Optical instruments — human eye, simple microscope, compound microscope, telescope: Worked examples

2.5.8 · D3 · Physics › Optics › Optical instruments — human eye, simple microscope, compound

Yeh page ek firing range hai: hum parent topic ke har tarah ke questions ko shoot karte hain, aur har ek ko fully worked example se knock down karte hain. Pehle hum target board — yaani scenario matrix — lay out karte hain, taaki aap dekh sako ki koi hidden cell nahi hai. Phir hum har cell ko work karte hain, compute karne se pehle answer forecast karte hue.

Shuru karne se pehle, teen reminders jo hum har example mein use karenge:

Definition Cartesian sign convention (HAR example par apply hoti hai, sirf tab nahi jab hum iska zikr karte hain)

Lens equation and sign conventions se: light left→right travel karti hai; distances jis direction mein light travel karti hai us direction mein measured karein toh positive hain, uske against wale negative hain. Toh far side par real image ka $v > 0$ ; near side par real object ka $u < 0$ ; object ke same side par virtual image ka $v < 0$ . Lens equation hai $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ jisme converging lens ka $f > 0$ hota hai. "Shortcut" formulas ( $M_{D} = 1 + D / f$ , $D / f$ , $L / f_{o} \cdot D / f_{e}$ , $f_{o} / f_{e}$ ) mein saari quantities already positive magnitudes mein likhi hain — signs ek baar unhe derive karne ke liye use hue (jaise hum explicitly Ex 4 mein dikhate hain) aur phir drop kar diye gaye. Toh Examples 1, 2, 3, 5, 6 sirf magnitudes use karte hain; Examples 4 aur 7 mein signed distances wapas aate hain jahan raw lens equation ki zaroorat hai.

Recall Master definition (pehli line se hi chahiye)

$M = \frac{θ _{with instrument}}{θ _{naked eye}}$ — angles ka ratio, kyunki retinal image size woh angle determine karta hai jo cheez subtend karti hai, uski real height nahi. Dekho Linear vs angular magnification.

Scenario matrix

Har optical-instrument problem inhi cells mein se kisi ek mein aata hai. Last column us worked example ka naam batata hai jo usse cover karta hai.

#	Cell class	Kya vary karta hai / twist	Covered by
A	Simple microscope, dono eye settings	image at $D$ vs at $\infty$	Ex 1
B	Simple microscope, design backwards	given $M$ , find $f$ (power in dioptres)	Ex 2
C	Compound microscope, relaxed eye	multiply, not add	Ex 3
D	Compound microscope, near-point eye + tube length	eyepiece uses $1 + D / f_{e}$ ; find $L$ from lens eqn	Ex 4
E	Telescope, dono eye settings	normal vs near-point adjustment	Ex 5
F	Telescope, word problem (real sky)	angle in, physical size out	Ex 6
G	Degenerate / limiting input	$u \to f$ ( $M \to \infty$ ), $f \to \infty$ ( $P \to 0$ ), object at $D$ , $f_{e} \to f_{o}$	Ex 7
H	Exam twist — galat formula pakdo	telescope disguised as a microscope	Ex 8

Sign convention (upar definition box mein stated) har example par apply hoti hai; jahan signed distances actually matter karte hain — Examples 4 aur 7 — wahan hum use apni jagah restate karte hain.

Example 1 — Simple microscope, dono eye settings (Cell A)

Forecast: Aage padhne se pehle guess karo — kya relaxed setting bada ya chhota number degi? (Hint: comfort usually costs magnification.)

Image at near point. $M_{D} = 1 + \frac{D}{f}$ use karo (yaad karo $M_{D}$ = image at $D$ , subscript box se). $M_{D} = 1 + \frac{25}{5} = 1 + 5 = 6.$ Yeh step kyun? Virtual image exactly $D$ par rakhna matlab eye ka sabse close focus, toh object lens ke sabse paas reh sakta hai → sabse bada angle → sabse bada $M$ .
Relaxed eye. $M_{\infty} = \frac{D}{f}$ use karo ( $M_{\infty}$ = image at infinity). $M_{\infty} = \frac{25}{5} = 5.$ Yeh step kyun? Object focus par $\Rightarrow$ rays parallel nikалти hain $\Rightarrow$ eye muscles relaxed. "+1" nahi kyunki image $\infty$ par hai, $D$ par nahi.
Difference. $M_{D} - M_{\infty} = 6 - 5 = 1.$ Yeh step kyun? Hamesha exactly $1$ hoga: $(1 + \frac{D}{f}) - \frac{D}{f} = 1$ . "+1" sabse sharp view ki "price" hai.

