Geometric optics — rectilinear propagation, reflection, refraction: Worked examples

2.5.1 · D3 · Physics › Optics › Geometric optics — rectilinear propagation, reflection, refr

Yeh page the parent topic par build karti hai aur Fermat's principle, Total internal reflection, aur Optical fibres pe rely karti hai.

The scenario matrix

Ray optics ka har problem kuch giney-chune case classes mein se ek hota hai. Neeche har row ek alag "shape" ka question hai, jo is baat se define hota hai ki kaun si quantity unknown hai aur usmein kya special/degenerate feature hai. Sabse daayaan column batata hai ki us cell ko kaun sa worked example cover karta hai.

#	Case class	Special/degenerate feature	Covered by
A	Rarer → denser refraction ( $n_{1} < n_{2}$ )	ray normal ki taraf jhukti hai, $θ_{2} < θ_{1}$	Ex 1
B	Denser → rarer refraction ( $n_{1} > n_{2}$ )	ray door jhukti hai, $θ_{2} > θ_{1}$	Ex 2
C	Normal incidence	$θ_{1} = 0$ (degenerate: koi bending nahi)	Ex 3
D	Grazing / critical angle	$θ_{2} = 90°$ limit, phir uske baad (TIR)	Ex 4
E	Slab with parallel faces	ray parallel nikalta hai, sirf shift hoti hai	Ex 5
F	Wavelength & speed inside medium	$f$ fixed rehta hai, $λ$ & $v$ change hote hain	Ex 6
G	Reflection geometry	$θ_{i} = θ_{r}$ , rotating mirror	Ex 7
H	Rectilinear / pinhole (real-world)	similar triangles, inversion	Ex 8
I	Exam twist (multi-step + limiting check)	TIR inside a fibre bent at entry face	Ex 9

Ab hum har row ka ek example solve karenge.

Ex 1 — Cell A: rarer → denser (normal ki taraf jhukna)

Worked example Light ka air se water mein jaana

Light air ( $n_{1} = 1.00$ ) se water ( $n_{2} = 1.33$ ) mein jaati hai, aur surface se $θ_{1} = 40°$ ke angle par normal se milti hai. Refraction angle $θ_{2}$ nikalo.

Forecast: pehle guess karo — kya $θ_{2}$ , $40°$ se bada hoga ya chhota?

Snell's law likho. $n_{1} sin θ_{1} = n_{2} sin θ_{2}$ . Yeh step kyun? Yahi ek equation hai jo boundary ke paar angles ko jodti hai; har refraction problem yahan se shuru hoti hai.
Unknown sine ke liye solve karo. $sin θ_{2} = \frac{n _{1} sin θ _{1}}{n _{2}} = \frac{1.00 \times sin 40°}{1.33} = \frac{0.643}{1.33} = 0.483$ . Yeh step kyun? Hum us cheez ko isolate karte hain jo hum nahi jaante; sine woh natural variable hai jo Snell hamen deta hai.
Sine ko undo karo. $θ_{2} = arcsin (0.483) = 28.9°$ . Yeh step kyun? $arcsin$ jawaab deta hai "kis angle ka yeh sine hai?" — yeh ratio ko wapas ek geometric angle mein convert karta hai.

Verify: $28.9° < 40°$ ✔ — ek denser medium mein jaane par ray normal ki taraf jhukti hai, bilkul waise jaisa forecast tha. Saath hi $n_{1} sin 40° = 0.643 = n_{2} sin 28.9°$ balance karta hai. ✔

Neeche wali figure bilkul yahi case draw karti hai. Dekho ki orange incident ray kaisi (halkay) air region mein dashed grey normal se $40°$ door hai, aur boundary ke neeche green refracted ray normal ke kitni kareeb hai $28.9°$ par — chhota green angle arc normal ki taraf jhukne ka visual signature hai.

Figure — Geometric optics — rectilinear propagation, reflection, refraction

Ex 2 — Cell B: denser → rarer (normal se door jhukna)

Worked example Light ka glass se air mein jaana

Glass ( $n_{1} = 1.50$ ) ke andar ek ray glass–air surface ( $n_{2} = 1.00$ ) se $θ_{1} = 25°$ par milti hai. $θ_{2}$ nikalo.

