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

2.5.2 · D3 · Physics › Optics › Mirrors — plane, concave, convex; mirror equation 1 - v + 1

Kuch bhi shuru karne se pehle, main tools ko plain words mein re-state karta hoon taaki kuch unexplained na rahe.

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

Neeche sab kuch bas yahi teen lines hain, alag-alag numbers ke saath.

Scenario matrix

Ek mirror problem ko teen cheezein choose karne ke roop mein socho: kaunsa mirror, object kahan hai, aur setup normal hai, degenerate hai, ya extreme hai. Neeche table har class of outcome enumerate karti hai. Har cell ek worked example se covered hai ("Ex" column). Yaad raho $C$ pole se signed distance $R = 2 f$ par hai.

#	Mirror	Object position	Image: real/virtual, erect/inverted, size	Ex
A0	Concave	Infinity par ( $u \to - \infty$ )	$F$ par real point image (limiting)	1b
A	Concave	$C$ se pare ( $∥ u ∥ > ∥ R ∥$ )	real, inverted, diminished	1
B	Concave	$C$ par ( $u = R$ )	real, inverted, same size (degenerate)	2
C	Concave	$C$ aur $F$ ke beech	real, inverted, magnified	3
D	Concave	$F$ par ( $u = f$ )	image at infinity (limiting)	4
E	Concave	$F$ ke andar ( $∥ u ∥ < ∥ f ∥$ )	virtual, erect, magnified	5
F	Convex	Kahin bhi real	hamesha virtual, erect, diminished	6
G	Plane	Kahin bhi ( $f \to \infty$ )	virtual, erect, same size (limiting $R \to \infty$ )	7
H	Real-world word problem	(convex car mirror)	field-of-view / safety-distance	8
I	Exam twist	image = object ka given multiple, $f$ find karo	two-case sign trap	9
J	Degenerate	object at pole ( $u = 0$ ) & virtual object	image at pole $m = 1$ ; virtual object case	10

Table ki spine concave mirror hai: jaise object door se (A) pole ke paas (E) chalti hai, image ka ek poora life-cycle hota hai. Woh walk is poore topic ki sabse zaroori picture hai — yahan hai (mirror ka saamna left hai, pole $P$ $x = 0$ par, aur object A se E tak slide karta hai):

Case A0 & A — Concave, object at infinity aur C se pare

Worked example Ex 1b (cell A0) — limiting case: object infinitely door

Concave $f = - 10 cm$ , object infinity par ( $u \to - \infty$ , jaise ek door ka tara). Image kahan hai?

Forecast: Infinitely door ki cheez se rays axis ke parallel aati hain. $F$ ki definition se hi, parallel rays $F$ par milti hain. Toh predict karo ki image exactly $F$ par hogi.

Limit lo: $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ . Jaise $u \to - \infty$ , $\frac{1}{u} \to 0$ , toh $\frac{1}{v} \to \frac{1}{f} \Rightarrow v \to f = - 10 cm$ . Yeh step kyun? $\frac{1}{u} \to 0$ bhejne se algebra "object infinitely door hai, incoming rays parallel hain" encode karta hai. Jo bachta hai, $\frac{1}{v} = \frac{1}{f}$ , kehta hai image focus par hai — exactly woh property jo $F$ ko define karti hai.
Magnification: $m = - \frac{v}{u} \to 0$ (finite $v$ divided by infinite $∣ u ∣$ ). Yeh step kyun? $m \to 0$ matlab image point mein shrink ho jaati hai — infinitely door object finite-height image nahi bana sakta. Isi tarah concave mirror telescope mein tare ki sharp point-image banata hai.

Verify: Bade $∣ u ∣ = 1 0^{6}$ ke saath: $\frac{1}{v} = \frac{1}{- 10} - \frac{1}{- 1 0 ^{6}}$ , deta hai $v \approx - 10.0001 cm \approx f$ . ✔

Worked example Ex 1 (cell A) — classic case

Concave mirror, $R = 20 cm$ , object $30 cm$ saamne. $v$ , $m$ find karo, aur image describe karo.

