Power of a lens, combination of lenses: Exercises — Hinglish deep dive

2.5.7 · D4 · Physics › Optics › Power of a lens, combination of lenses

Pehli line se pehle, do reminders jo har problem mein kaam aate hain:

Kisi bhi problem ko touch karne se pehle, ensure karo ki tum dono combination rules ko padh sako, sirf ratt ke nahi. Agla box dono ko ek hi picture se build karta hai taaki is page par kuch bhi black box na rahe.

Intuition Dono combination rules kahaan se aate hain (ek baar padho, phir har jagah reuse karo)

Contact rule ( $1/ F = 1/ f_{1} + 1/ f_{2}$ ). Ek object ko lens 1 se guzaaro; woh $v_{1}$ par ek image banaata hai. Kyunki dono lenses touch karte hain, woh image turant lens 2 ke liye object ban jaati hai. Dono lenses ke liye Thin Lens Equation likhkar aur unhe add karke internal $v_{1}$ cancel ho jaata hai:

\;\Rightarrow\; \frac{1}{v}-\frac{1}{u}=\frac{1}{f_1}+\frac{1}{f_2}.$$ $\tfrac{1}{v}-\tfrac{1}{u}=\tfrac{1}{F}$ se compare karne par $\tfrac{1}{F}=\tfrac{1}{f_1}+\tfrac{1}{f_2}$ milta hai, yaani $P=P_1+P_2$. **WHY it works:** bina gap ke do bends ek bend mein stack ho jaate hain jo unka *sum* hota hai. **Separation rule ($P=P_1+P_2-dP_1P_2$).** Ab lenses ko $d$ door kheeenchो. Lens 1 se bend hone ke baad ray gap $d$ travel karta hai aur lens 2 tak ek **alag height** par pahunchta hai jis height par usne lens 1 chhoda tha. Lens 2 ka bend us height par depend karta hai, isliye dono bends cleanly add nahi hote; height-change ko algebra mein carry karne par exactly correction $-d\,P_1P_2$ milta hai. **Sanity check:** $d=0$ (touching) rakho aur yeh term gayab ho jaata hai, $P=P_1+P_2$ wapas milta hai. ✓ Height change ko *dekhne* ke liye neeche figure dekho. ![[deepdives/dd-physics-2.5.07-d4-s01.png]]

Hum image positions find karne ke liye Thin Lens Equation bhi use karenge, isliye ab iska sign convention fix karo.

Level 1 — Recognition

(Kya tum definition ko sahi tarike se read kar sakte ho aur units convert kar sakte ho?)

Exercise 1.1

Ek convex lens ki focal length $f = 50 cm$ hai. Iska power dioptres mein batao.

Recall Solution 1.1

WHAT: $P = 1/ f$ apply karo. WHY pehla step unit conversion hai: dioptre defined hai metres ke liye, isliye divide karne se pehle convert karna zaroori hai. $f = 50 cm = 0.50 m$ $P = \frac{1}{0.50} = + 2 D$ Positive isliye kyunki lens convex (converging) hai. Answer: $P = + 2 D$ .

Exercise 1.2

Ek lens par $- 4 D$ likha hai. Kya yeh convex hai ya concave, aur iska focal length cm mein kya hai?

Recall Solution 1.2

WHAT sign batata hai: $P < 0 \Rightarrow f < 0 \Rightarrow$ concave (diverging) lens. WHY invert karein: $P = 1/ f \Rightarrow f = 1/ P$ . $f = \frac{1}{- 4} = - 0.25 m = - 25 cm$ Answer: concave lens, $f = - 25 cm$ .

Exercise 1.3

$+ 2.5 D$ ki power ko metres aur centimetres dono mein focal length mein convert karo.

Recall Solution 1.3

WHY invert karein: power aur focal length reciprocals hain ( $P = 1/ f$ ), isliye diye gaye $P$ se $f$ recover karne ke liye hum simply relation ko flip karte hain, $f = 1/ P$ . Yahi ek algebra hai jo reciprocal ko undo karta hai. $f = \frac{1}{P} = \frac{1}{2.5} = 0.40 m = 40 cm$ Positive power $\Rightarrow$ converging lens. Answer: $f = 0.40 m = 40 cm$ .

Level 2 — Application

(Contact aur separation formulas mein plug karo.)

Exercise 2.1

Ek convex lens $P_{1} = + 6 D$ ko ek concave lens $f_{2} = - 20 cm$ ke saath contact mein rakha gaya hai. Net power aur equivalent focal length nikalo.

