Sign convention for mirrors and lenses: Worked examples

2.5.3 · D3 · Physics › Optics › Sign convention for mirrors and lenses

Shuru karne se pehle, aao un do tools ko dobara samjhein jo baar baar use honge, taaki koi bhi symbol bina explanation ke na rahe.

The scenario matrix

Is topic ka har problem is table ka ek cell hai. Last column us example ka naam batata hai jo use clear karta hai. (Definition se yaad karo: $C$ = distance $2∣ f ∣$ wala point, $f$ ke same sign ke saath.)

#	Element	Case class (the "cell")	Kya stress-test ho raha hai	Example
A	Concave mirror	Object beyond $C$ (yaani $	u	>2
B	Concave mirror	Object $F$ aur pole ke beech ($	u	<
C	Concave mirror	Object exactly at $F$ ( $u = f$ )	degenerate: image at infinity	Ex 3
D	Convex mirror	Koi bhi real object	hamesha virtual, erect, diminished; $v > 0$	Ex 4
E	Plane mirror	Limiting case $f \to \infty$	$v = - u$ , virtual, same size	Ex 5
F	Convex lens	Object beyond $2 f$	real, inverted, diminished	Ex 6
G	Convex lens	Object inside $f$ (magnifier)	virtual, erect, magnified; $v < 0$	Ex 7
H	Convex lens	Object exactly at $F$ ( $u = f$ )	degenerate: image at infinity	Ex 8
I	Concave lens	Koi bhi real object	hamesha virtual, erect, diminished	Ex 9
J	Word problem	Real-world (rear-view mirror)	words → signs mein translate karo	Ex 10
K	Exam twist	Two-lens combination + magnification	$v$ ko chain karo → next $u$	Ex 11

Cells A–K milake sign ke har quadrant, dono mirror aur lens ke degenerate/limiting inputs (C, E, H), ek word problem (J) aur ek exam twist (K) cover karte hain.

Mirror examples

Figure — Sign convention for mirrors and lenses

Figure 1 padhna. Horizontal chalk line principal axis hai (hamara $+ x$ , right ki taraf). Right par tall curved chalk stroke concave mirror hai, jiske saath pole P (white dot) hai jahan woh axis se milti hai. P ke left mein do marked points hain: yellow dot focus $F$ hai aur blue dot centre of curvature $C$ hai (the $2 f$ point). Top par pink arrow incident light left→right (+x) travel karte dikhata hai. Teen upright object arrows pole ke left mein khade hain — isi liye neeche ke har $u$ negative hai:

yellow arrow, beyond $C$ = Ex 1 (Cell A),

blue arrow, $F$ aur P ke beech = Ex 2 (Cell B),

pink arrow, $F$ par planted = Ex 3 (Cell C), jiske reflected rays parallel nikalte hain (image at infinity).

Neeche ke labels yaad dilaate hain: P ke left mein sab kuch negative hai, P ke peeche (right) mein sab kuch positive hai.

Worked example Ex 1 — Cell A: concave mirror, object beyond C

Ek candle ek concave mirror ke focal length 15 cm ke saamne 40 cm door hai. Image locate karo aur describe karo.

Forecast: Figure mein yeh yellow arrow hai, $C$ ( $2 f = 30$ cm point) ke left mein baitha hai. Guess: real, inverted, object se chhota. $v$ kahaan land karega?

Signs assign karo. $u = - 40$ cm, $f = - 15$ cm. Yeh step kyun? Figure mein yellow object arrow pole ke left mein hai → incident light ke against → negative; concave focus $F$ bhi pole ke left mein drawn hai → negative.
Mirror formula mein substitute karo. $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{- 15} - \frac{1}{- 40} = - \frac{1}{15} + \frac{1}{40}$ . Yeh step kyun? Hum upar se solved-for- $v$ mirror form $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ use karte hain.
Compute karo. Common denominator 120: $- \frac{8}{120} + \frac{3}{120} = - \frac{5}{120} = - \frac{1}{24}$ , toh $v = - 24$ cm. Yeh step kyun? Dono fractions ko ek mein combine karo, phir invert karke $v$ nikalo.
Sign interpret karo. $v < 0$ ⇒ (mirror rule) image $- x$ mein hai, mirror ke saamne = real — yellow object ke same left side.
Magnification. $m = - \frac{v}{u} = - \frac{- 24}{- 40} = - 0.6$ ⇒ inverted, $0.6 \times$ size. Yeh step kyun? Mirror magnification rule apply karo; iska sign orientation batata hai, size scaling.

