Sign convention for mirrors and lenses: Exercises — Hinglish deep dive

2.5.3 · D4 · Physics › Optics › Sign convention for mirrors and lenses

Kuch aur karne se pehle, us origin se milo jahan se har measurement shuru hoti hai, aur un chaar letters se bhi jo is page ke har problem mein aate hain. Ye letters lengths nahi hain — ye ek axis par signed coordinates hain.

Ab shared picture. Har solution neeche isi ki taraf point karta hai, isliye ise ek baar dhyan se padho.

Figure — Sign convention for mirrors and lenses

Apni ungli se trace karo: beech mein kala dot pole (mirror) / optical centre (lens) hai — woh origin jahan se sab distances shuru hoti hain. Blue arrow incident light hai jo left→right ja rahi hai — yahi chosen direction $+ x$ hai. Orange arrow ek real object hai, upstream (left) baitha hua — isliye $u < 0$ hai har problem mein neeche. Red half-line negative (upstream) region hai; green half-line positive (downstream) region hai. Right side ke do vertical black arrows height rule dikhate hain: axis ke upar ka point $h > 0$ hai, neeche ka $h < 0$ — yahi fact analysis problems mein $m$ ka sign decide karta hai. Jab bhi koi solution "upstream ⇒ negative" ya "axis ke upar ⇒ $h > 0$ " kehta hai, woh seedha is figure ko padh raha hai.

Level 1 — Recognition

Ye sirf test karte hain: kya tum kisi quantity ko sahi sign de sakte ho? Abhi koi formula nahi. Hum deliberately charon elements plus ek refracting surface aur ek height sab cover karte hain, taaki koi bhi sign untested na rahe.

Recall Solution — L1·Q1

Har quantity ko upar di gayi shared figure se padho (sab distances pole se).

Object left par hai, upstream (red region) ⇒ $u = - 25 cm$ .
Concave mirror ka focus mirror ke saamne (left) hota hai ⇒ upstream ⇒ $f = - 12 cm$ .
Object upar point kar raha hai, axis ke upar ⇒ $h = + 3 cm$ .

Answer: $u = - 25 cm$ , $f = - 12 cm$ , $h = + 3 cm$ .

Recall Solution — L1·Q2

(a) Convex lens converging hoti hai: light through jaati hai aur rays second focus par downstream (right, green region) milti hain ⇒ $f = + 20 cm$ .
(b) Concave lens diverging hoti hai: rays spread hoti hain aur sirf upstream (left, red region) ek focus se aati huyi dikhti hain ⇒ $f = - 20 cm$ .

Answer: convex $f = + 20$ cm; concave $f = - 20$ cm.

Recall Solution — L1·Q3

$R$ sirf curvature ke centre $C$ ki position hai, usi axis se padhi gayi.

(a) $C$ downstream (right, green region) ⇒ $R > 0$ (jaise $+ 20$ cm).
(b) $C$ upstream (left, red region) ⇒ $R < 0$ (jaise $- 20$ cm).

Answer: (a) $R > 0$ ; (b) $R < 0$ .

Level 2 — Application

Ab signs ko $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ (mirror) ya $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ (lens) mein plug karo. Dekhna hai ki ye kahan se aate hain toh Mirror formula derivation aur Lens formula and lensmaker's equation dekho.

Recall Solution — L2·Q1

Step 1 (signs): $u = - 30$ , $f = - 10$ , $h = + 2$ cm. $u$ aur $f$ dono upstream (figure ka red region) ⇒ dono negative. Step 2 (formula): $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{- 10} - \frac{1}{- 30} = - \frac{1}{10} + \frac{1}{30} = - \frac{3}{30} + \frac{1}{30} = - \frac{2}{30} .$ Step 3: $v = - 15 cm$ . Step 4 (nature sign se): $v < 0$ ⇒ image left (saamne) hai ⇒ real. Dekho Real and virtual images. Step 5 (orientation): $m = - \frac{v}{u} = - \frac{- 15}{- 30} = - 0.5$ . $m < 0$ ⇒ inverted; $∣ m ∣ < 1$ ⇒ diminished. Image height $h^{'} = m h = (- 0.5) (+ 2) = - 1$ cm (axis ke neeche, inverted confirm hota hai). Answer: $v = - 15 cm$ , real, inverted, half-size ( $h^{'} = - 1$ cm), mirror ke saamne.

