Huygens' principle — wavefront propagation: Exercises

2.5.10 · D4 · Physics › Optics › Huygens' principle — wavefront propagation

Saare problems parent note pe build karte hain. Use hone wale symbols:

$v$ = kisi medium mein wave speed (metres per second).
$t$ = beeta hua time (seconds).
$r = v t$ = time $t$ ke baad secondary wavelet ki radius.
$n = c / v$ = refractive index ( $c$ = vacuum mein light ki speed $\approx 3.0 \times 1 0^{8} m/s$ ).
$v_{1}, v_{2}$ = medium 1 (jis medium se light shuru hoti hai) aur medium 2 (jis medium mein light enter karti hai) mein wave speed.
$n_{1}, n_{2}$ = medium 1 aur medium 2 ka refractive index, jahan $n_{1} = c / v_{1}$ aur $n_{2} = c / v_{2}$ .
$i$ = angle of incidence, $r_{ref}$ = angle of refraction, dono normal se measure kiye jaate hain — woh line jo surface pe jahan light strike karti hai, wahan perpendicular ( $9 0^{\circ}$ pe) drawn hoti hai.
$λ$ = wavelength, $f$ = frequency. Yeh $λ = v / f$ se jude hain.

LEVEL 1 — Recognition

L1.1

Wavefront ko un points ka locus define kiya jaata hai jo ek instant pe kaunsi physical quantity share karte hain?

Recall Solution

Same phase. Wavefront woh surface hai jo wave ke un saare points ko jodti hai jo ek step mein oscillate kar rahe hain (apne upar-neeche cycle mein same point pe — same phase, jaisa upar define kiya gaya hai) us instant pe.

L1.2

Ek chote point source se bahut door, wavefronts kaisi shape ke lagte hain, aur kyun?

Recall Solution

Plane (flat). Source ke paas wavefronts spheres hote hain. Ek bade sphere ke chhote se patch ko dekho — jaise Earth pe khade hone par woh flat lagti hai. Toh door jaane par curvature negligible ho jaata hai aur wavefront effectively ek plane ban jaata hai. Sphere→plane picture ke liye parent dekho.

L1.3

Time $t$ ke baad, ek wavefront ke point se secondary wavelet radius $r$ tak spread ho gayi hai. $r$ ko $v$ aur $t$ mein likho.

Recall Solution

$r = v t$ Kyunki wavelet medium speed $v$ se time $t$ tak travel karti hai, aur distance = speed $\times$ time.

LEVEL 2 — Application

L2.1

Light vacuum mein $c = 3.0 \times 1 0^{8} m/s$ pe travel karti hai aur kisi khaas glass mein $2.0 \times 1 0^{8} m/s$ tak slow ho jaati hai. Glass ka refractive index $n$ find karo.

Recall Solution

$n = \frac{c}{v} = \frac{3.0 \times 1 0 ^{8}}{2.0 \times 1 0 ^{8}} = 1.5$ Divide kyun karte hain? $n$ measure karta hai ki light medium ke andar vacuum se kitni zyada slow hai. $n = 1.5$ ka matlab hai ki light glass mein $1.5 \times$ slow hai.

L2.2

L2.1 ke glass mein, ek point source se wavelet $t = 5.0 \times 1 0^{- 9} s$ ke liye expand karti hai. Uski radius kya hai?

Recall Solution

$r = v t = (2.0 \times 1 0^{8}) (5.0 \times 1 0^{- 9}) = 1.0 m$

L2.3

Air ( $v_{1} = 3.0 \times 1 0^{8} m/s$ ) mein ek plane wavefront $i = 3 0^{\circ}$ incidence par glass surface se takraata hai. Glass ka refractive index $n = 1.5$ hai. Refraction ka angle $r_{ref}$ find karo.

Neeche figure (s01): air mein incident ray (magenta), glass mein refracted ray (violet), dotted normal, aur do angles — iska kaam woh bend dikhana hai normal ki taraf jo tum abhi compute karne wale ho.