Verify: $- 24$ cm, $F$ (15) aur $C$ (30) ke beech land karta hai jaisa beyond $C$ wale object ke liye hona chahiye — consistent. Plug back: $\frac{1}{- 24} + \frac{1}{- 40} = - \frac{5}{120} - \frac{3}{120} = - \frac{8}{120} = - \frac{1}{15} = \frac{1}{f}$ . ✔

Worked example Ex 2 — Cell B: concave mirror, object inside F

Ek tooth dentist ke concave mirror ke focal length 10 cm ke saamne 6 cm door hai. Image nikalo.

Forecast: Figure mein yeh blue arrow hai, $F$ aur pole ke beech daba hua. Concave mirror + object inside $F$ = "magnifying mirror" regime → guess virtual, erect, enlarged, toh $v > 0$ .

Signs. $u = - 6$ cm, $f = - 10$ cm. Yeh step kyun? Figure mein blue object arrow aur focus dono pole ke left mein hain → dono negative.
Substitute. $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{- 10} - \frac{1}{- 6} = - \frac{1}{10} + \frac{1}{6}$ . Yeh step kyun? Same solved-for- $v$ mirror form $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ .
Compute. Denominator 30: $- \frac{3}{30} + \frac{5}{30} = \frac{2}{30} = \frac{1}{15}$ , toh $v = + 15$ cm. Yeh step kyun? Ek fraction mein merge karo aur invert karo; positive result already side ka switch signal karta hai.
Interpret. $v > 0$ ⇒ (mirror rule) image $+ x$ mein hai, mirror ke peeche = virtual (figure mein positive/right zone).
Magnification. $m = - \frac{v}{u} = - \frac{15}{- 6} = + 2.5$ ⇒ erect, $2.5 \times$ magnified. Yeh step kyun? $m$ ka sign "erect" confirm karta hai, dentist-mirror expectation se match karta hai.

Verify: Positive $m > 1$ aur erect dentist-mirror ke purpose se match karta hai. Plug back: $\frac{1}{15} + \frac{1}{- 6} = \frac{2}{30} - \frac{5}{30} = - \frac{3}{30} = - \frac{1}{10} = \frac{1}{f}$ . ✔

Worked example Ex 3 — Cell C: concave mirror, object AT the focus (degenerate)

Object exactly ek concave mirror ke focus par rakha gaya, $f = - 12$ cm, toh $u = - 12$ cm.

Forecast: Figure mein yeh pink arrow hai, bilkul $F$ par planted. Yeh "real, beyond" aur "virtual, inside" regimes ke beech ki boundary hai. Guess: image kisi finite jagah nahi — rays parallel nikalti hain.

Signs. $u = - 12$ cm, $f = - 12$ cm. Yeh step kyun? Object aur focus coincide karte hain, dono pole ke left → dono same negative value lete hain.
Substitute. $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{- 12} - \frac{1}{- 12} = 0$ . Yeh step kyun? Jab $u = f$ hota hai toh do identical fractions exactly cancel ho jaate hain, $1/ v$ zero ho jaata hai.
Interpret. $\frac{1}{v} = 0 \Rightarrow v = \pm \infty$ : reflected rays parallel hain, image infinity par banta hai. Yeh step kyun? $1/ v \to 0$ "no crossing point" ka mathematical face hai — pink object ke parallel outgoing rays se match karta hai.

Verify: Yeh "parallel rays $F$ par converge hoti hain" ka reverse hai: $F$ par object hone se parallel rays nikalni chahiye (light ki reversibility). $\frac{1}{u} = \frac{1}{f}$ ⇒ $\frac{1}{v} = 0$ exactly. ✔ Degenerate case handled.

Worked example Ex 4 — Cell D: convex mirror (hamesha virtual)

Ek convex security mirror ka $f = + 20$ cm hai. Ek insaan 60 cm door khada hai.

Forecast: Convex mirrors hamesha virtual, erect, diminished image mirror ke peeche deti hain → $v > 0$ , $0 < m < 1$ .