Forecast: Object $C$ se pare hai (kyunki $∣ R ∣ = 20 < 30$ ). Computing se pehle image type guess karo — s01 picture kehti hai A kahan land karta hai?

Inputs sign karo. $f = \frac{R}{2}$ use karo; concave ⇒ $R < 0$ , toh $R = - 20 cm$ aur $f = - 10 cm$ . Real object saamne ⇒ $u = - 30 cm$ . Yeh step kyun? Equation tab hi sahi hai jab signed numbers use karo, aur $f = R /2$ given radius ko focal length mein convert karta hai. Signs skip karo → bakwaas milega.
$v$ solve karo: $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{- 10} - \frac{1}{- 30} = - \frac{3}{30} + \frac{1}{30} = - \frac{2}{30}$ $v = - 15 cm$ Yeh step kyun? $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ ko rearrange karke ek unknown isolate hota hai.
Sign interpret karo: $v < 0$ ⇒ image saamne hai ⇒ real.
Magnification: $m = - \frac{v}{u} = - \frac{- 15}{- 30} = - 0.5$ . Negative ⇒ inverted; $∣ m ∣ < 1$ ⇒ diminished.

Verify: $\frac{1}{- 15} + \frac{1}{- 30} = - \frac{2}{30} - \frac{1}{30} = - \frac{3}{30} = - \frac{1}{10} = \frac{1}{f}$ . ✔ Aur $∣ v ∣ = 15$ , $F$ ( $10$ ) aur $C$ ( $20$ ) ke beech hai — exactly "real, inverted, diminished" cell A ne predict kiya tha.

Case B — Concave, object exactly C par (degenerate: same size)

Case C — Concave, object C aur F ke beech (real, magnified)

Case D — Concave, object F par (limiting: image at infinity)

Worked example Ex 4 (cell D) — woh limit jo searchlight banata hai

Concave $f = - 10 cm$ , object exactly focus par, $u = - 10 cm$ . Image kahan hai?

Forecast: Yeh "parallel rays $F$ par milti hain" ka reverse hai. Toh rays parallel jaani chahiye — image at infinity. Dekho algebra gracefully kaise break karta hai.

Solve karo: $\frac{1}{v} = \frac{1}{- 10} - \frac{1}{- 10} = 0 \Rightarrow v = \infty$ . Yeh step kyun? $\frac{1}{v} = 0$ algebra ka tarika hai yeh kehne ka "koi finite meeting point nahi" — reflected rays kabhi converge nahi hoti, parallel jaati hain. Yeh real (C) aur virtual (E) ke beech limiting case hai: exactly woh frontier jahan image sides change karti hai.
Magnification: $m = - \frac{v}{u} \to \infty$ (infinitely stretched, undefined size). Physically image "at infinity" hai. Yeh step kyun? Finite $u$ par infinite $v$ magnification ko blow up kar deta hai — parallel outgoing rays tabhi "milti" hain jab infinitely far away, toh koi finite-size image exist nahi karti. Yeh exactly case A0 ka reverse hai.

Verify: Jaise $u \to f^{-}$ (object just $F$ se pare, jaise $u = - 10 - ε$ ), $\frac{1}{v} = \frac{1}{- 10} - \frac{1}{- 10 - ε} \to 0^{-}$ , toh $v \to - \infty$ (real, saamne); jaise $u \to f^{+}$ (object just $F$ ke andar, jaise $u = - 10 + ε$ ), $\frac{1}{v} \to 0^{+}$ toh $v \to + \infty$ (virtual, peeche). Image genuinely infinity se hoke swing karti hai aur sides change karti hai. Isiliye torch mirror ke focus par bulb rakhne se parallel beam banti hai. ✔

Case E — Concave, object F ke andar (virtual, erect, magnified)

Case F — Convex mirror (hamesha virtual, hamesha shrunk)

Worked example Ex 6 (cell F) — prove karo ki yeh kabhi aur nahi kar sakta

Convex $f = + 15 cm$ , object $u = - 20 cm$ . $v$ , $m$ find karo, aur argue karo ki result forced hai.