Recall Solution — L2·Q2

Step 1: $u = - 20$ , $f = + 10$ . Step 2 (lens = minus): $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} + \frac{1}{- 20} = \frac{2}{20} - \frac{1}{20} = \frac{1}{20} .$ Step 3: $v = + 20 cm$ . Nature: $v > 0$ ⇒ image right (green region) par ⇒ real (lens ke liye, real images downstream hoti hain). Object $2 f$ par ⇒ image $2 f$ par. ✔

Level 3 — Analysis

Yahan answer ka sign interpret karna padta hai, aur magnification $m$ (upar define kiya gaya $h^{'} / h$ ke roop mein) aata hai — dekho Magnification in optics. Hum deliberately yahan dono lens edge cases include karte hain: ek convex lens jisme object focus ke andar hai, aur ek concave lens — yahi do configurations lenses ko unki characteristic virtual images deti hain.

Recall Solution — L3·Q1

Step 1: $u = - 20$ , $f = + 15$ (convex focus mirror ke peeche hota hai, downstream, green region ⇒ positive), $h = + 4$ cm. Step 2: $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{15} - \frac{1}{- 20} = \frac{1}{15} + \frac{1}{20} = \frac{4}{60} + \frac{3}{60} = \frac{7}{60} .$ Step 3: $v = \frac{60}{7} = + 8.571 cm$ . Positive ⇒ mirror ke peeche ⇒ virtual. Step 4 (magnification, mirror): $m = - \frac{v}{u} = - \frac{60/7}{- 20} = + \frac{60}{140} = + 0.4286.$ Step 5 (image height): $h^{'} = m h = (+ 0.4286) (+ 4) = + 1.71 cm$ . Positive ⇒ axis ke upar ⇒ erect, aur 4 cm se chhota ⇒ diminished. Interpret: $m > 0$ ⇒ erect; $∣ m ∣ < 1$ ⇒ diminished — exactly wahi jo ek convex mirror hamesha produce karta hai. ✔ Answer: $v \approx + 8.57$ cm, virtual, erect, diminished, $h^{'} \approx + 1.71$ cm ( $m \approx + 0.43$ ).

Recall Solution — L3·Q2

Step 1: $u = - 8$ , $f = - 12$ (object focus ke andar hai), $h = + 2$ cm. Step 2: $\frac{1}{v} = \frac{1}{- 12} - \frac{1}{- 8} = - \frac{1}{12} + \frac{1}{8} = - \frac{2}{24} + \frac{3}{24} = \frac{1}{24} .$ Step 3: $v = + 24 cm$ . Positive ⇒ mirror ke peeche ⇒ virtual (concave mirror tab virtual image deta hai jab object $f$ ke andar ho — jaise paas se shaving mirror). Step 4: $m = - \frac{v}{u} = - \frac{24}{- 8} = + 3.$ Step 5 (image height): $h^{'} = m h = (+ 3) (+ 2) = + 6 cm$ . Positive ⇒ axis ke upar ⇒ erect, aur lamba ⇒ magnified. ✔ Answer: $v = + 24$ cm, virtual, erect, magnified $3 \times$ , $h^{'} = + 6$ cm.

Recall Solution — L3·Q3

Step 1: $u = - 6$ , $f = + 10$ , $h = + 2$ cm. Note karo $∣ u ∣ = 6 < f = 10$ : object focal length ke andar hai. Step 2 (lens = minus): $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} + \frac{1}{- 6} = \frac{3}{30} - \frac{5}{30} = - \frac{2}{30} = - \frac{1}{15} .$ Step 3: $v = - 15 cm$ . Negative ⇒ image left par, object ki same side ⇒ virtual (lens ke liye, virtual image upstream hoti hai). Step 4 (magnification, lens): $m = + \frac{v}{u} = \frac{- 15}{- 6} = + 2.5.$ Step 5 (image height): $h^{'} = m h = (+ 2.5) (+ 2) = + 5 cm$ . Positive ⇒ axis ke upar ⇒ erect, aur lamba ⇒ magnified. Interpret: exactly aisa hi magnifying glass kaam karta hai — $f$ ke andar object hone par virtual, erect, magnified image object ki side par milti hai. ✔ Answer: $v = - 15$ cm, virtual, erect, magnified $2.5 \times$ , $h^{'} = + 5$ cm.

