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

2.5.2 · D2 · Physics › Optics › Mirrors — plane, concave, convex; mirror equation 1 - v + 1

Hum sirf do facts use karte hain: light is tarah bounce karti hai ki ==angle in = angle out (dekho Reflection of Light — Laws & Normal), aur do triangles jinke same angles hon unki sides proportional hoti hain== (similar triangles). Neeche sab kuch inhi do facts ka baar baar istemal hai.

Step 0 — Woh stage jis par hum draw karte hain

KYA. Koi bhi ray aane se pehle, hum ek concave mirror ka fixed furniture rakh dete hain.

KYUN. Final formula mein har distance ek khaas point (the pole) se measure hoti hai. Agar hum woh origin aur "positive" direction pin down nahi karte, toh signs bakwaas ban jayenge.

PICTURE. Figure dekho. Curved arc mirror hai. Uske through seedhi line principal axis hai — hamari number line. Jahan axis mirror se milti hai woh point hai pole $P$ (hamara origin, "zero"). Dot $C$ hai centre of curvature — us sphere ka centre jiska yeh mirror ek tukda hai. Unke beech mein baitha hai $F$ , yani focus.

Incident light left → right travel karti hai. $P$ ke left mein jo kuch bhi hai (light ke against) woh negative count hota hai; right mein, positive. Height upar positive hai. Concave mirror ke liye $F$ aur $C$ , $P$ ke left mein hote hain, toh signed coordinates mein $f < 0$ aur $R < 0$ — bilkul parent note ka convention. Unki ruler lengths $∣ f ∣, R$ zaroor positive hain; dono ko alag rakkho.

Step 1 — Ek parallel ray, aur woh normal jo use control karta hai

KYA. Ek akela ray bilkul seedha, axis ke parallel, mirror par point $M$ par strike karate hue bhejo — thoda sa $P$ ke upar.

KYUN. Axis ke parallel ray hi definition hai us cheez ki jo focus banati hai. Use bounce karne ke liye hume $M$ par normal chahiye (woh line jo "surface se seedha bahar" ki direction bataye). Sphere ke liye woh normal easy hai: woh $C$ ki taraf point karta hai, kyunki radius hamesha sphere se right angle par milti hai.

PICTURE. Blue ray flat aati hai. $M$ par orange line normal $M C$ hai. Reflected green ray $M$ se nikalti hai aur axis par $F$ par cross karti hai.

Step 2 — Focus exactly (almost) halfway kyun baitha hai ( $f = R /2$ )

KYA. Triangle $M F C$ mein teen angles $θ$ track karo aur prove karo ki $F$ , $P C$ ka midpoint hai — near-axis limit mein.

KYUN. $R$ ke terms mein $F$ kahan hai yeh jaanna zaroori hai, $f$ ke baare mein kuch bhi bolne se pehle. Yeh pehla number hai jo mirror hume free mein deta hai.

PICTURE. Incoming blue ray axis ke parallel hai, aur $M C$ ek transversal hai jo do parallel lines (ray aur axis) ko cut karta hai. Toh $C$ par angle, $∠ M C F$ , $θ$ ke barabar hai (alternate angles). Bounce se hume pehle se $∠ F M C = θ$ mila hai. Do equal base angles wala triangle isosceles hota hai, toh unke opposite sides equal hote hain: $∣ F M ∣ = ∣ F C ∣$ .

Ab is page ki ek aur akeli approximation — paraxial (near-axis, small-angle) limit:

Woh limit lete hue, taaki $∣ F M ∣ \to ∣ F P ∣$ , aur $∣ F M ∣ = ∣ F C ∣$ ke saath combine karke:

$= ∣ f ∣ ∣ F P ∣ = = R - ∣ f ∣ ∣ F C ∣ ⟹ ∣ f ∣ = R - ∣ f ∣ ⟹ ∣ f ∣ = \frac{R}{2}$

Yahan $∣ F P ∣$ ruler focal length $∣ f ∣$ hai; $∣ F C ∣$ woh hai jo radius mein $∣ f ∣$ ke baad bachta hai, yani $R - ∣ f ∣$ (kyunki $F$ , $P$ aur $C$ ke beech mein hai). Solve karne par halfway result milta hai. Step 0 ke signs restore karne par (concave ke liye dono $f$ aur $R$ negative) neat $f = R /2$ milta hai.

Step 3 — Ek real object rakho; woh do rays dhundo jo uski image banate hain

KYA. Axis par, $C$ se aage, height $h$ ka ek arrow khado karo. Do rays draw karo jinke milne ka point image tip locate karta hai.

KYUN. Ek ray ek direction batati hai, lekin ek point ke liye do lines cross honi chahiye. Hum woh do rays chunte hain jinki geometry hum pehle se samajh chuke hain: parallel ray (Step 1) aur seedha pole par jaane wali ray.