Forecast: Convex ke liye $f > 0$ aur real object $u < 0$ deta hai. Kya $v$ kabhi negative (real) aa sakta hai? Computing se pehle no prove karne ki koshish karo.

Solve karo: $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{15} - \frac{1}{- 20} = \frac{1}{15} + \frac{1}{20} = \frac{4 + 3}{60} = \frac{7}{60}$ . $v = \frac{60}{7} \approx + 8.57 cm$ Yeh step kyun? $\frac{1}{f}$ (positive, convex) aur $- \frac{1}{u}$ (positive, kyunki $u < 0$ ) dono add hote hain, $\frac{1}{v} > 0$ force karte hain — yeh algebra humein interpret karne se pehle hi dikhata hai ki image mirror ke peeche honi hi chahiye.
Magnification: $m = - \frac{v}{u} = - \frac{60/7}{- 20} = \frac{60}{140} = \frac{3}{7} \approx + 0.43$ . Erect, diminished. Yeh step kyun? $0 < m < 1$ aur $m > 0$ matlab har convex-mirror image choti aur upright hoti hai.
Forced argument. Real object ko distance $d = ∣ u ∣ > 0$ par rakho, toh $u = - d$ aur $f > 0$ . Phir $\frac{1}{v} = \frac{1}{f} + \frac{1}{d} > 0 \Rightarrow v > 0 hamesha (image virtual, peeche).$ Size ke liye, $\frac{1}{v} = \frac{1}{f} + \frac{1}{d} > \frac{1}{d}$ , toh $v < d = ∣ u ∣$ , deta hai $∣ m ∣ = \frac{v}{∣ u ∣} < 1$ hamesha (diminished). $∣ f ∣$ ke baare mein koi claim nahi chahiye — bas $f > 0$ aur $u < 0$ .

Verify: $\frac{7}{60} + \frac{1}{- 20} = \frac{7}{60} - \frac{3}{60} = \frac{4}{60} = \frac{1}{15} = \frac{1}{f}$ . ✔ Exactly isiliye car side-mirror (convex) ek wide, shrunken duniya dikhata hai: woh kabhi badi scary real image nahi bana sakta.

Case G — Plane mirror as the $R \to \infty$ limit

Worked example Ex 7 (cell G) — plane mirror ek mirror hai jiska

$f = \infty$ hai Plane mirror ko ek spherical mirror manno jiska $R \to \infty$ hai, toh $f = \frac{R}{2} \to \infty$ . Object $u = - 25 cm$ . $v$ aur $m$ find karo.

Forecast: Flat mirrors tumhari image "utni hi door peeche" dikhate hain jitne tum saamne ho, same size, upright. Kya equation yahi reproduce karta hai?

Limit lo: $\frac{1}{f} \to 0$ , toh $\frac{1}{v} + \frac{1}{u} = 0 \Rightarrow \frac{1}{v} = - \frac{1}{u} \Rightarrow v = - u$ . Yeh step kyun? $\frac{1}{f} = 0$ set karna "no curvature" ( $R \to \infty$ ) ka mathematical meaning hai. Yeh mirror equation ko pole ke baare mein ek mirror-image mein collapse kar deta hai.
Plug karo: $v = - u = - (- 25) = + 25 cm$ . Positive ⇒ mirror ke peeche ⇒ virtual, aur $∣ v ∣ = 25 = ∣ u ∣$ ⇒ equal distance. Yeh step kyun? Actual object distance substitute karna general relation $v = - u$ ko concrete "25 cm peeche" fact mein turn karta hai, aur positive sign humein batata hai ki yeh equal distance par virtual image hai — plane mirror ki defining property.
Magnification: $m = - \frac{v}{u} = - \frac{25}{- 25} = + 1$ . Erect, same size. Yeh step kyun? $m = + 1$ "upright aur identical in size" ka numerical fingerprint hai, confirm karta hai ki plane-mirror behaviour general equation se seedha nikalta hai.

Verify: $\frac{1}{25} + \frac{1}{- 25} = 0 = \frac{1}{f}$ . ✔ Virtual, erect, same size, equidistant — textbook plane-mirror facts, general equation ke limit ke roop mein recover kiye gaye.