Signs. $u = - 60$ cm, $f = + 20$ cm. Yeh step kyun? Object abhi bhi left par hai → $u < 0$ ; lekin convex focus mirror ke peeche hai = light ke downstream → $f > 0$ .
Substitute. $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{20} - \frac{1}{- 60} = \frac{1}{20} + \frac{1}{60}$ . Yeh step kyun? Solved-for- $v$ mirror form; negative $u$ ko subtract karna addition ban jaata hai.
Compute. Denominator 60: $\frac{3}{60} + \frac{1}{60} = \frac{4}{60} = \frac{1}{15}$ , toh $v = + 15$ cm. Yeh step kyun? Ek fraction, phir invert; positive $v$ confirm karta hai ki image mirror ke peeche hai.
Interpret. $v > 0$ ⇒ (mirror rule) image $+ x$ mein, mirror ke peeche = virtual.
Magnification. $m = - \frac{v}{u} = - \frac{15}{- 60} = + 0.25$ ⇒ erect, quarter size. Yeh step kyun? Magnification rule orientation (erect) aur shrinkage ( $0.25 \times$ ) pin karta hai.

Verify: $0 < v < f$ aur $0 < m < 1$ — har convex mirror ka fingerprint. Plug back: $\frac{1}{15} + \frac{1}{- 60} = \frac{4}{60} - \frac{1}{60} = \frac{3}{60} = \frac{1}{20} = \frac{1}{f}$ . ✔

Worked example Ex 5 — Cell E: plane mirror as the limit

$f \to \infty$ Ek plane mirror infinite radius wala mirror hota hai, toh $f \to \infty$ aur $\frac{1}{f} \to 0$ . Object $u = - 25$ cm par.

Forecast: Sab jaante hain ki plane mirror "jitna object saamne, utna hi peeche" image deta hai, same size, erect. Formula se prove karte hain.

Limit lo. $\frac{1}{f} = 0$ , toh mirror formula ban jaata hai $\frac{1}{v} + \frac{1}{u} = 0$ . Yeh step kyun? Flat mirror spherical family ka $R \to \infty$ end hai, toh iska focal term vanish ho jaata hai.
Solve karo. $\frac{1}{v} = - \frac{1}{u} \Rightarrow v = - u = - (- 25) = + 25$ cm. Yeh step kyun? Focal term hatne se $v$ exactly $u$ ka mirror-partner ban jaata hai.
Interpret. $v > 0$ ⇒ (mirror rule) image 25 cm mirror ke peeche = virtual.
Magnification. $m = - \frac{v}{u} = - \frac{25}{- 25} = + 1$ ⇒ erect, same size. Yeh step kyun? $m = + 1$ "identical, upright mirror image" ka algebraic statement hai.

Verify: $v = - u$ aur $m = + 1$ exactly roz ki "mirror image" rule hai. Limiting case confirmed. ✔

Lens examples

Figure 2 padhna. Horizontal chalk line phir se principal axis (+x) hai. Centre mein double-curved chalk shape ek convex lens hai, jiske optical centre O axis par hai. Do yellow dots dono foci $F$ (left) aur $F^{'}$ (right) mark karte hain. Pink top arrow incident light (+x) hai. Importantly, page ab do shaded bands mein divide hai, jo upar "meaning of $v$ " rule se seedha drawn hain:

left par blue band $v < 0$ zone hai — lens ke liye yeh virtual side hai (object ke same side). Ex 7 ka answer yahaan land karta hai.

right par yellow band $v > 0$ zone hai — lens ke liye yeh real (transmitted) side hai. Ex 6 ka answer yahaan land karta hai.

Blue object arrow left par khada hai ( $u < 0$ ). Kyunki light lens se guzarti hai, real side right hai — mirror case ka mirror image, aur exactly isi liye lens formula mein minus aata hai. Ex 8 mein object $F$ par push hota hai aur image right edge se infinity par chala jaata hai.

Worked example Ex 6 — Cell F: convex lens, object beyond 2f

Ek slide projector lens ke $f = + 10$ cm ke saath 30 cm left mein hai. Image nikalo.

