Sign convention for mirrors and lenses: Question bank

2.5.3 · D5 · Physics › Optics › Sign convention for mirrors and lenses

Definition Do reference points, phir paanch signed quantities

Sab kuch principal axis ke saath measure kiya jaata hai (wo seedhi line jo mirror/lens ke centre se hokar guzarti hai, uski surface ke perpendicular). Pehle fix karo kahan se measure karte ho:

Pole $P$ — mirror ke liye, woh point jahan principal axis reflecting surface se milti hai (mirror ka dead centre).
Optical centre $O$ — thin lens ke liye, lens ka middle point jahan se ray seedha bina deviate hue guzarti hai.

$P$ (mirrors) ya $O$ (lenses) origin hai, koi distance nahi. Us origin se hum paanch signed quantities padhte hain:

$u$ = object distance — $P / O$ se object ka coordinate.
$v$ = image distance — $P / O$ se image ka coordinate.
$f$ = focal length — $P / O$ se focus ka coordinate (jahan parallel rays milti hain).
$R$ = radius of curvature — $P$ se centre of curvature $C$ ka coordinate (us sphere ka centre jis se mirror kaat ke nikala gaya hai); aur mirror ke liye $f = R /2$ .
$m$ = magnification = $\frac{h ^{'}}{h}$ = (signed image height) ÷ (signed object height) — ek pure ratio jiska sign orientation batata hai.

Figure — Sign convention for mirrors and lenses

Figure s01 — master picture: origin $P / O$ par, orange arrow = incident-light direction (hamara $+ x$ ), magenta region upstream $-$ hai, violet region downstream $+$ hai. Is page par har sign bas "yeh point kis coloured region mein hai?" ka jawab hai.

Figure s02 — MIRROR. Reflection light ko wapas bhejta hai, isliye real object (left) aur uski real image (bhi left) $P$ ke same side par hoti hain. Paraxial similar triangles magnitude law dete hain $\frac{1}{a} + \frac{1}{b} = \frac{1}{F}$ . Ab sign karo: object left $\Rightarrow u = - a$ ; real image left $\Rightarrow v = - b$ ; concave focus left $\Rightarrow f = - F$ . $a = - u, b = - v, F = - f$ substitute karo $\frac{1}{a} + \frac{1}{b} = \frac{1}{F}$ mein aur har reciprocal same $- 1$ pick up karta hai, jo cancel ho jaata hai, aur milta hai $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ — ek plus, kyunki dono points ek side share karte hain.

Figure s03 — LENS. Light paar ho jaati hai, isliye real object (left) aur uski real image (right) $O$ ke opposite sides par hoti hain. Magnitude law abhi bhi $\frac{1}{a} + \frac{1}{b} = \frac{1}{F}$ hai. Sign karo: object left $\Rightarrow u = - a$ isliye $a = - u$ ; real image right $\Rightarrow v = + b$ isliye $b = + v$ ; converging focus right $\Rightarrow f = + F$ isliye $F = + f$ . Substitute karo: $\frac{1}{- u} + \frac{1}{v} = \frac{1}{f}$ , yaani $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ — ek minus, jo purely image aur object ke opposite sides par hone ki wajah se forced hai.

True or false — justify karo

Ek concave mirror ka focal length hamesha negative hota hai.

True. Uska focus incident-light side (left, mirror ke saamne) par hota hai, jo light direction ke against hai, isliye

f < 0

hamesha, chahe object kahan bhi rakha ho.

Ek convex mirror aur ek convex lens dono converging hain, isliye inka focal-length sign same hoga.

Premise aur conclusion dono galat hain. Convex mirror diverge karta hai; convex lens converge karta hai. Inका sign

f > 0

zaroor same hota hai, lekin unrelated geometric reasons ki wajah se (mirror focus peeche = downstream; lens focus right = downstream), "converging" ki wajah se nahi.

Real object ke liye object distance

u

hamesha negative hota hai.

True. Real object pole/optical centre ke upstream (left) par hota hai, incident-light direction ke opposite, isliye uska coordinate

u < 0

har baar.

Positive image distance ka matlab hamesha real image hota hai.

False. Lens ke liye

v > 0

ka matlab real hai (image downstream, right par). Mirror ke liye

v > 0

ka matlab virtual hai (image mirror ke peeche). Same sign alag padha jaata hai kyunki mirror light ko peeche bhejta hai.

Magnification

m > 0

guarantee karta hai ki image virtual hai.

False.

m > 0

ka matlab sirf yeh hai ki image erect hai (image height aur object height ka sign same). Erect images zyaadatar virtual hoti hain, lekin

m

ka sign orientation report karta hai, reality nahi — reality ke liye

v

ka sign check karo.

Ek concave mirror ki radius of curvature

R

negative hoti hai.

True. Centre of curvature object ke same side (front, left) par hota hai, light ke upstream, isliye

R < 0

, jo

f = R /2 < 0

se consistent hai.

Jis side object hai woh swap karne se saare standard signs flip ho jaayenge.