Case H — Real-world word problem (convex car mirror)

Worked example Ex 8 (cell H) — "objects are closer than they appear"

Car ke convex side mirror ka $R = 2.0 m$ hai. Ek motorcycle car ke $8.0 m$ peeche hai (yaani mirror ke saamne). Image position aur magnification/diminution find karo.

Forecast: Convex ⇒ image virtual, shrunken, aur mirror ke bahut paas. Kya $∣ m ∣$ tiny hoga?

Inputs sign karo. Convex ⇒ $R > 0$ , toh $f = \frac{R}{2} = + 1.0 m$ . Object saamne ⇒ $u = - 8.0 m$ . Yeh step kyun? Word problems signs ko words mein chhupate hain; unhe sahi se extract karna (aur $f = R /2$ use karna) aadhi battle hai.
Solve karo: $\frac{1}{v} = \frac{1}{1.0} - \frac{1}{- 8.0} = 1 + \frac{1}{8} = \frac{9}{8} \Rightarrow v = \frac{8}{9} \approx + 0.889 m$ . Yeh step kyun? Yeh mirror equation apna ek kaam kar raha hai — jaane hue $f$ aur $u$ se image position nikalna; positive result pehle se hi mirror ke paas virtual image flag kar raha hai.
Magnification: $m = - \frac{v}{u} = - \frac{8/9}{- 8} = \frac{1}{9} \approx + 0.111$ . Yeh step kyun? Real-world question ka jawab dene ke liye size chahiye. $m \approx 0.11$ matlab bike $\sim 9 \times$ choti dikhti hai — aankhein "chota" ko "door" padhti hain, isiliye etched warning kehti hai objects are closer than they appear.

Verify: $\frac{9}{8} + \frac{1}{- 8} = \frac{8}{8} = 1 = \frac{1}{f}$ . ✔ Virtual ( $v > 0$ ), heavily diminished — safe wide-angle view, cell H.

Case I — Exam twist: image object ka given multiple hai, f find karo

Worked example Ex 9 (cell I) — do answers, ek sign trap

Ek concave mirror ek real object jo usse $20 cm$ door rakha hai, uski 3 guni badi image banata hai. Focal length find karo. (Do valid setups hain — dono nikalo.)

Forecast: "3 guna" mein erect ya inverted nahi bola! Concave do tarike se magnified image de sakta hai: real inverted ( $m = - 3$ ) ya virtual erect ( $m = + 3$ ). Do focal lengths expect karo.

Magnification set up karo. $u = - 20 cm$ . $m = - \frac{v}{u}$ se $v = - m u$ milta hai. Yeh step kyun? Mirror equation touch karne se pehle size clue ko $v$ aur $u$ ke beech relation mein convert karte hain, kyunki mirror equation ko numeric $v$ chahiye.
Case I-a, real image ( $m = - 3$ ): $v = - (- 3) (- 20) = - 60 cm$ . Phir $\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{- 60} + \frac{1}{- 20} = - \frac{1}{60} - \frac{3}{60} = - \frac{4}{60} = - \frac{1}{15}$ $f = - 15 cm$ Yeh step kyun? Concave mirror se real magnified image inverted hoti hai, toh $m$ negative hai; iske $v$ ko equation mein plug karna pehla focal length deta hai. $f < 0$ concave confirm karta hai — problem ke consistent hai.
Case I-b, virtual image ( $m = + 3$ ): $v = - (+ 3) (- 20) = + 60 cm$ . Phir $\frac{1}{f} = \frac{1}{60} + \frac{1}{- 20} = \frac{1}{60} - \frac{3}{60} = - \frac{2}{60} = - \frac{1}{30}$ $f = - 30 cm$ Yeh step kyun? Virtual magnified image erect hoti hai, toh $m$ positive aur $v > 0$ hai; equation phir doosra focal length deta hai, phir $f < 0$ (concave), toh dono setups physically valid hain.