Forecast: Object $2 f$ (jo 20 cm hai) se aage → real, inverted, diminished, image figure ke yellow right-hand zone mein $f$ aur $2 f$ ke beech land karega.

Signs. $u = - 30$ cm, $f = + 10$ cm. Yeh step kyun? Blue object arrow lens ke left mein hai → $u < 0$ ; convex lens rays ko right mein yellow zone mein converge karta hai → focus positive.
Lens formula. $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} + \frac{1}{- 30}$ . Yeh step kyun? Hum solved-for- $v$ lens form $\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$ use karte hain (summary se) ( $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ mein $\frac{1}{u}$ cross karne par plus ban gaya).
Compute. Denominator 30: $\frac{3}{30} - \frac{1}{30} = \frac{2}{30} = \frac{1}{15}$ , toh $v = + 15$ cm. Yeh step kyun? Ek fraction mein combine karo aur invert karo; positive $v$ image ko yellow (real) zone mein rakhta hai.
Interpret. $v > 0$ ⇒ (lens rule) image $+ x$ mein transmitted right side par = real.
Magnification. $m = + \frac{v}{u} = \frac{15}{- 30} = - 0.5$ ⇒ inverted, half size. Yeh step kyun? Lens magnification rule orientation (inverted) aur scale (half) fix karta hai.

Verify: $15$ cm $f = 10$ aur $2 f = 20$ ke beech hai jaisa beyond $2 f$ wale object ke liye chahiye. Plug back: $\frac{1}{15} - \frac{1}{- 30} = \frac{2}{30} + \frac{1}{30} = \frac{3}{30} = \frac{1}{10} = \frac{1}{f}$ . ✔

Worked example Ex 7 — Cell G: convex lens as a magnifier (object inside f)

Ek stamp ek magnifying lens ke $f = + 10$ cm ke saath 6 cm left mein rakha hai.

Forecast: Converging lens ke focus ke andar object → "magnifying glass" mode: virtual, erect, enlarged, image object ke same side par, yaani figure ke blue $v < 0$ zone mein wapas.

Signs. $u = - 6$ cm, $f = + 10$ cm. Yeh step kyun? Object left par → $u < 0$ ; lens convex hai → $f > 0$ . Object ab $F$ aur lens ke beech baitha hai.
Substitute. $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} + \frac{1}{- 6} = \frac{1}{10} - \frac{1}{6}$ . Yeh step kyun? Same solved-for- $v$ lens form $\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$ ; kyunki $∣ u ∣ < f$ hai toh doosra term dominate karta hai, toh sign flip ke liye watch karo.
Compute. Denominator 30: $\frac{3}{30} - \frac{5}{30} = - \frac{2}{30} = - \frac{1}{15}$ , toh $v = - 15$ cm. Yeh step kyun? Merge aur invert karo; negative result batata hai ki image left (blue) zone mein jump kar gayi — virtual image.
Interpret. $v < 0$ ⇒ (lens rule) image $- x$ mein object ki side (left) par = virtual.
Magnification. $m = + \frac{v}{u} = \frac{- 15}{- 6} = + 2.5$ ⇒ erect, $2.5 \times$ enlarged. Yeh step kyun? Lens magnification rule erect aur enlarged confirm karta hai — exactly wahi jo magnifying glass karta hai.

Verify: Virtual + erect + enlarged precisely wahi hai jo magnifying glass karta hai. Plug back: $\frac{1}{- 15} - \frac{1}{- 6} = - \frac{2}{30} + \frac{5}{30} = \frac{3}{30} = \frac{1}{10} = \frac{1}{f}$ . ✔

Worked example Ex 8 — Cell H: convex lens, object AT the focus (degenerate)

Ek collimator ek convex lens $f = + 10$ cm ke saath use karta hai aur object exactly uske focus par rakha hai, toh $u = - 10$ cm.

Forecast: Yeh Ex 3 ka lens twin hai. Figure mein object arrow $F$ par hi slide ho jaata hai; guess hai ki emerging rays parallel hain aur image infinity tak chali jaati hai — Ex 6 (real, right) aur Ex 7 (virtual, left) ke beech ki boundary.