True. Poora convention "incident light left→right travel karti hai" se anchor hai. Agar tumne (unusually) light right→left bheji, toh upar ke har

+

aur

-

invert ho jaayenge. Signs chosen light direction ke relative hain, absolute nahi.

Ek plane mirror ka

f = 0

hota hai.

False. Plane mirror ki radius of curvature infinite hoti hai, isliye

f = R /2 = \infty

, zero nahi.

f \to \infty

\frac{1}{v} + \frac{1}{u} = \frac{1}{f} = 0

mein plug karo toh

v = - u

milta hai: image utni hi door peeche hai jitna object saamne hai.

Error dhundho

"Object concave mirror se 30 cm door hai, toh main

u = + 30

cm use karunga."

Galat sign. Object pole ke upstream hai →

u = - 30

cm.

u

ko plain positive length treat karna ek sabse common convention slip hai;

u

ek coordinate hai, distance nahi.

"Convex lens hai, toh main

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

likhta hoon."

Galat equation. Lenses mein minus hota hai:

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

. Reason figure s03 mein derive hai: object left (

u = - a

) aur real image right (

v = + b

)

O

ke opposite sides par hain, isliye magnitude law

\frac{1}{a} + \frac{1}{b} = \frac{1}{F}

mein substitute karne par

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

milta hai. Mirror plus rakhta hai (figure s02) kyunki reflection object aur image ko same side par rakhta hai.

"Mujhe lens ke liye

v = - 15

cm mila, toh image virtual hai aur object ki side par hai — lekin main ise real report karunga kyunki lenses real images banate hain."

Report sign ko contradict karti hai. Lens ke liye

v < 0

genuinely virtual image ka matlab hai jo object ki same side par hai (jaise converging lens ke focus ke andar object). Sign ko memorised habit se kabhi override mat karo.

"Concave mirror strongly converge karta hai, toh uski focal length positive honi chahiye."

"Strength/converging = positive" wali instinct trap hai. Sign geometry se set hoti hai ki focus kahan hai, rays converge hoti hain ya nahi isse nahi. Concave mirror focus upstream hai ⇒

f < 0

"Convex mirror, real object, toh main expect karunga

v < 0

jaise usual."

Convex mirror hamesha apne peeche virtual image banata hai ⇒

v > 0

. "Usual" front-image sign expect karna yeh bhool jaata hai ki yeh mirror real object ki kabhi real image nahi bana sakta.

"Magnification ke liye maine mirror ke liye

m = + v / u

use kiya."

Mirror ke liye galat sign. Mirrors mein

m = - v / u

hota hai; lenses mein

m = + v / u

. Mirror ke liye extra minus pole par similar triangles ki reflection geometry se aata hai.

Why questions

Ek single signed equation case-by-case formulas ka poora dhera kyun replace kar deta hai?

Kyunki

u

v

f

ke signs encode karte hain ki har object/image/focus kis side par hai. Jab side-information sign mein aa jaata hai, ek algebraic relation concave/convex, real/virtual, inside/outside focus sab cover kar leta hai — koi re-drawing nahi chahiye.

Mirror formula "

+

" kyun hoti hai lekin lens formula "

-

Mirror reflect karta hai, isliye object aur image same side par hote hain (figure s02) aur magnitude law ko sign karne par plus bachta hai. Lens transmit karta hai, isliye image aur object opposite sides par hote hain (figure s03), jo minus force karta hai. Connecting sign same-side vs opposite-side geometry ka direct fingerprint hai.

Answer

v

ka sign image ki nature free mein kyun bata deta hai?

Kyunki

v

ek coordinate hai: uska sign bataata hai ki image surface ke kis side par land karti hai, aur har side ek nature hai (front/real vs behind/virtual mirrors ke liye; opposite/real vs same-side/virtual lenses ke liye). Algebra image ko place karta hai; sign use label karta hai.

Mirror formula

u

aur

v

mein symmetric kyun honi chahiye?

Kyunki light paths reversible hote hain — object aur image ko trade karo aur ray diagram unchanged rahta hai. Formula

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

precisely usi physical reversibility ko honour karne ke liye symmetric hai.

Convex mirror ka focus positive kyun liya jaata hai jab uska object abhi bhi

u < 0

deta hai?

Object upstream (left) par hai →

u < 0

, lekin convex mirror ka focus mirror ke peeche hai, jo reflected light ke downstream hai →

f > 0

. Do alag points do alag sides par, isliye opposite signs.

Sirf "converging" ya "diverging" shabd se

f

ka sign decide kyun nahi kar sakte?

Kyunki "converging" rays ke behavior ko describe karta hai, focus ki location ko light ke relative nahi. Sirf focus ki side (upstream/downstream) sign fix karti hai, aur woh side mirrors aur lenses ke beech alag hoti hai.

Edge cases

Concave mirror ke focus par exactly rakha object — formula kya kehta hai?

u = f

set karo: phir

\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = 0

, isliye

v \to \infty

. Reflected rays parallel niklaati hain aur sirf infinity par milti hain — koi finite image nahi banti.

Concave mirror mein object focal length ke andar (

∣ u ∣ < ∣ f ∣

) — real ya virtual image?