Recall Solution 2.1

Step 1 — $P_{2}$ nikalo. $f_{2} = - 0.20 m$ , isliye $P_{2} = \frac{1}{- 0.20} = - 5 D .$ Step 2 — add karo (woh touch karte hain, isliye powers add hoti hain): $P = P_{1} + P_{2} = 6 + (- 5) = + 1 D .$ WHY addition legal hai: contact mein koi gap nahi hai, isliye ray lens 1 chhod kar lens 2 mein essentially same height par enter karti hai — do bends ek sum mein stack ho jaate hain (upar derivation box dekho). Step 3 — equivalent focal length: $F = \frac{1}{P} = \frac{1}{1} = + 1 m = + 100 cm .$ Positive $\Rightarrow$ combination phir bhi converge karta hai, lekin sirf weakly. Answer: $P = + 1 D$ , $F = + 1 m$ .

Exercise 2.2

Do convex lenses $f_{1} = 10 cm$ aur $f_{2} = 40 cm$ contact mein rakhe gaye hain। $F$ nikalo.

Recall Solution 2.2

WHY reciprocals add hote hain: dono lenses touch karte hain, isliye (jaise upar derive kiya gaya) unki powers add hoti hain — aur power $1/ f$ hai। $P = P_{1} + P_{2}$ likhna $P = 1/ F$ , $P_{1} = 1/ f_{1}$ , $P_{2} = 1/ f_{2}$ ke saath literally yeh statement hai $\frac{1}{F} = \frac{1}{f _{1}} + \frac{1}{f _{2}}$ । Isliye hum directly reciprocals ke saath kaam kar sakte hain। Metres mein: $\frac{1}{F} = \frac{1}{0.10} + \frac{1}{0.40} = 10 + 2.5 = 12.5 D .$ $F = \frac{1}{12.5} = 0.08 m = 8 cm .$ Notice: combined $F$ ( $8 cm$ ) dono individual $f$ se chhota hai। Do converging pushes ek stronger, shorter-focus lens banate hain। Answer: $F = 8 cm$ ( $P = 12.5 D$ ).

Exercise 2.3

Do convex lenses $f_{1} = 20 cm$ , $f_{2} = 25 cm$ ko $d = 5 cm$ se alag rakha gaya hai। Equivalent power aur focal length nikalo।

Recall Solution 2.3

WHY separation formula: $d \neq = 0$ , isliye ray lens 2 par alag height par land karti hai — $- d P_{1} P_{2}$ correction zaroori hai (upar top box mein derive kiya gaya; figure s01 dekho)। Metres mein: $P_{1} = 1/0.20 = 5 D$ , $P_{2} = 1/0.25 = 4 D$ , $d = 0.05 m$ । $P = P_{1} + P_{2} - d P_{1} P_{2} = 5 + 4 - (0.05) (5) (4) = 9 - 1 = 8 D .$ $F = \frac{1}{8} = 0.125 m = 12.5 cm .$ Answer: $P = 8 D$ , $F = 12.5 cm$ .

Level 3 — Analysis

(Ulta sochо, ya result ko interpret karo।)

Exercise 3.1

Ek optician ke cabinet mein ek $+ 3 D$ lens hai। Pair ko plane glass plate (zero net power) jaisa behave karaane ke liye tumhe contact mein kaunsa single lens lagaana hoga?

Recall Solution 3.1

WHAT "plane glass" ka matlab hai: naa converge karta hai naa diverge, isliye $P = 0$ । WHY contact addition: contact mein, $P = P_{1} + P_{2} = 0$ । $0 = 3 + P_{2} \Rightarrow P_{2} = - 3 D .$ $f_{2} = \frac{1}{- 3} = - 0.3333 m \approx - 33.3 cm .$ Tumhe == $- 3 D$ ka concave lens== chahiye ( $f \approx - 33.3 cm$ )। Iska diverging push exactly converging push ko cancel karta hai। Answer: $P_{2} = - 3 D$ , $f_{2} = - \frac{1}{3} m \approx - 33.3 cm$ .

Exercise 3.2

Ek convex ( $+ 5 D$ ) aur ek concave ( $- 8 D$ ) lens ka combination contact mein use kiya jaata hai। Kya combination converging hai ya diverging? $F$ nikalo।

Recall Solution 3.2

$P = 5 + (- 8) = - 3 D .$ $P < 0 \Rightarrow$ concave lens jeetta hai; pair diverging hai। $F = \frac{1}{- 3} = - 0.3333 m \approx - 33.3 cm .$ Interpretation: ek lens converge karta hai fir bhi, lekin stronger diverging power sum ko dominate karta hai। Answer: diverging, $F = - \frac{1}{3} m \approx - 33.3 cm$ .