Forecast: is baar $25°$ se bada hoga ya chhota?

Snell's law. $1.50 sin 25° = 1.00 sin θ_{2}$ . Yeh step kyun? Wahi master equation — travel ki direction nahi badlati ki kaun sa law apply hoga.
Isolate karo. $sin θ_{2} = 1.50 \times 0.4226 = 0.634$ . Yeh step kyun? Ab multiplying factor $n_{1} / n_{2} = 1.5 > 1$ hai, isliye $sin θ_{2} > sin θ_{1}$ — angle barna chahiye.
Sine undo karo. $θ_{2} = arcsin (0.634) = 39.4°$ .

Verify: $39.4° > 25°$ ✔ — dense medium se nikalne par ray normal se door jhukti hai. Kyunki $sin θ_{2} = 0.634 < 1$ hai, ek real refracted ray abhi bhi exist karti hai (hum abhi critical angle tak nahi pahunche — Ex 4 dekho). ✔

(Yeh sirf Ex 1 figure ka mirror-image hai jisme media swap ho gaye hain — green ray normal se door jati hai uske paas aane ki jagah, isliye alag diagram ki zaroorat nahi hai.)

Ex 3 — Cell C: normal incidence (degenerate case)

Worked example Seedhi andar jaane wali ray

Light normal ke saath hi water surface ( $n_{2} = 1.33$ ) se air mein aati hai, yaani $θ_{1} = 0°$ . $θ_{2}$ nikalo.

Forecast: kya seedhi andar jaane wali ray bilkul bhi jhukti hai?

Snell's law. $1.00 sin 0° = 1.33 sin θ_{2}$ . Yeh step kyun? Hum "intuition se reason" nahi kar sakte — $θ_{1} = 0$ ko actual law mein plug karo aur use decide karne do.
Evaluate karo. $sin 0° = 0$ , isliye $1.33 sin θ_{2} = 0 \Rightarrow sin θ_{2} = 0$ . Yeh step kyun? Ek product zero hota hai tabhi jab koi factor zero ho; $n_{2} \neq = 0$ , isliye $sin θ_{2} = 0$ .
Undo karo. $θ_{2} = arcsin (0) = 0°$ .

Verify: ray seedhi, bina jhuke nikal jaati hai — yeh refraction ka woh akela case hai jisme koi bending nahi hoti. Isliye pool mein seedha neeche dekhne par tumhare bilkul neeche ka bottom distort nahi dikhta. ✔ (Speed phir bhi change hoti hai; sirf direction same rehti hai.)

(Ex 1 figure ki kalpana karo jisme dono rays dashed normal line par collapse ho jaati hain — ek seedha arrow — isliye is degenerate case ke liye koi figure nahi bani.)

Ex 4 — Cell D: critical angle aur usse aage (limiting behaviour)

Worked example Critical angle ke paas aana aur usse cross karna

Ek glass–air surface ke liye ( $n_{1} = 1.50$ , $n_{2} = 1.00$ ): (a) critical angle $θ_{c}$ nikalo; (b) $θ_{1} = 50°$ par kya hota hai?

Forecast: kis incidence par refracted ray gayab ho jaati hai?

Refracted ray ko graze karne par set karo. Critical angle par ray $θ_{2} = 90°$ par jhukti hai (possible maximum refraction). $1.50 sin θ_{c} = 1.00 sin 90° = 1$ . Yeh step kyun? $θ_{2} = 90°$ "ek refracted ray exist karti hai" aur "nahi kar sakti" ke beech ki boundary hai — wahi exact edge jo hum dhundh rahe hain.
Solve karo. $sin θ_{c} = \frac{1.00}{1.50} = 0.667 \Rightarrow θ_{c} = arcsin (0.667) = 41.8°$ . Yeh step kyun? $arcsin$ index ratio ko threshold angle mein convert karta hai.
$θ_{1} = 50°$ test karo. $sin θ_{2} = 1.50 \times sin 50° = 1.50 \times 0.766 = 1.149 > 1$ . Yeh step kyun? Kyunki $50° > 41.8°$ hai, Snell ek aise sine ki demand karta hai jo 1 se upar ho — kisi real angle ke liye impossible.
Interpret karo. Koi refracted ray exist nahi kar sakti → total internal reflection: saari light wapas $θ_{r} = 50°$ par bounce hoti hai (yahan $θ_{r}$ reflection angle hai).