Figure — Huygens' principle — wavefront propagation

Figure s01 — Refraction air→glass: violet ray, magenta ray se zyada normal ke paas jhukti hai kyunki glass slower medium hai.

Recall Solution

Parent note mein Huygens refraction result se (medium 1 = air, medium 2 = glass), $\frac{s i n i}{s i n r _{ref}} = \frac{v _{1}}{v _{2}} = \frac{n _{2}}{n _{1}} = \frac{1.5}{1} = 1.5$ Toh $sin r_{ref} = \frac{s i n 3 0 ^{\circ}}{1.5} = \frac{0.5}{1.5} = 0.3333$ $r_{ref} = arcsin (0.3333) = 19.4 7^{\circ}$ Yeh kaisa dikhta hai (figure s01): wavefront ka end $A$ pehle slow medium mein enter karta hai, isliye woh lag jaata hai; door wala end $B$ abhi bhi air mein speed kar raha hai. Wavefront tilt hoti hai aur normal ki taraf bend karti hai — exactly wahi jo ek chhota angle ( $19. 5^{\circ} < 3 0^{\circ}$ ) matlab hai.

LEVEL 3 — Analysis

L3.1

$λ_{1} = 600 nm$ wavelength ki yellow light aur frequency $f$ air mein $n = 1.5$ wale glass mein enter karti hai. (a) Glass mein uski wavelength $λ_{2}$ find karo, aur (b) batao uski frequency ka kya hota hai.

Recall Solution

(b) pehle — frequency unchanged rehti hai. Wavefronts boundary par continuously judi rehti hain; crests pile up ya vanish nahi ho sakti, isliye har second crests ka count ( $f$ ) conserved rehta hai. (a) Kyunki $λ = v / f$ aur $f$ fixed hai, $λ \propto v$ . Speed $n$ factor se girta hai: $v_{2} = v_{1} / n$ . Toh $λ_{2} = \frac{λ _{1}}{n} = \frac{600}{1.5} = 400 nm$ Wavelength slower medium mein shrink hoti hai; frequency same rehti hai.

L3.2

Ek plane wavefront $A B$ ek mirror se reflect hoti hai. $A$ wavefront ka woh end hai jo pehle mirror ko touch karta hai; $C$ mirror pe woh spot hai jahan doosra end $B$ abhi pahunchna baaki hai; aur $E$ us wavelet ka leading edge hai jo $A$ ne medium mein emit ki jab $B$ abhi travel kar raha tha. End $B$ , $B C = v t$ distance travel karta hai surface $C$ tak pahunchne ke liye; usi time mein $A$ ki wavelet radius $A E = v t$ spread karti hai. Geometrically show karo ki angle of incidence, angle of reflection ke barabar kyun hota hai.

Neeche figure (s02): yeh $A$ , $B$ , $C$ , $E$ , incident front $A B$ , reflected front $C E$ , do equal legs $B C = A E = v t$ , aur shared hypotenuse $A C$ mark karta hai — poora congruence argument wahan drawn hai.

Figure s02 — Huygens se Reflection: triangles $A B C$ aur $A E C$ hypotenuse $A C$ share karte hain aur equal legs $B C = A E = v t$ rakhte hain, jo $∠ i = ∠ r$ force karta hai.

Recall Solution

Points $A$ , $C$ mirror pe hain; $B$ incoming front ka lagging end hai; $E$ wahan hai jahan $A$ ki wavelet pahunch gayi hai. Triangles $A B C$ aur $A E C$ dekho (figure s02):

Dono right-angled hain ( $A B ⊥ B C$ incoming front pe; $A E ⊥$ apne reflected front $C E$ ke).
Dono hypotenuse $A C$ share karte hain.
$B C = A E = v t$ (equal radii, kyunki same speed $v$ ne same time $t$ ke liye kaam kiya).