Exercise 3.3

Do lenses $d$ distance se alag hain; kaunsi separation par combination afocal ho jaata hai ( $P = 0$ , i.e. $F \to \infty$ )? $f_{1} = 15 cm$ , $f_{2} = 30 cm$ lo (dono convex)।

Recall Solution 3.3

WHAT afocal ka matlab hai: parallel rays in $\to$ parallel rays out; equivalent power zero hoti hai। Yahi ek telescope ki geometry hai (dekho Microscope and Telescope)। Separation formula mein $P = 0$ set karo: $0 = P_{1} + P_{2} - d P_{1} P_{2} \Rightarrow d = \frac{P _{1} + P _{2}}{P _{1} P _{2}} = \frac{1}{P _{2}} + \frac{1}{P _{1}} = f_{1} + f_{2} .$ WHAT yeh kaisa dikhta hai — neeche figure padho। Ek horizontal ray lens 1 mein enter karti hai (figure ke upar) aur lens 1 ke focus tak neeche bend hoti hai, red mein mark kiya gaya hai। Kyunki separation $d = f_{1} + f_{2}$ choose ki gayi hai, wahi red point bhi lens 2 ka front focus hai, isliye us focus se nikalti aur lens 2 se takraati ray dubara horizontal nikal jaati hai। Dono focal points ek mein merge ho gaye hain — woh ek shared focus hi woh poora reason hai jis se system afocal hai। Red dot trace karo: yahi woh hinge hai jo "parallel in" ko "parallel out" banaata hai। $d = f_{1} + f_{2} = 0.15 + 0.30 = 0.45 m = 45 cm .$ Answer: $d = f_{1} + f_{2} = 45 cm$ .

Figure — Power of a lens, combination of lenses

Level 4 — Synthesis

(Power ko magnification, ya lensmaker's relation ke saath combine karo।)

Exercise 4.1

Ek convex lens ( $f_{1} = 12 cm$ ) aur ek concave lens ( $f_{2} = - 18 cm$ ) contact mein hain। Ek object pair ke $24 cm$ aage rakha gaya hai। Image position aur magnification nikalo।

Recall Solution 4.1

Step 1 — equivalent lens। Contact mein, focal lengths (metres) use karo: $\frac{1}{F} = \frac{1}{0.12} + \frac{1}{- 0.18} = 8.333 - 5.556 = 2.778 D .$ $F = \frac{1}{2.778} = 0.36 m = 36 cm .$ WHY ek lens maano: contact mein, pair ek hi equivalent lens hai focal length $F$ ka, isliye hum ordinary Thin Lens Equation ek baar use kar sakte hain। Step 2 — thin lens equation। Upar fix kiye gaye sign convention ka use karke (real object aage $\Rightarrow u < 0$ ), $u = - 24 cm$ , $F = + 36 cm$ lo: $\frac{1}{v} - \frac{1}{u} = \frac{1}{F} \Rightarrow \frac{1}{v} = \frac{1}{36} + \frac{1}{- 24} = \frac{2 - 3}{72} = - \frac{1}{72} .$ $v = - 72 cm .$ Negative $v$ $\Rightarrow$ image object ke same side par hai: virtual, erect image। Step 3 — magnification। $m = v / u = (- 72) / (- 24) = + 3$ । Answer: $v = - 72 cm$ (virtual, same side), $m = + 3$ (erect, $3 \times$ magnified).

Exercise 4.2

Ek biconvex lens refractive index $n = 1.5$ wale glass se bani hai aur dono curvature radii magnitude mein $20 cm$ hain। Sign convention $R_{1} = + 20 cm$ , $R_{2} = - 20 cm$ use karke, iska power nikalo। Phir power nikalo agar use immerse kiya jaaye taaki effectively sirf $n - 1$ aadha ho jaaye।

Recall Solution 4.2

Part A — lensmaker's power (from Lensmaker's Equation): $P = (n - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}}), R in metres .$ $P = (1.5 - 1) (\frac{1}{0.20} - \frac{1}{- 0.20}) = 0.5 (5 + 5) = 0.5 \times 10 = + 5 D .$ Toh $f = 1/5 = 0.20 m = 20 cm$ । Part B — $(n - 1)$ factor aadha karo। Power directly $(n - 1)$ ke proportional hai; ise aadha karne par power aadhi ho jaati hai: $P^{'} = \frac{1}{2} \times 5 = + 2.5 D (f^{'} = 40 cm) .$ WHY: surface curvatures nahi badle; sirf bending strength per surface badli, jo poori tarah $(n - 1)$ factor mein hai। Answer: $P = + 5 D$ ; immersed $P^{'} = + 2.5 D$ .

