PICTURE. Wahi diagram, ab neeche ek signed ruler ke saath: $P$ ke baayi taraf sab kuch negative padhta hai, daayeni taraf positive. Har length label $a, b, F$ ko cross out karke $- u, - v, - f$ se replace kiya gaya hai.

Step 7 — Har case test karo: kya ek equation sach mein sab cover karta hai?

KYA. Hum signed formula ko un chaar situations ke against stress-test karte hain jiske baare mein parent ne warning di thi, bina ek bhi triangle redraw kiye — bas signs plug in karo.

KYUN. Contract yeh hai: reader ko aisa koi scenario nahi mileega jo humne nahi dikhaya. Isliye hum concave (real & virtual) aur convex tour karte hain, plus degenerate flat-mirror limit.

Case	Signs in	$\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$	Result	Nature
Concave, object beyond $F$	$u = - 30, f = - 10$	$- \frac{1}{10} + \frac{1}{30} = - \frac{2}{30}$	$v = - 15$	real, inverted (left)
Concave, object inside $F$	$u = - 6, f = - 10$	$- \frac{1}{10} + \frac{1}{6} = + \frac{1}{15}$	$v = + 15$	virtual, erect (behind)
Convex mirror	$u = - 20, f = + 15$	$\frac{1}{15} + \frac{1}{20} = \frac{7}{60}$	$v = + 8.57$	virtual, erect, small
Flat mirror ( $f \to \infty$ )	$u = - 20, \frac{1}{f} = 0$	$0 + \frac{1}{20} = \frac{1}{20}$	$v = + 20$	virtual, same size, behind

Dekho $v$ ka sign har baar khud nature bata deta hai: $v < 0$ ⇒ real aur saamne; $v > 0$ ⇒ virtual aur peeche. Flat mirror sirf limit $∣ f ∣ \to \infty$ hai jahan $\frac{1}{f} \to 0$ , giving $v = - u$ (image utni hi door peeche jitna object saamne hai) — exactly tumhara bathroom mirror. Koi nayi formula nahi chahiye. "Real vs virtual" ka matlab jaanne ke liye, dekho Real and virtual images.

Ek-picture summary

Sab kuch iss par collapse hota hai: axis draw karo, $P$ ko zero mark karo, positive ko us taraf point karo jis taraf light jaati hai, signed coordinates $u, v, f$ read off karo, aur ek equation baaki sab karta hai. Lenses ke liye sister equation $+$ ko $-$ se swap karti hai kyunki light bounce karne ki jagah transmit hoti hai — same tarah se Lens formula and lensmaker's equation mein derive ki gayi, Refraction at spherical surfaces mein uski spherical-surface roots ke saath aur Combination of thin lenses mein stacking ke saath.

Recall Feynman retelling — plain words mein wapas bolo

Maine ek line draw ki aur daayeni end par ek mirror lagaya. Maine kaha "jis taraf light jaati hai — left to right — woh meri positive direction hai, aur mera zero woh pole hai jahan axis mirror ko touch karti hai." Maine baayi taraf ek object rakha; kyunki woh zero ke baayi taraf hai, uski distance $u$ negative aati hai — yeh koi trick nahi, yeh bas ruler par wahan hai. Maine do rays chhodi jo mujhe pehle se pata hain: ek axis ke parallel jo focus se bounce karti hai, ek centre se jo seedha wapas bounce karti hai. Jahan woh cross huin woh image hai. Phir maine do triangles dhunde jiske same angles the, isliye unke sides same ratio mein hain; us ratio ne object height, image height, aur distances ko link kiya. Dono ratios ko equal set karke aur clean up karke mujhe plain positive lengths mein $\frac{1}{a} + \frac{1}{b} = \frac{1}{F}$ mila. Finally maine har length ko uske signed coordinate se swap kiya — $- 1$ se multiply karo, $- a = u$ , $- b = v$ , $- F = f$ substitute karo — aur $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ nikal aaya. Sundar part yeh hai: yeh ek equation ab concave, convex, real image, virtual image, even flat mirror ko "infinite focus" limit ke roop mein handle karta hai, aur answer ka sign mujhe bina kisi extra thought ke bata deta hai ki image real hai ya virtual.

Recall Quick self-check

Object hamesha $u < 0$ kyun deta hai? ::: Woh pole ke baayi taraf hai, chosen positive (incident-light) direction ke against, isliye ruler par uska coordinate negative hai. Step 6 mein minus sign kahan se aata hai? ::: Humne positive-length equation ko $- 1$ se multiply kiya taaki lengths $a, b, F$ signed coordinates $u = - a, v = - b, f = - F$ ban jayein. Mirror ke liye $v > 0$ ka matlab kya hai? ::: Image mirror ke peeche hai — virtual aur erect. Flat mirror ek special case kaise hai? ::: $∣ f ∣ \to \infty$ lo taaki $\frac{1}{f} \to 0$ , $v = - u$ bachta hai: image utni hi door peeche jitna object saamne, same size.