Verify: Units: $D$ aur $f$ dono cm mein, toh $D / f$ pure number hai — magnification dimensionless hai ✓. Sanity: $M_{D} > M_{\infty}$ ✓, difference exactly $1$ hai ✓.

Example 2 — Design backwards: given $M$ , find $f$ aur power (Cell B)

Forecast: Ek stronger magnifier — kya uska focal length Ex 1 ke $5 cm$ lens se lamba ya chhota hoga?

Relaxed formula invert karo. $M_{\infty} = \frac{D}{f} \Rightarrow f = \frac{D}{M}$ . $f = \frac{25}{10} = 2.5 cm .$ Yeh step kyun? Hume $M$ diya hai aur $f$ chahiye; relaxed formula dono ke beech ka sabse simple link hai, toh hum ise $f$ ke liye solve karte hain.
Power mein convert karo. Power of a lens and dioptres se, $P = \frac{1}{f ( in metres )}$ . $f = 2.5 cm = 0.025 m \Rightarrow P = \frac{1}{0.025} = 40 D .$ Yeh step kyun? Power dioptres mein define hoti hai, jisme $f$ metres mein chahiye, toh pehle convert karo.

Verify: Plug back karo: $D / f = 25/2.5 = 10 = M$ ✓. Ex 1 ke $5 cm$ se chhota $f$ (jaise forecast kiya — stronger lens fatter/shorter $f$ wala hota hai) ✓. Power positive hai kyunki converging lens hai ✓.

Example 3 — Compound microscope, relaxed eye (Cell C)

Forecast: Order of magnitude guess karo — tens? hundreds? Aur abhi decide karo: dono stages add hote hain ya multiply?

Objective (linear) magnification. $m_{o} \approx \frac{L}{f _{o}}$ . $m_{o} = \frac{20}{1} = 20.$ Yeh step kyun? Object $f_{o}$ ke just bahar baithta hai, toh $u_{o} \approx f_{o}$ aur $v_{o} \approx L$ , jisse $m_{o} = v_{o} / u_{o} \approx L / f_{o}$ milta hai.
Eyepiece magnification (relaxed). $m_{e} = \frac{D}{f _{e}}$ . $m_{e} = \frac{25}{2.5} = 10.$ Yeh step kyun? Eyepiece sirf ek simple microscope ki tarah intermediate image par act kar raha hai, relaxed setting → $D / f_{e}$ use karo.
Combine — multiply karo. $M = m_{o} m_{e} = 20 \times 10 = 200.$ Yeh step kyun? Eyepiece pehle se magnified intermediate image ko aur magnify karta hai, toh magnifications do photocopies ki tarah compound hoti hain (×2 phir ×3 = ×6), kabhi add nahi hoti.

Verify: $20 \times 10 = 200$ ; galat "add" answer hota $30$ . ×2-then-×3 copier ×6 deta hai ×5 nahi, jisse multiplication confirm hoti hai ✓.

Example 4 — Compound microscope, near-point eye + tube length find karo (Cell D)

Figure — Optical instruments — human eye, simple microscope, compound microscope, telescope

Forecast: Near-point eyepiece $D / f_{e}$ ki jagah $1 + D / f_{e}$ use karta hai — toh kya yeh $M$ relaxed version se bada ya chhota hoga?

Pehle upar ka figure padho. Yeh tube ko axis par lay out karta hai: objective position $0$ par hai, real intermediate image uske right par $v_{o} = 12 cm$ par form hoti hai (orange arrow), aur eyepiece thoda aur right par hai. Eyepiece us intermediate image ko apna object maanta hai jo uske left mein $∣ u_{e} ∣$ distance par hai, aur virtual final image ko poore $D = 25 cm$ tak (plum, dashed, far left) throw karta hai. Neeche bracket dikhata hai ki $L = v_{o} + ∣ u_{e} ∣$ kyun hai: tube physically objective se eyepiece tak span karni chahiye, yaani objective→image ( $v_{o}$ ) plus image→eyepiece ( $∣ u_{e} ∣$ ).