Signs. $u = - 10$ cm, $f = + 10$ cm. Yeh step kyun? Object left par → $u < 0$ ; convex lens → $f > 0$ ; yahaan unke magnitudes equal hain kyunki object $F$ par baitha hai.
Substitute. $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} + \frac{1}{- 10} = 0$ . Yeh step kyun? Solved-for- $v$ lens form; do equal-and-opposite fractions exactly cancel ho jaate hain, $1/ v$ zero ho jaata hai.
Interpret. $\frac{1}{v} = 0 \Rightarrow v = \pm \infty$ : transmitted rays parallel nikalti hain, image infinity par banta hai (figure ke right edge se bahar). Yeh step kyun? $1/ v \to 0$ phir se "no crossing point" signature hai — exactly aise hi ek collimator point source ko parallel beam mein badalta hai.

Verify: Reversibility: $F$ par source hone se converging lens se parallel beam nikalti hai, "parallel rays $F^{'}$ par focus hoti hain" ka inverse. $\frac{1}{u} = - \frac{1}{f}$ ⇒ $\frac{1}{f} + \frac{1}{u} = 0$ exactly. ✔ Lens degenerate case handled.

Worked example Ex 9 — Cell I: concave lens (hamesha virtual)

Ek concave lens ka $f = - 20$ cm hai. Ek object uske 30 cm left mein hai.

Forecast: Diverging lens → hamesha virtual, erect, diminished, object ki side par → $v < 0$ , $0 < m < 1$ .

Signs. $u = - 30$ cm, $f = - 20$ cm. Yeh step kyun? Object left par → $u < 0$ ; concave-lens focus upstream hai (lens ke left) → $f < 0$ .
Substitute. $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{- 20} + \frac{1}{- 30}$ . Yeh step kyun? Solved-for- $v$ lens form; dono terms negative hain, toh image guaranteed left par land karegi.
Compute. Denominator 60: $- \frac{3}{60} - \frac{2}{60} = - \frac{5}{60} = - \frac{1}{12}$ , toh $v = - 12$ cm. Yeh step kyun? Combine aur invert; negative $v$ object ki side par virtual image confirm karta hai.
Interpret. $v < 0$ ⇒ (lens rule) image $- x$ mein, object ke same side = virtual.
Magnification. $m = + \frac{v}{u} = \frac{- 12}{- 30} = + 0.4$ ⇒ erect, diminished. Yeh step kyun? Lens rule orientation (erect) aur shrinkage ( $0.4 \times$ ) fix karta hai.

Verify: $∣ v ∣ < ∣ u ∣$ aur $0 < m < 1$ — har diverging lens ka signature. Plug back: $\frac{1}{- 12} - \frac{1}{- 30} = - \frac{5}{60} + \frac{2}{60} = - \frac{3}{60} = - \frac{1}{20} = \frac{1}{f}$ . ✔

Word problem aur exam twist

Worked example Ex 10 — Cell J: real-world (car rear-view mirror)

Ek car ke convex rear-view mirror ka $f = + 2.0$ m hai. Ek vehicle 10 m peeche hai (yaani mirror ke saamne 10 m). Agar real vehicle 1.5 m tall hai, toh uski image kitni tall hai aur kya woh upright hai?

Forecast: Convex mirror → chhoti, upright virtual image ("objects are closer than they appear").

Signs mein translate karo. "10 m saamne" ⇒ $u = - 10$ m; convex ⇒ $f = + 2$ m; height $h = + 1.5$ m. Yeh step kyun? Mirror ke saamne object/left side hai → negative distance; convex focus peeche hai → positive.
$v$ nikalo. $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{2} - \frac{1}{- 10} = \frac{1}{2} + \frac{1}{10} = \frac{5}{10} + \frac{1}{10} = \frac{6}{10}$ , toh $v = + \frac{10}{6} \approx 1.667$ m. Yeh step kyun? Solved-for- $v$ mirror form; positive $v$ = mirror ke peeche image (virtual).
Magnification. $m = - \frac{v}{u} = - \frac{1.667}{- 10} = + 0.1667$ . Yeh step kyun? Magnification rule dono distances ko orientation + scale mein convert karta hai.
Image height. $h^{'} = m h = 0.1667 \times 1.5 = + 0.25$ m, yaani 25 cm, positive ⇒ upright. Yeh step kyun? $h^{'} = m h$ abstract magnification ko ek real measurable height mein convert karta hai.