Recall Solution — L3·Q4

Step 1: $u = - 30$ , $f = - 20$ (concave lens ⇒ focus upstream ⇒ negative), $h = + 3$ cm. Step 2 (lens = minus): $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{- 20} + \frac{1}{- 30} = - \frac{3}{60} - \frac{2}{60} = - \frac{5}{60} = - \frac{1}{12} .$ Step 3: $v = - 12 cm$ . Negative ⇒ image left (object ki same side) ⇒ virtual. Step 4 (magnification, lens): $m = + \frac{v}{u} = \frac{- 12}{- 30} = + 0.4.$ Step 5 (image height): $h^{'} = m h = (+ 0.4) (+ 3) = + 1.2 cm$ . Positive ⇒ erect, aur chhota ⇒ diminished. Interpret: ek diverging lens hamesha virtual, erect, diminished image deti hai, kisi bhi real object position ke liye — yahi iski pehchaan hai. ✔ Answer: $v = - 12$ cm, virtual, erect, diminished, $h^{'} = + 1.2$ cm ( $m = + 0.4$ ).

Level 4 — Synthesis

Multi-step: do elements chain karo, ya magnification se ulta solve karo. Dekho Combination of thin lenses. Pehle, ek idea jo Level-1 ne cover nahi kiya:

Figure yahi dikhati hai: converging rays (blue) ek point $P$ ki taraf right par ja rahi hain; second lens (green) $P$ se pehle baith gayi hai, isliye us lens ke liye, $P$ ek virtual object hai jiska $u > 0$ hai.

Recall Solution — L4·Q1

Idea: lens 1 se bani image, lens 2 ka object ban jaati hai. Dono ko ek ek karke handle karo, har ek ke saath lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ .

Lens 1: $u_{1} = - 30$ , $f_{1} = + 10$ . $\frac{1}{v _{1}} = \frac{1}{10} + \frac{1}{- 30} = \frac{3}{30} - \frac{1}{30} = \frac{2}{30} \Rightarrow v_{1} = + 15 cm$ (lens 1 ke 15 cm right par).

Transfer: lens 2, lens 1 ke 5 cm right par hai, isliye yeh image lens 2 ke $15 - 5 = 10$ cm right par banti — yaani iske aage. Rays abhi bhi usi taraf converge kar rahi hain jab lens 2 se milti hain, isliye yeh ek virtual object hai (upar di gayi definition): $u_{2} = + 10 cm$ .

Lens 2: $f_{2} = - 15$ . $\frac{1}{v _{2}} = \frac{1}{f _{2}} + \frac{1}{u _{2}} = \frac{1}{- 15} + \frac{1}{10} = - \frac{2}{30} + \frac{3}{30} = \frac{1}{30} \Rightarrow v_{2} = + 30 cm .$ Answer: final image concave lens ke 30 cm right par hai, real.

Recall Solution — L4·Q2

Step 1 (words ko signs mein translate karo): real inverted image ⇒ $m = - 0.5$ (inverted ⇒ $m < 0$ ; half size ⇒ $∣ m ∣ = 0.5$ ). Mirror ke liye $m = - \frac{v}{u}$ , toh $- \frac{v}{u} = - 0.5 \Rightarrow v = 0.5 u$ . Step 2 (formula): $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ with $f = - 20$ . $v = 0.5 u$ substitute karo: $\frac{1}{0.5 u} + \frac{1}{u} = \frac{2}{u} + \frac{1}{u} = \frac{3}{u} = \frac{1}{- 20} .$ Step 3: $u = 3 \times (- 20) = - 60 cm$ . Answer: object mirror ke 60 cm saamne hai. (Check karo: $v = 0.5 u = - 30$ cm, real, saamne. ✔)

Level 5 — Mastery

Recall Solution — L5·Q1

Step 1 (signs): object focus par, object side par baitha hai, isliye $u = - 18$ ; concave ⇒ $f = - 18$ . Yahan $u = f$ hai. Step 2: $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{- 18} - \frac{1}{- 18} = 0.$ Step 3: $\frac{1}{v} = 0 \Rightarrow v \to \infty$ . Interpret: jab object focus par hota hai, reflected rays parallel nikalti hain, isliye kabhi milti nahi — image infinity par hai (yahi searchlight/torch principle hai ulta chala ke). Ek degenerate lekin bilkul consistent case: jab bhi $u = f$ hoga, formula force karta hai $v \to \infty$ . ✔ Answer: $v \to \infty$ (reflected rays parallel, koi finite image nahi).