Objective, exact lens equation. $\frac{1}{v _{o}} - \frac{1}{u _{o}} = \frac{1}{f _{o}}$ jisme $v_{o} = + 12$ , $f_{o} = + 2$ . $\frac{1}{u _{o}} = \frac{1}{v _{o}} - \frac{1}{f _{o}} = \frac{1}{12} - \frac{1}{2} = \frac{1 - 6}{12} = - \frac{5}{12} \Rightarrow u_{o} = - 2.4 cm .$ Yeh step kyun? Exact $m_{o}$ ke liye sahi object distance chahiye; jab value directly mili ho toh approximate mat karo. Negative $u_{o}$ sirf confirm karta hai ki object incoming-light side par hai, jaise convention demand karta hai.
Objective magnification. $m_{o} = \frac{v _{o}}{u _{o}} = \frac{12}{2.4} = 5.$ Yeh step kyun? Real intermediate image ka linear magnification $v_{o} / u_{o}$ hai; hum size factor ke liye magnitude lete hain.
Eyepiece, near-point. $m_{e} = 1 + \frac{D}{f _{e}} = 1 + \frac{25}{5} = 6.$ Yeh step kyun? Final image $D$ par hai ( $\infty$ par nahi) → "+1" form use karo, exactly jaise near point par simple microscope.
Total. $M = m_{o} m_{e} = 5 \times 6 = 30.$ Yeh step kyun? Ex 3 jaise hi multiply rule.
Tube length — geometric argument jo figure se tied hai. Figure mein eyepiece ka object intermediate image hai; uski virtual final image us object ke same side par hai (dono eyepiece ke left mein), toh sign convention se $v_{e} = - 25$ , $f_{e} = + 5$ ke saath. Eyepiece object distance $u_{e}$ ke liye lens equation solve karo: $\frac{1}{v _{e}} - \frac{1}{u _{e}} = \frac{1}{f _{e}} \Rightarrow \frac{1}{u _{e}} = \frac{1}{v _{e}} - \frac{1}{f _{e}} = - \frac{1}{25} - \frac{1}{5} = - \frac{6}{25} \Rightarrow u_{e} = - \frac{25}{6} \approx - 4.17 cm .$ Negative sign bolta hai ki object (intermediate image) eyepiece ke $4.17 cm$ left mein hai — exactly woh gap jo figure mein drawn hai. Tube ko objective→image aur image→eyepiece dono cover karne hain: $L = objective \to image v_{o} + image \to eyepiece ∣ u_{e} ∣ = 12 + 4.17 = 16.17 cm .$ Yeh step kyun? Hum dono physical gaps add karte hain kyunki objective aur eyepiece exactly us total distance se separated hain; image unके beech mein hai.

Verify: $m_{o}$ : $∣12/ (- 2.4) ∣ = 5$ ✓. Objective ke liye lens eqn check karo: $1/12 - 1/ (- 2.4) = 1/12 + 5/12 = 6/12 = 1/2 = 1/ f_{o}$ ✓. $m_{e} = 1 + 5 = 6$ ✓. $M = 30$ ✓. $∣ u_{e} ∣ = 25/6 \approx 4.17$ ✓, $L \approx 16.17 cm$ ✓.

Example 5 — Telescope, dono eye settings (Cell E)

Forecast: Kaun sa adjustment bada $M$ deta hai — relaxed (normal) ya near-point? (Telescopes yahan microscopes ki same direction mein behave karte hain.)

Pehle figure padho. Upar picture mein, orange rays far-left se chhote real star-angle $α$ par enter karti hain optical axis ke against. Woh objective (teal lens) cross karti hain aur common focal plane (dashed line) mein ==image height $h$ == par milti hain. Phir teal rays eyepiece (plum lens) se dubara parallel nikalti hain, lekin ab bahut bade angle $β$ par tilted hain. Dono shaded angles $α$ (orange, objective par) aur $β$ (teal, eyepiece exit par) exactly woh dono angles hain jo $M = β / α$ mein hain — figure hi derivation hai.