Verify: I-a: $\frac{1}{- 60} + \frac{1}{- 20} = - \frac{4}{60} = - \frac{1}{15} = \frac{1}{f}$ . ✔ I-b: $\frac{1}{60} + \frac{1}{- 20} = - \frac{2}{60} = - \frac{1}{30} = \frac{1}{f}$ . ✔ Trap yeh hai ki sirf ek jawab dena; $m$ ka sign kabhi pin down nahi hua tha.

Case J — Object at the pole aur virtual-object edge case

Worked example Ex 10 (cell J) — degenerate aur virtual-object cases

(J-a) Object at the pole, $u = 0$ . Mirror equation kya kehta hai?

Forecast: Agar object mirror surface par hi hai, toh uski image bhi whin honi chahiye — magnify karne ko kuch nahi.

Limit $u \to 0^{-}$ lo: $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ . Jaise $u \to 0^{-}$ , $\frac{1}{u} \to - \infty$ , toh $\frac{1}{v} \to + \infty$ , matlab $v \to 0$ . Yeh step kyun? Finite $\frac{1}{f}$ bade $\frac{1}{u}$ term se dab jaata hai, toh image object ke saath pole par collapse ho jaati hai. Yeh dikhata hai ki $u = 0$ ek genuine boundary hai, normal image-forming setup nahi.
Magnification: limit ke paas $\frac{1}{v} \approx - \frac{1}{u}$ hone se, $v \approx u$ , toh $m = - \frac{v}{u} \to + 1$ . Image pole par, same size. Yeh step kyun? Limit mein ratio work karna (blindly $- 0/0$ likhne ki bajay) dikhata hai ki size ratio $+ 1$ ki taraf tend karta hai — object aur image mirror par saath baithe hain, upright aur identical. Yeh matrix ka degenerate edge hai.

(J-b) Virtual object ( $u > 0$ ). Agar mirror ke peeche kisi point ki taraf ja rahi converging light pehle intercept ho jaaye, woh point ek virtual object ki tarah behave karta hai jiska $u > 0$ hai. Mirror equation phir bhi hold karta hai — bas positive $u$ feed karo. Example: concave $f = - 10$ , virtual object $u = + 30$ :

Solve karo: $\frac{1}{v} = \frac{1}{- 10} - \frac{1}{30} = - \frac{3}{30} - \frac{1}{30} = - \frac{4}{30} \Rightarrow v = - 7.5 cm$ (real, saamne). Yeh step kyun? Positive $u$ feed karna ek hi change hai; equation khud kabhi break nahi hoti, dikhata hai ki same formula converging-beam inputs ko multi-mirror systems se cover karta hai.

Verify: J-b: $\frac{1}{- 7.5} + \frac{1}{30} = - \frac{4}{30} + \frac{1}{30} = - \frac{3}{30} = - \frac{1}{10} = \frac{1}{f}$ . ✔ Equation kabhi break nahi hoti; bas positive $u$ ki interpretation change hoti hai. (Virtual objects multi-mirror/lens systems mein milte hain — Magnification & Image Formation dekho.)

Recall

Recall Kaunsa cell hai? (pehle forecast, phir check)

Concave, object at infinity → image kahan? ::: focus $F$ par ( $v = f$ ), point-size ( $m \to 0$ ) Concave, object $2 f$ par (yaani $C$ par) → same-size real inverted (cell B) ::: $u = v = R$ , $m = - 1$ Concave, object focus par → image at infinity (cell D) ::: $\frac{1}{v} = 0$ Convex, object kahin bhi real → cell F ::: hamesha virtual, erect, diminished Plane mirror as a limit → $f = ?$ ::: $f = R /2 \to \infty$ , toh $v = - u$ , $m = + 1$ Object exactly pole par ( $u = 0$ ) → image kahan, size? ::: pole par, $m = + 1$ (degenerate) "3× magnified" bina orientation ke diya → kitne answers? ::: do ( $m = \pm 3$ )

Yeh bhi dekho: Magnification & Image Formation, Spherical Aberration, Reflection of Light — Laws & Normal, aur transmission analogue ke liye Refraction & Lenses — Lens Maker's Equation.