Recall Solution — L5·Q2

Step 1 (signs): $u = - 40$ , $R = + 20$ (curvature ka centre downstream, right, green region — exactly L1·Q3 case a), $n_{1} = 1, n_{2} = 1.5$ . Step 2: $\frac{n _{2}}{v} = \frac{n _{2} - n _{1}}{R} + \frac{n _{1}}{u} = \frac{1.5 - 1}{20} + \frac{1}{- 40} = \frac{0.5}{20} - \frac{1}{40} = \frac{1}{40} - \frac{1}{40} = 0.$ Step 3: $\frac{1.5}{v} = 0 \Rightarrow v = \infty$ . Interpret: object exactly surface ke first focal point par hai, isliye refracted rays parallel nikalti hain — image infinity par hai. Ek degenerate lekin bilkul consistent case, L5·Q1 jaisa. ✔ Answer: $v \to \infty$ (rays parallel nikalti hain).

Recall Solution — L5·Q3

Step 1: $u = - 30$ , $R = + 20$ , $n_{1} = 1$ , $n_{2} = 1.5$ . Step 2: $\frac{1.5}{v} = \frac{0.5}{20} + \frac{1}{- 30} = \frac{1}{40} - \frac{1}{30} .$ Common denominator 120: $\frac{3}{120} - \frac{4}{120} = - \frac{1}{120} .$ Step 3: $\frac{1.5}{v} = - \frac{1}{120} \Rightarrow v = 1.5 \times (- 120) = - 180 cm .$ Interpret: $v < 0$ ⇒ image object ki same side (left, incoming hawa mein) ⇒ virtual. Kyunki object ab first focal length ke andar hai, refracted rays diverge karti hain aur sirf ek point se 180 cm peeche se aati huyi dikhti hain. ✔ Answer: $v = - 180$ cm, virtual, object ki side par.

Recall check

Recall Origin kahan hai (jahan se

$u, v, f$ measure hote hain)? Mirror ke liye pole (reflecting surface ka centre) ya lens ke liye optical centre (beech ka woh point jisse ray bina deviate hue guzarti hai). Dono $x = 0$ par rakhte hain.

Recall Object distance

$u$ — normal real object ke liye sign? $u < 0$ . Real object upstream (left, figure ka red region) hota hai, incident-light direction ke ulta, isliye iska coordinate negative hota hai.

Recall Focal length

$f$ — charon elements ke liye signs? Concave mirror $f < 0$ (focus saamne, left). Convex mirror $f > 0$ (focus peeche, right). Convex lens $f > 0$ (rays downstream converge hoti hain, right). Concave lens $f < 0$ (rays upstream se aati huyi dikhti hain, left).

Recall Convex lens, object focus ke andar (

$∣ u ∣ < f$ ) — kaisi image? Virtual, erect, magnified, object ki side par ( $v < 0$ , $m > 0$ , $∣ m ∣ > 1$ ) — magnifying-glass case.

Recall Concave (diverging) lens — kisi bhi real object ke liye kaisi image?

Hamesha virtual, erect, diminished, object ki side par ( $v < 0$ , $0 < m < 1$ ).

Recall Image distance

$v$ — iska sign image ke baare mein kya batata hai? Mirror: $v < 0$ ⇒ real (saamne, left); $v > 0$ ⇒ virtual (peeche). Lens/surface: $v > 0$ ⇒ real (downstream, right); $v < 0$ ⇒ virtual (object ki same side).

Recall Magnification

$m$ aur image height $h^{'}$ — kya batate hain? $m = \frac{h ^{'}}{h}$ ; mirror $m = - \frac{v}{u}$ , lens $m = + \frac{v}{u}$ . Phir $h^{'} = m h$ . $m > 0$ ⇒ erect ( $h^{'}$ axis ke upar); $m < 0$ ⇒ inverted ( $h^{'}$ axis ke neeche); $∣ m ∣ > 1$ ⇒ magnified.

Recall Kya hota hai jab

$u = f$ (mirror) ya object surface ke focus par ho? $\frac{1}{v} \to 0$ , isliye $v \to \infty$ : outgoing rays parallel ho jaati hain aur koi finite image nahi banti.

Recall Refracting surface ka radius

$R$ — sign rule? $R > 0$ agar curvature ka centre downstream (right) ho; $R < 0$ agar upstream (left) ho. $u, v, f$ jaisa hi rule.

Recall Virtual object — kya hota hai aur kaunsa sign?

Ek downstream point ( $u > 0$ ) jissi taraf incoming rays already converge kar rahi hoti hain jab element unhe intercept karta hai. Tab hota hai jab chain ke 2nd lens par, jab 1st image us se aage banti.

Parent: Sign convention for mirrors and lenses.