Normal adjustment. Figure mein dono right triangles padho, same image height $h$ objective focus par $α \approx h / f_{o}$ aur eyepiece focus par $β \approx h / f_{e}$ subtend karta hai. Toh $M = \frac{β}{α} = \frac{h / f _{e}}{h / f _{o}} = \frac{f _{o}}{f _{e}} = \frac{100}{2} = 50.$ Yeh step kyun? $h$ cancel ho jaata hai — star ka real size kabhi enter nahi karta, sirf angles. Isliye telescope angular magnification quote karta hai.
Tube length (normal). $L = f_{o} + f_{e} = 100 + 2 = 102 cm .$ Yeh step kyun? Normal adjustment mein objective ka focus aur eyepiece ka focus coincide karte hain (figure mein dashed line), toh tube dono focal lengths ka sum hai.
Near-point adjustment — algebra sketch karo. Ab eyepiece pull in kiya gaya hai taaki uska virtual output image $\infty$ par nahi $D$ par ho. Eyepiece phir height $h$ ki image par ek simple microscope ki tarah act karta hai, toh uska angular magnification near-point form ban jaata hai $m_{e} = 1 + \frac{D}{f _{e}}$ angle par apply hote hue. $M = β / α$ scratch se build karo: $α \approx \frac{h}{f _{o}}, β = \frac{h}{f _{e}} (1 + \frac{f _{e}}{D}) (eyepiece near-point angle) .$ Divide karo: $M = \frac{β}{α} = \frac{f _{o}}{f _{e}} (1 + \frac{f _{e}}{D}) .$ Ab numbers plug karo: $M = 50 (1 + \frac{2}{25}) = 50 \times \frac{27}{25} = 54.$ Yeh step kyun? Eye ko strain karke eyepiece output $D$ par focus karna ( $\infty$ ki jagah) extra factor $(1 + \frac{f _{e}}{D})$ squeeze out karta hai — wahi "comfort ke badle magnification milti hai" idea jaise microscope mein.

Verify: $100/2 = 50$ ✓; $L = 102$ ✓; $50 \times 27/25 = 1350/25 = 54$ ✓. Near-point > normal, jaise forecast kiya ✓.

Example 6 — Telescope word problem: real sky (Cell F)

Forecast: Poora $0. 5^{\circ}$ 50× blow up hoga... kya answer kuch degrees hoga, ya tens of degrees (yaani view ka bada hissa)?

$M$ ki definition apply karo. $M = \frac{β}{α} \Rightarrow β = M α .$ Yeh step kyun? $M$ hi apparent-to-real angle ka ratio hai — instrument ka poora point yahi hai. Note karo hum directly $α, β$ angles se kaam karte hain; Ex 5 ka image height $h$ yahan chahiye nahi kyunki woh $M$ mein pehle hi cancel ho gaya tha.
Numbers plug karo. $β = 50 \times 0. 5^{\circ} = 2 5^{\circ} .$ Yeh step kyun? Direct substitution; dono angles degrees mein hain, toh koi conversion nahi chahiye.

Verify: $50 \times 0.5 = 25$ ✓. Sanity: $2 5^{\circ}$ sky ka ek bada hissa hai (arm's length par muththi ~ $1 0^{\circ}$ hoti hai), jo "telescope mein Moon bahut bada dikhta hai" experience se match karta hai ✓. Units: $M$ dimensionless × degrees = degrees ✓.

Example 7 — Degenerate & limiting inputs (Cell G)

Forecast: Har limit guess karo: kya magnification blow up hoti hai, vanish hoti hai, ya 1 par settle hoti hai?

(a) $f \to \infty$ . $M_{\infty} = \frac{D}{f} \to \frac{25}{\infty} = 0$ ; aur $M_{D} = 1 + \frac{D}{f} \to 1.$ Yeh step kyun? Flat lens ( $P = 0$ , dekho Power of a lens and dioptres) kuch bend nahi karta, toh koi angle enlarge nahi kar sakta. Near-point form $M_{D} = 1$ par floor karta hai — literally "naked eye se $D$ par koi help nahi."
(b) Koi lens nahi, object $D$ par. Instrument angle $= θ_{0} = h / D$ = naked-eye angle, toh $M = \frac{θ _{0}}{θ _{0}} = 1.$ Yeh step kyun? Consistency check: "no instrument" by definition $M = 1$ dena chahiye. Yeh (a) mein $M_{D}$ ke $f \to \infty$ limit se agree karta hai ✓.
(c) $f_{e} \to f_{o}$ . $M = \frac{f _{o}}{f _{e}} \to \frac{f _{o}}{f _{o}} = 1.$ Yeh step kyun? Equal focal lengths mein koi angular gain nahi milta — eyepiece angle utna hi spread karta hai jitna objective ne shrink kiya tha. Magnify karne ke liye zaroor $f_{o} > f_{e}$ chahiye.
(d) Object focus par, $u \to f$ (classic $M \to \infty$ edge). Signed lens equation $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ use karo real object $u = - f$ ke saath (object focal point par, incoming side): $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{f} + \frac{1}{- f} = 0 \Rightarrow v \to \infty.$ Linear magnification $m = \frac{v}{u} \to \frac{\infty}{f} \to \infty.$ Yeh step kyun? Jab object exactly focus par baithe, image infinity par race kar jaati hai, toh uska size (aur linear magnification) blow up ho jaata hai. Practice mein aap ise dekh nahi sakte — image infinitely far aur infinitely large hai — isliye usable magnifier object ko $f$ ke just andar rakhta hai, jo Ex 1 ka finite $M_{D} = 1 + D / f$ deta hai. Yeh woh boundary case hai jis par finite formulas approach karti hain.