Verify: $41.8°$ parent note ke critical-angle example se match karta hai ✔. $sin θ_{2} = 1.149 > 1$ ki demand TIR ka mathematical fingerprint hai — Total internal reflection aur Optical fibres dekho. ✔

Figure "beyond critical" case ko $θ_{1} = 50°$ par dikhata hai. Dhyaan do ki koi green refracted arrow air mein cross nahi kar raha — balki red arrow glass mein wapas usi $50°$ par bounce ho raha hai. Plot par caption text "no refracted ray" spell out karta hai taaki impossibility visible ho.

Ex 5 — Cell E: parallel-sided slab (parallel nikalta hai, shift hota hai)

Worked example Glass slab se ray

Air ( $n = 1.00$ ) mein ek ray glass slab ( $n = 1.50$ , parallel faces) se $θ_{1} = 45°$ par milti hai. Dusre face se air mein wapas nikalne ke baad exit angle nikalo.

Forecast: kya exit ray, entry ray se zyada steep, zyada shallow, ya same slant hogi?

Top face par refract in karo. $1.00 sin 45° = 1.50 sin θ_{2} \Rightarrow sin θ_{2} = \frac{0.7071}{1.5} = 0.4714$ , isliye $θ_{2} = 28.1°$ . Yeh step kyun? Air→glass ek rarer→denser refraction hai (cell A); hum slab ke andar ka angle nikaalte hain.
Andar ka angle bottom face se milta hai. Kyunki faces parallel hain, normal same direction mein hai, isliye bottom face par incidence angle $θ_{2} = 28.1°$ ke barabar hai. Yeh step kyun? Parallel faces ek shared normal rakhte hain; internal ray us shared normal ke relative apna slant maintain karta hai.
Bottom face par refract out karo. $1.50 sin 28.1° = 1.00 sin θ_{3} \Rightarrow sin θ_{3} = 1.50 \times 0.4714 = 0.7071$ , isliye $θ_{3} = 45°$ . Yeh step kyun? Yeh glass→air hai (cell B); hum final exit angle solve karte hain.

Verify: $θ_{3} = 45° = θ_{1}$ ✔ — exit ray, entry ray ke parallel hai; slab use sirf sideways shift karta hai, direction kabhi nahi badlata. Do $n$ -factors cancel ho jaate hain kyunki hum same pair of media ko reverse mein cross karte hain. ✔

Figure mein teen arrows trace karo: orange entry ray $45°$ par, blue ray slab ke andar $28.1°$ par normal ke kareeb tilted, aur green exit ray jo $45°$ par wapas aati hai. Do dashed grey lines top aur bottom faces par (parallel) normals ko mark karti hain — green ray orange ray ke parallel hai lekin sideways push ho gayi hai, jo lateral shift hai.

Ex 6 — Cell F: medium ke andar speed, wavelength aur frequency

Worked example Glass ke andar kya change hota hai?

Vacuum wavelength $λ_{0} = 500 nm$ ki green light $n = 1.50$ wale glass mein enter karti hai. Glass ke andar uski speed $v$ , wavelength $λ_{m e d}$ , aur frequency $f$ nikalo. $c = 3.00 \times 1 0^{8} m/s$ lo.

Forecast: teen mein se kaun sa ( $v$ , $λ$ , $f$ ) same rehta hai?