Verify: Positive chhota $m$ , upright, diminished ⇒ real mirror warning label se match karta hai. Units: metres throughout, $h^{'}$ metres mein. ✔

Worked example Ex 11 — Cell K: exam twist, two thin lenses in a row

Do thin convex lenses $L_{1}$ ( $f_{1} = + 10$ cm) aur $L_{2}$ ( $f_{2} = + 20$ cm) 40 cm apart rakhe gaye hain. Ek object $L_{1}$ ke 15 cm left mein hai. Final image nikalo. (Dekho Combination of thin lenses.)

Forecast: $L_{1}$ ek intermediate image banata hai; woh image $L_{2}$ ke liye object ban jaata hai. Signs carefully chase karo — $L_{2}$ ke liye $u$ uske apne optical centre se measure hota hai.

Lens 1. $u_{1} = - 15$ , $f_{1} = + 10$ : $\frac{1}{v _{1}} = \frac{1}{f _{1}} + \frac{1}{u _{1}} = \frac{1}{10} + \frac{1}{- 15} = \frac{3}{30} - \frac{2}{30} = \frac{1}{30}$ , toh $v_{1} = + 30$ cm (real, $L_{1}$ ke 30 cm right mein). Yeh step kyun? Pehle $L_{1}$ alone ko solved-for- $v$ lens form se solve karo taaki intermediate image locate ho.
Lens 2 ke liye object. Intermediate image $L_{1}$ ke 30 cm right mein hai; lenses 40 cm apart hain, toh woh $L_{2}$ ke $40 - 30 = 10$ cm left mein hai ⇒ $u_{2} = - 10$ cm. Yeh step kyun? $L_{2}$ ke liye har distance uske apne centre se measure hota hai; abhi bhi $L_{2}$ ke left par, yeh normal real object ki tarah kaam karta hai → negative.
Lens 2. $f_{2} = + 20$ : $\frac{1}{v _{2}} = \frac{1}{f _{2}} + \frac{1}{u _{2}} = \frac{1}{20} + \frac{1}{- 10} = \frac{1}{20} - \frac{2}{20} = - \frac{1}{20}$ , toh $v_{2} = - 20$ cm. Yeh step kyun? Final image position $L_{2}$ se nikalne ke liye lens formula dobara apply karo.
Interpret. $v_{2} < 0$ ⇒ (lens rule) final image $L_{2}$ ke left mein 20 cm = virtual (for $L_{2}$ ).
Total magnification. $m = m_{1} m_{2} = \frac{v _{1}}{u _{1}} \cdot \frac{v _{2}}{u _{2}} = \frac{30}{- 15} \cdot \frac{- 20}{- 10} = (- 2) \times (2) = - 4$ ⇒ inverted, $4 \times$ . Yeh step kyun? Lens chain ke magnifications multiply hote hain; product ka sign final orientation deta hai.

Verify: $m_{1} = - 2$ (inverted), $m_{2} = + 2$ , product $- 4$ . Har stage $1/ f$ reproduce karta hai: stage 1 $\frac{1}{30} - \frac{1}{- 15} = \frac{1}{10}$ ; stage 2 $\frac{1}{- 20} - \frac{1}{- 10} = \frac{1}{20}$ . ✔

Recall Fast self-test: kaun sa cell?

Ek erect, magnified, virtual image ek single curved mirror se — kaun sa element aur object kahaan hai? ::: Concave mirror, object focus ke andar (Cell B). Single lens se real, diminished, inverted image — kaun sa lens aur kahaan? ::: Convex lens, object $2 f$ se aage (Cell F). Tumhare algebra se mirror ke liye $\frac{1}{v} = 0$ nikla — physically kya hua? ::: Object focus par baitha tha; image at infinity, rays parallel (Cell C). Convex lens ke liye $\frac{1}{v} = 0$ nikla — yeh kaun sa device hai? ::: Ek collimator: object $F$ par, parallel beam bahar, image at infinity (Cell H). Lens ke liye $v > 0$ aur mirror ke liye $v > 0$ — same physical meaning? ::: Nahi: $v > 0$ ka hamesha matlab "image $+ x$ mein" hai, lekin lens ke liye woh real (right, transmitted) hai jabki mirror ke liye virtual (peeche) hai.