PICTURE. Red arrow = object, height $h$ , $P$ se $∣ u ∣$ left par. Green ray uski tip se parallel jaati hai, $F$ se hoke bounce karti hai. Blue ray uski tip se pole $P$ par jaati hai aur axis se bounce karti hai. Jahan wo cross karte hain, image tip banti hai: $∣ v ∣$ distance par height $h^{'}$ ka ek ulta arrow.

Step 4 — Pole ray magnification deta hai $m = - v / u$

KYA. Sirf blue pole ray aur woh do right triangles zoom in karo jo woh axis ke saath banati hai.

KYUN. Pole ray axis ko apna normal manke reflect karti hai, toh woh axis ke upar aur neeche equal angles banati hai. Equal angles ⇒ object triangle aur image triangle similar hain ⇒ unki sides same ratio mein hain. Wahi ratio "magnification" ka matlab hai.

PICTURE. Left triangle: object height $∣ h ∣$ over base $∣ u ∣$ . Right triangle: image height $∣ h^{'} ∣$ over base $∣ v ∣$ . $P$ par shared vertex angle dono sides par equal hai (law of reflection), toh triangles similar hain.

Ab signed coordinates mein translate karo. Step 0 ke convention mein image neeche point karti hai ( $h^{'} < 0$ ) jabki object upar hai ( $h > 0$ ), aur dono $u, v < 0$ . $∣ h^{'} ∣ = - h^{'}$ , $∣ v ∣ = - v$ , $∣ u ∣ = - u$ substitute karke:

$\frac{- h ^{'}}{h} = \frac{- v}{- u} ⟹ \frac{h ^{'}}{h} = - \frac{v}{u}$

$m = \frac{h ^{'}}{h} = - \frac{v}{u}$

Minus sign decoration nahi hai — yeh geometry hai jo bata rahi hai ki real image flip ho jaati hai.

Step 5 — Parallel ray ek doosra ratio deta hai

KYA. Ab green parallel ray use karo. Do aur similar triangles dikhte hain, dono focus $F$ share karte hain.

KYUN. Hamare paas ek equation hai jo $h^{'} / h$ ko distances se link karti hai. Heights eliminate karke pure $u, v, f$ relation paane ke liye hume ek doosri, independent equation chahiye. Parallel-through- $F$ ray woh deti hai.

PICTURE — mirror-height $h$ ke barabar kyun hai. Green ray object tip se exactly horizontally nikalti hai (axis ke parallel). Ek horizontal line har $x$ par same height rakhti hai. Toh jab woh mirror tak pahunchti hai bilkul $P$ ke paas (jo paraxial limit mein essentially axis par hai), uski height wahan bhi object ki height $∣ h ∣$ hi hoti hai. Figure mein dekho green segment object tip se mirror tak bilkul flat hai — dono ends par equal heights bas "ek horizontal line level hoti hai" hai. Woh level segment ek leg hai; mirror se neeche $F$ se hote hue image tip tak drop doosra leg hai.

Toh $P$ ke paas mirror height $∣ h ∣$ carry karta hai. Wahaan se ray seedha $F$ se hoke image tip tak jaati hai height $∣ h^{'} ∣$ par. Isse do right triangles bante hain jo $F$ par vertex angle share karte hain:

Triangle A — base $∣ f ∣$ ( $P$ se $F$ tak axis par), upright side $∣ h ∣$ (mirror height).
Triangle B — base $∣ v ∣ - ∣ f ∣$ ( $F$ se image foot tak), upright side $∣ h^{'} ∣$ (image height).

Same apex angle at $F$ ⇒ similar ⇒ matching sides proportion mein.

Ab signed coordinates mein translate karo Step 0 use karke (concave: $u, v, f$ sab negative; $h^{'} < 0$ , $h > 0$ ). $∣ h^{'} ∣ = - h^{'}$ , $∣ h ∣ = h$ , $∣ v ∣ = - v$ , $∣ f ∣ = - f$ substitute karke:

$\frac{- h ^{'}}{h} = \frac{( - v ) - ( - f )}{- f} = \frac{- v + f}{- f} = \frac{v - f}{f}$

toh signed doosra ratio hai

$\frac{h ^{'}}{h} = - \frac{v - f}{f} .$

$v - f$ = image aur focus ke beech signed gap; $f$ = signed focal length. Yeh same height ratio ka hamara doosra expression hai — ab usi signed language mein likha gaya hai jaise Step 4 mein.

Step 6 — Dono ratios equal set karo aur solve karo

KYA. Step 4 aur Step 5 dono $h^{'} / h$ describe karte hain. Unhe equal karo; heights gaayab ho jaati hain; sirf algebra kaam khatam karta hai.

KYUN. Koi naya physics nahi bacha — reflection ne apna kaam Steps 1–5 mein kar diya. Jo bacha hai woh ek consistent sign language mein bookkeeping hai.

PICTURE. Figure dono triangle-pairs overlay karta hai taaki aap literally dekh sako ki dono $h^{'} / h$ ratios same image tip par milte hain.