Andar ki speed. $v = \frac{c}{n} = \frac{3.00 \times 1 0 ^{8}}{1.50} = 2.00 \times 1 0^{8} m/s$ . Yeh step kyun? $n = c / v$ index ki definition hai; light dense medium mein slow ho jaati hai.
Andar ki wavelength. $λ_{m e d} = \frac{λ _{0}}{n} = \frac{500}{1.50} = 333 nm$ . Yeh step kyun? Wave "squeeze" ho jaati hai kyunki woh slow ho jaati hai lekin uski frequency change nahi ho sakti.
Frequency (vacuum). $f = \frac{c}{λ _{0}} = \frac{3.00 \times 1 0 ^{8}}{500 \times 1 0 ^{- 9}} = 6.00 \times 1 0^{14} Hz$ . Yeh step kyun? Frequency source se fix hoti hai; hum ise ek baar compute karte hain aur yeh waise hi rehti hai.
Frequency (glass mein) cross-check. $\frac{v}{λ _{m e d}} = \frac{2.00 \times 1 0 ^{8}}{333 \times 1 0 ^{- 9}} = 6.00 \times 1 0^{14} Hz$ . Yeh step kyun? $v = f λ$ glass ke andar bhi hold karna chahiye — yeh prove karta hai ki $f$ unchanged hai.

Verify: vacuum aur glass dono mein frequency identical hai ( $6.00 \times 1 0^{14} Hz$ ) ✔; sirf $v$ aur $λ$ factor $n$ se gire. Units: $m/s \div m = 1/s = Hz$ ✔. Isliye colour ( $f$ se tied) nahi badlata — Dispersion dekho. ✔

Ex 7 — Cell G: reflection geometry (rotating mirror)

Worked example Mirror ghuma do, ray dekho

Ek ray mirror se normal se $θ_{i} = 20°$ par milti hai. (a) Reflected angle batao. (b) Mirror ab $10°$ rotate ho jaata hai (incidence $30°$ ho jaata hai). Reflected ray kitna mudi?

Forecast: kya reflected ray $10°$ , $20°$ , ya kuch aur ghoomti hai?

Reflection ka law apply karo. $θ_{r} = θ_{i} = 20°$ . Yeh step kyun? Reflection hamesha angle ko normal ke paar mirror karta hai — koi exception nahi.
Incident aur reflected ray ke beech angle (pehla case). Woh normal ke symmetrically baithe hain, isliye unke beech ka angle $θ_{i} + θ_{r} = 20° + 20° = 40°$ hai. Yeh step kyun? Har ray normal se opposite sides par $θ$ door hai; inhe add karo.
Naya case. Incidence $30° \Rightarrow$ reflected $30° \Rightarrow$ rays ke beech angle $= 60°$ . Yeh step kyun? Wahi rule, bada angle.
Reflected-ray direction mein change. Between-rays angle $60° - 40° = 20°$ change hua, yaani reflected ray mirror ke $10°$ rotation se double ghoom gayi. Yeh step kyun? Ek hi incident ray ke liye dono configurations compare karo.

Verify: reflected ray $20° = 2 \times 10°$ swing karti hai ✔ — classic "reflected ray mirror rotation se double ghoomti hai." ✔

Ex 8 — Cell H: rectilinear propagation (pinhole word problem)

Worked example Pinhole camera

$H = 20 cm$ height ki ek mombatti pinhole ke saamne $u = 50 cm$ par khadi hai. Ek screen uske peeche $v = 15 cm$ hai. Image height $h$ aur uski orientation nikalo.

Forecast: mombatti se lamba ya chhota, seedha ya ulta?

Similar triangles use karo. Ek pinhole vertex se seedhi rays $\frac{h}{H} = \frac{v}{u}$ deti hain. Yeh step kyun? Rectilinear propagation rays ko bilkul seedha banata hai, isliye do triangles (object side, image side) similar hain.
$h$ ke liye solve karo. $h = H \times \frac{v}{u} = 20 \times \frac{15}{50} = 6 cm$ . Yeh step kyun? Known lengths ko proportion mein plug karo.
Orientation. Top rays hole se hokar neeche cross karti hain → image inverted hai. Yeh step kyun? Har seedhi ray single hole se guzarti hai, top aur bottom swap karte hue.

Verify: magnification $m = v / u = 15/50 = 0.30$ , aur $h = 0.30 \times 20 = 6 cm$ ✔; kyunki $v < u$ hai, image chhoti hai, aur inverted ✔. Units: $cm \times (cm / cm) = cm$ ✔.