Verify: (a) $lim_{f \to \infty} 25/ f = 0$ , $lim 1 + 25/ f = 1$ ✓. (b) koi bhi $h, D$ ratio $1$ deta hai ✓. (c) $f_{o} / f_{o} = 1$ ✓. (d) $u = - f$ par, $1/ v = 0 \Rightarrow v = \infty$ , toh $m \to \infty$ ✓. Saare limits ya toh finite hain ya sensibly diverge karte hain — koi scenario chhuta nahi.

Example 8 — Exam twist: galat formula pakdo (Cell H)

Forecast: Kya ek $80 cm$ objective wala telescope sirf ×6 magnify karega? Yahan apna gut trust karo.

Error identify karo. Student ne eyepiece-as-magnifier formula $D / f_{e}$ use ki. Yeh object ko near point $D$ par treat karta hai. Yeh step kyun? Reference angle choose karna hi poora game hai — pehle poochho "object kahan hai?" Telescope ka object infinity par hai, $D$ par nahi.
Sahi reference angle. Star ka real angle $α = h / f_{o}$ hai, $h / D$ nahi. Toh $M = \frac{f _{o}}{f _{e}}$ use karo. $M = \frac{80}{4} = 20.$ Yeh step kyun? Object infinity par hone se naked-eye angle actual $α$ hai, jisse $M = β / α = f_{o} / f_{e}$ milta hai.
Galti theek kyun lagti hai. Dono instruments mein eyepiece simple magnifier ki tarah act karta hai, toh $D / f_{e}$ familiar lagta hai — lekin yeh sirf tab apply hota hai jab naked eye se dekhi jaane wali cheez $D$ par ho, jo ek star ke liye false hai.

Verify: Sahi $M = 80/4 = 20$ ; galat value $25/4 = 6.25$ approximately ×3 chhoti hai ✓. Sanity: bada objective ⇒ bada magnification, toh ×20 ≫ ×6.25 sahi direction mein hai ✓.

Recall One-line self-test

Answers cover karo. Kaun si cell "+1" use karti hai? ::: Near-point settings — simple ( $1 + D / f$ ) aur compound eyepiece; telescope ki jagah multiply karta hai $f_{o} / f_{e}$ ko $(1 + f_{e} / D)$ se. Telescope ka object kahan hai ::: infinity par, toh reference angle $α = h / f_{o}$ hai, $h / D$ nahi. Jab $f_{e} \to f_{o}$ ho toh telescope ka $M \to$ ::: $1$ (koi magnification nahi). Jab object focus tak pahunche ( $u \to f$ ) toh simple lens ka linear $m \to$ ::: $\infty$ (image infinity par race karti hai). No-instrument magnification hamesha ::: exactly $1$ hoti hai. Full telescope near-point formula? ::: $M = \frac{f _{o}}{f _{e}} (1 + \frac{f _{e}}{D})$ — $f_{o} / f_{e}$ times bracket, sirf bracket nahi.

Connections

Built on Optical Instruments — Eye, Microscope, Telescope
Uses Lens equation and sign conventions (Ex 4, Ex 7)
Uses Power of a lens and dioptres (Ex 2, Ex 7)
Compare with Linear vs angular magnification (Ex 3 vs Ex 5)
Further reading: Resolving power and diffraction limit, Defects of vision — myopia, hypermetropia, Reflecting telescope (Cassegrain) vs refracting