Virtual. Lo

u = - 6

cm,

f = - 10

cm:

\frac{1}{v} = \frac{1}{- 10} - \frac{1}{- 6} = + \frac{1}{15}

, isliye

v = + 15

> 0

— mirror ke liye positive

v

ka matlab image mirror ke peeche hai, yaani virtual, erect, magnified. Yeh shaving/make-up-mirror regime hai, aur yeh dikhata hai ki concave mirror hamesha real image nahi deta.

Diverging (concave) lens mein real object — woh hamesha kaisi image banata hai?

Hamesha virtual, erect, reduced, object ki side par. Lo

u = - 20

cm,

f = - 10

cm:

\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{- 10} + \frac{1}{- 20} = - \frac{3}{20}

, isliye

v = - \frac{20}{3} \approx - 6.67

< 0

(object ki same side ⇒ virtual), aur

m = v / u = (- 20/3) / (- 20) = + \frac{1}{3} > 0

(erect) aur

< 1

(reduced). Concave lens real object ki kabhi real image nahi bana sakta.

Concave mirror ke centre of curvature par object (

u = R = 2 f

) —

m

ke sign ka kya hota hai?

v = u = 2 f

, isliye

m = - v / u = - 1

: image real hai, same size, aur inverted (negative

m

), object ki side par

C

par wapas.

Infinity par object (parallel incident rays) — mirror aur lens ke liye image kahan banti hai?

Jaise

u \to - \infty

\frac{1}{u} \to 0

, isliye formula collapse ho ke

\frac{1}{v} = \frac{1}{f}

ban jaata hai, yaani

v = f

. Image exactly focus par banti hai: concave mirror ke liye

f < 0

par (real, saamne), convex lens ke liye

f > 0

par (real, far side par). Yahi focal point ki definition hai.

Virtual object (ek lens par converging light jo focus hone se pehle hit karti hai) — uska sign kaise alag hota hai?

Virtual object surface ke downstream hota hai (right par), light direction ke saath, isliye

u > 0

— ordinary real object ke opposite sign. Yeh Combination of thin lenses mein aata hai jahan ek lens converging rays doosre ko deta hai.

Jaise concave mirror plane ki taraf flatten hota hai (

R \to \infty

f

aur

v

ka kya hota hai?

f = R /2 \to - \infty

, isliye

\frac{1}{f} \to 0

aur

\frac{1}{v} + \frac{1}{u} = 0

v = - u

milta hai: image plane-mirror result ki taraf approach karti hai — utni hi door peeche, virtual, erect, same size.

Convex lens mein object

2 f

par — answer sign kyun is baare mein neutral hai ki kaunsa "

2 f

u = - 2 f

aur

f > 0

ke saath,

v = + 2 f

milta hai: far side par real image, inverted, unit magnification. Do

2 f

points ki symmetry formula ke structure ka direct consequence hai, koi coincidence nahi.

Object mirror ke bilkul paas rakha ho (

u \to 0^{-}

) — image kahan hai, proper limit mein?

Rearrange karo

v = \frac{f u}{u - f}

. Limit

u \to 0

lete hain: numerator

f u \to 0

jabki denominator

u - f \to - f \neq = 0

, isliye

v \to \frac{0}{- f} = 0

. Image surface par hai

m = - v / u \to 1

ke saath (unit, erect) — object aur image pole par coincide karte hain. Yeh ek idealised limit hai; thin-element/paraxial model itna close reality describe karna chhod deta hai.

Recall

Recall Memorisation se aage — har ek ko reason karo

Sirf $u < 0$ aur $f < 0$ diya ho, bina diagram ke, decide karo ki concave-mirror image kabhi virtual ho sakti hai. ::: Haan — jab $∣ u ∣ < ∣ f ∣$ ho toh algebra $v > 0$ deta hai, jo mirror ke liye ek virtual image hai uske peeche; sign convention shaving-mirror case ko bina drawing ke predict karta hai. $v = f$ kyun pop out hota hai jab bhi object infinity par ho, mirror aur lens dono ke liye? ::: Kyunki $\frac{1}{u} \to 0$ object term ko khatam kar deta hai, $\frac{1}{v} = \frac{1}{f}$ bachta hai; "parallel rays focus par milti hain" literally yahi $f$ ko define karta hai. Ek student lens ke liye $m = + 2$ aur $v > 0$ report karta hai — consistent hai? ::: Nahi. $m > 0$ erect maangta hai (single lens ke liye virtual), lekin lens ke liye $v > 0$ ka matlab real hai (inverted, $m < 0$ ). Dono claims contradict karte hain; ek sign galat hai. Tum mirror-vs-lens ko sirf bare formula se bina labels ke kaise identify kar sakte ho? ::: Connecting sign se: $1/ v$ aur $1/ u$ terms ke beech " $+$ " mirror hai (same-side reflection); " $-$ " lens hai (opposite-side transmission).

Cross-links: Mirror formula derivation, Lens formula and lensmaker's equation, Magnification in optics, Real and virtual images, Refraction at spherical surfaces.