Level 5 — Mastery

(Multi-step design aur limiting-case reasoning।)

Exercise 5.1

Ek student exactly $+ 4 D$ power ka converging system chahta hai lekin uske paas sirf do convex lenses hain $f_{1} = 25 cm$ aur $f_{2} = 50 cm$ । Woh unhe distance $d$ se separate kar sakta hai। Har woh value of $d$ (cm mein) nikalo jo $P = + 4 D$ de, aur check karo ki har ek physically sensible hai।

Recall Solution 5.1

Setup। $P_{1} = 1/0.25 = 4 D$ , $P_{2} = 1/0.50 = 2 D$ । $P = P_{1} + P_{2} - d P_{1} P_{2} = 4 + 2 - d (4) (2) = 6 - 8 d .$ $6 - 8 d = 4$ solve karo: $8 d = 2 \Rightarrow d = 0.25 m = 25 cm .$ Physicality check: $d = 25 cm$ positive aur finite hai — lenses $25 cm$ apart hain, perfectly buildable। WHY sirf ek answer: $P$ yahan $d$ mein linear hai, isliye equation $6 - 8 d = 4$ ka exactly ek root hai। ( $d = 0$ par $6 D$ milega; $d$ badhne par power girati hai; yeh $4 D$ se ek baar guzarti hai $d = 25 cm$ par, aur $d = f_{1} + f_{2} = 75 cm$ par $0$ ho jaati hai।) Answer: $d = 25 cm$ (unique).

Exercise 5.2

Teen thin lenses contact mein hain jinki powers $+ 10 D$ , $- 5 D$ , aur $+ 2 D$ hain। (a) Net power aur focal length nikalo। (b) Ek myopic eye ko $- 3 D$ corrective power chahiye; is set mein se kaunsa single lens (ya kaunsa pair contact mein) yeh provide karta hai?

Recall Solution 5.2

(a) Contact mein, saari powers add hoti hain: $P = 10 + (- 5) + 2 = + 7 D .$ $F = \frac{1}{7} = 0.1429 m \approx 14.3 cm (converging) .$ (b) Hume $- 3 D$ chahiye। Koi single lens yeh nahi deta। Contact mein pairs try karo: $10 + (- 5) = + 5, 10 + 2 = + 12, (- 5) + 2 = - 3 ✓$ $- 5 D$ aur $+ 2 D$ lenses contact mein exactly $- 3 D$ dete hain — woh diverging power jo ek short-sighted eye ko chahiye (dekho Defects of Vision)। Answer: (a) $+ 7 D$ , $F \approx 14.3 cm$ ; (b) $- 5 D$ aur $+ 2 D$ lenses together.

Exercise 5.3 (limiting case)

Separation formula mein, do convex lenses ke liye equivalent power $P$ ka kya hota hai jab $d \to \infty$ ? Physically interpret karo।

Recall Solution 5.3

$P = P_{1} + P_{2} - d P_{1} P_{2} .$ Jab $d \to \infty$ , $- d P_{1} P_{2}$ term dominate karta hai aur (do positive powers ke liye) $P \to - \infty$ । $F = \frac{1}{P} \to 0^{-} .$ Physical reading: jab doosra lens bahut door ho, lens 1 se bend hone ke baad ray jab tak lens 2 tak pahunchti hai tab tak enormously off-axis ho jaati hai, isliye lens 2 use drastically deflect karta hai — system ek strong (formally diverging) system ban jaata hai। Formula kabhi "break" nahi hota; woh bas yeh bata raha hai ki ek baar $d$ bahut bada ho jaaye toh two-lens model useful single equivalent lens nahi raha। Isliye clean $P = P_{1} + P_{2}$ rule contact ( $d = 0$ ) ke liye reserved hai, aur isliye real instruments $d$ ko modest rakhte hain, focal lengths ke comparable (dekho Microscope and Telescope)। Answer: $P \to - \infty$ , $F \to 0^{-}$ ; combination formally strongly diverging ho jaata hai।

Connections

Power of a lens, combination of lenses — woh parent note jise yeh exercises test karti hain।
Thin Lens Equation — §4.1 mein image locate karne ke liye aur contact rule derive karne ke liye use hua।
Lensmaker's Equation — §4.2 mein radii aur $n$ se powers।
Magnification of Lenses — §4 mein "multiply" rule।
Microscope and Telescope — afocal systems (§3.3, §5.3)।
Defects of Vision — dioptres mein corrective powers (§5.2)।
Resistors in Parallel — wohi reciprocal-addition mathematics।

Exercises — Power of a lens, combination of lenses