Figure mein do blue rays follow karo: dono baayi taraf orange mombatti ki tip aur base se nikalti hain, single grey pinhole dot se guzarti hain, aur cross hoti hain — isliye top ray axis ke neeche land karti hai, daayein taraf chhoti green inverted image banate hue. Hole par crossing exactly wahi reason hai jis se image flip hoti hai.

Ex 9 — Cell I: exam twist (fibre entry face + TIR check)

Worked example Kya light fibre mein trapped rahegi?

Light air ( $n_{0} = 1.00$ ) se ek optical fibre ke flat end mein $θ_{in} = 20°$ par enter karti hai. Fibre core ka $n_{1} = 1.50$ hai, jo air cladding ( $n_{2} = 1.00$ ) se ghira hua hai. (a) Entry ke baad ray fibre axis se kitna angle banati hai? (b) Woh side wall se kis angle par milti hai, aur decide karo ki woh totally internally reflect hogi ya nahi. Wall par critical angle $θ_{c} = 41.8°$ hai (Ex 4 se).

Yahan hum $θ_{f}$ likhte hain fibre ke andar refracted angle ke liye (entry-face normal se measure kiya hua, jo fibre axis ke saath point karta hai). Hum yahan deliberately $r$ letter se bachte hain, kyunki is book mein $θ_{r}$ already reflection angle matlab rakhta hai — jo ek alag cheez hai.

Forecast: kya yeh ray side se escape karti hai, ya bounce hokar fibre mein chalti hai?

Flat entry face par refract karo. $1.00 sin 20° = 1.50 sin θ_{f} \Rightarrow sin θ_{f} = \frac{0.3420}{1.50} = 0.2280$ , isliye $θ_{f} = 13.2°$ . Yeh step kyun? Entry face normal refraction hai; $θ_{f}$ face ke normal se measure hota hai, jo fibre axis ke saath point karta hai.
Side wall par angle. Side wall ka normal axis ke perpendicular hai, isliye wall incidence angle $90° - θ_{f} = 90° - 13.2° = 76.8°$ hai. Yeh step kyun? Right angles par do normals ka matlab hai do angles complementary hain — axis ke parallel ray, wall par steep hogi.
Critical angle se compare karo. $76.8° > 41.8° = θ_{c}$ . Yeh step kyun? TIR tab hota hai jab wall incidence critical angle se zyada ho.
Conclude karo. Kyunki $76.8° > θ_{c}$ hai, light wall par totally internally reflect hoti hai aur trapped rehti hai, fibre mein zig-zag karti hui.

Verify: $θ_{f} = 13.2°$ Snell se ✔; wall angle $76.8°$ usse complement karta hai ✔; $76.8° > 41.8°$ TIR guarantee karta hai ✔. Yahi Optical fibres aur Total internal reflection ka principle hai. ✔

Active recall

Recall Yeh kaun sa cell hai? (answers cover karo)

Air → water at $30°$ , $θ_{2}$ nikalo ::: Cell A (rarer→denser); $sin θ_{2} = sin 30°/1.33$ . Ek ray seedhi surface mein jaati hai, $θ_{1} = 0$ ::: Cell C; degenerate, $θ_{2} = 0$ , koi bending nahi. Glass→air at $50°$ with $n = 1.5$ ::: Cell D; $sin θ_{2} > 1$ ⇒ total internal reflection. Parallel slab se ray ::: Cell E; parallel nikalta hai, sirf laterally shift hota hai. Medium mein $v$ , $λ$ , $f$ mein se kaun unchanged rahta hai? ::: Frequency $f$ ; $v$ aur $λ$ factor $n$ se shrink karte hain. Mirror $10°$ rotate ho, reflected ray kitna ghoomegi...? ::: $20°$ — mirror rotation se double.

Connections

Fermat's principle — upar use kiye gaye har angle law ki origin least time se hai.
Total internal reflection, Optical fibres — Ex 4 aur Ex 9.
Dispersion — kyun frequency-fixed refraction (Ex 6) colours ko split karta hai.
Mirrors and Lenses — Ex 7 ki reflection geometry curved surfaces tak extend hoti hai.