Dono signed expressions $\frac{h ^{'}}{h}$ ke barabar hain, toh unhe equal karo:

$Step 4 - \frac{v}{u} = Step 5 - \frac{v - f}{f}$

Dono minus signs cancel ho jaate hain, bilkul clean start milta hai koi ambiguity nahi:

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

Cross-multiply (dono sides ko $u f$ se multiply karo): $v f = u (v - f) = uv - u f$ .
$uv$ term collect karo: $v f + u f = uv$ .
Har term ko $uv f$ se divide karo: $\frac{v f}{uv f} + \frac{u f}{uv f} = \frac{uv}{uv f} ⟹ \frac{1}{u} + \frac{1}{v} = \frac{1}{f}$

Upar har move signed quantities par ordinary algebra hai — koi sign haath se flip nahi hua; signs simply Step 0 se saath chale aaye. Isliye yahi equation convex aur virtual cases ke liye bhi bina kisi badlav ke kaam karegi.

Step 7 — Edge & degenerate cases (har ek walk karo)

KYA. Check karo ki picture boundaries par bhi theek behave kare: object $F$ par, object $C$ par, aur ek convex mirror.

KYUN. Ek derivation jis par aap trust karo use apne extremes survive karne chahiye. Agar koi case toot jaaye, formula galat hai.

PICTURE. Teen mini-scenes — parallel-out (object at $F$ ), self-return (object at $C$ ), aur convex mirror ki always-behind image.

Object at focus ( $u = f$ ): $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = 0 \Rightarrow v \to \infty$ . Rays parallel nikalti hain — koi image nahi, image "at infinity." Step 5 ke dono triangles degenerate ho jaate hain (zero base at $F$ ), bilkul match karta hai.
Object at centre ( $u = 2 f$ , kyunki $R = 2 f$ ): algebra se $v = 2 f$ bhi milta hai. Image $C$ par wapas baitti hai, same size, inverted — mirror $C$ ko apne aap map kar deta hai. $m = - v / u = - 1$ . ✔
Convex mirror: yahan $f > 0$ . Kisi bhi real object ke liye ( $u < 0$ ), $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ do positives ka sum hai ⇒ $v > 0$ hamesha ⇒ image hamesha peeche, virtual, erect, shrunk. Parallel ray ab $F$ se, mirror ke peeche se, aati hui lagti hai; same similar-triangle skeleton dashed virtual rays ke saath chalta hai. Figure mein always-virtual image dekho.

Ek-picture summary

Upar sab kuch ek single diagram mein collapse hota hai: ek object tip se do rays (pole ray blue, parallel ray green), chaar labelled distances, do pairs of similar triangles, aur woh boxed equation jo woh force karte hain.

Recall Feynman retelling (zor se bolo, koi symbols nahi)

Maine ek curved mirror ke saamne ek arrow rakha. Uski tip se do rays trace karta hoon: ek seedhi mirror ke middle mein jaati hai jo equal angle par bounce karti hai, aur ek flat chalti hai jo focus se hoke bounce karti hai. Jahan woh cross karti hain wahan image tip land karti hai. Flat ray ka bounce, sirf "ek radius sphere se right angle par milti hai" aur "angle in equals angle out" use karke, prove karta hai ki focus sphere ke centre ke halfway par baitha hai — jab tak hum axis ke paas rahein. Middle ray do look-alike triangles banati hai, jo mujhe batati hai ki image dono distances ke ratio se stretch hui hai (ek flip ke saath). Flat ray do aur look-alike triangles banati hai jo usi stretch ko doosre tarike se deti hai. Dono ratios plain ruler lengths se build karta hoon, phir har ek ko signed number line mein translate karta hoon, taaki dono ek hi language mein baat karein. Unhe equal karne par, heights cancel ho jaati hain, aur distances ke product se divide karne ke baad bachi hai: one-over-image plus one-over-object equals one-over-focus. Maine ise kabhi memorize nahi kiya — maine ise draw kiya.

Recall Forecast-then-verify

Concave, $f = - 10$ cm, object at $C$ ( $u = - 20$ ): predict then compute $v, m$ . ::: $\frac{1}{v} = \frac{1}{- 10} - \frac{1}{- 20} = - \frac{1}{20} \Rightarrow v = - 20$ cm, $m = - 1$ : same size, inverted, real — object apne aap par map ho gaya. ✔ Object exactly at focus ( $u = f = - 10$ ): kaunsi image? ::: $\frac{1}{v} = \frac{1}{- 10} - \frac{1}{- 10} = 0 \Rightarrow v = \infty$ : reflected rays parallel, koi finite image nahi. Convex $f = + 15$ , $u = - 30$ : compute karne se pehle $v$ ka sign? ::: Positive — do positive reciprocals ka sum — toh virtual, mirror ke peeche. ( $v = + 10$ , $m = + \frac{1}{3}$ .)

Related builds: Magnification & Image Formation Step 4 ka $m = - v / u$ use karta hai; Refraction & Lenses — Lens Maker's Equation lenses ke liye wahi similar-triangle machinery chalata hai; Step 2 ki paraxial caveat Spherical Aberration ki root hai.