Equal hypotenuse aur ek equal leg wale do right triangles congruent hote hain (RHS congruence). Isliye $∠ B A C = ∠ E C A .$ Yeh do equal angles exactly angle of incidence aur angle of reflection hain jo surface se, aur isliye normal se, measure kiye jaate hain: $∠ i = ∠ r$ Equal radii hi poora engine kyun hai: agar $v$ ya $t$ alag hote, $B C \neq = A E$ , triangles congruent nahi hote, aur reflected front ek alag angle se tilt karti. Equal speed symmetry force karta hai.

L3.3

Light glass ( $n_{1} = 1.5$ ) se water ( $n_{2} = 1.33$ ) mein $i = 2 0^{\circ}$ incidence par jaati hai. Kya woh normal ki taraf ya door bend karti hai? $r_{ref}$ compute karo.

Recall Solution

Glass, water se denser hai, isliye light water mein enter karte waqt speed up karti hai ⇒ woh normal se door bend karti hai ⇒ expect karo $r_{ref} > 2 0^{\circ}$ . $n_{1} sin i = n_{2} sin r_{ref}$ $sin r_{ref} = \frac{n _{1}}{n _{2}} sin i = \frac{1.5}{1.33} sin 2 0^{\circ} = (1.1278) (0.3420) = 0.3858$ $r_{ref} = arcsin (0.3858) = 22.6 9^{\circ}$ Indeed $22. 7^{\circ} > 2 0^{\circ}$ — normal se door bend hua, jaisa predict kiya tha.

LEVEL 4 — Synthesis

L4.1 — Total internal reflection threshold

Light glass ke andar ( $n_{1} = 1.5$ ) glass–air boundary ( $n_{2} = 1.0$ ) ki taraf travel karti hai. (a) Critical angle $θ_{c}$ find karo jis par refracted ray surface ko graze karti hai ( $r_{ref} = 9 0^{\circ}$ ). (b) Huygens' wavelet picture use karke explain karo ki $θ_{c}$ se aage koi forward wavefront kyun nahi ban sakti.

Neeche figure (s03): $A$ se fast air wavelet (orange), refracted front jo critical angle par surface graze karne tak flat ho gayi hai, aur $θ_{c}$ par incident ray — yeh dikhata hai kyun tangent room se bahar ho jaata hai.

Figure s03 — $θ_{c}$ par air-side wavelet itni badi hoti hai ki sirf tangent surface ke saath flat lie karti hai ( $r = 9 0^{\circ}$ ); steep incidence par koi forward tangent nahi hota.

Recall Solution

(a) Critical angle par, $r_{ref} = 9 0^{\circ}$ toh $sin r_{ref} = 1$ : $n_{1} sin θ_{c} = n_{2} sin 9 0^{\circ} = n_{2}$ $sin θ_{c} = \frac{n _{2}}{n _{1}} = \frac{1.0}{1.5} = 0.6667$ $θ_{c} = arcsin (0.6667) = 41.8 1^{\circ}$ (b) Huygens picture (figure s03): air mein $A$ ki wavelet faster spread hoti hai ( $v_{2} > v_{1}$ ), toh uski radius $A D = v_{2} t$ , end $B$ ka jo gap $B C = v_{1} t$ abhi cover karna baaki hai, usse bada hota hai. Refracted front draw karne ke liye tumhe $C$ se bade air-wavelet tak tangent lagani hogi. Jaise $i$ badhta hai, required tangent tilt hoti hai jab tak, $θ_{c}$ par, tangent surface ke saath flat nahi ho jaati ( $r_{ref} = 9 0^{\circ}$ ). Uske baad, $A D$ itna bada hota hai ki medium 2 mein koi tangent exist hi nahi karti — geometry ka koi solution nahi — toh saari energy peeche reh jaati hai: total internal reflection.

L4.2 — Observed bend se speed nikalna

Ek plane wavefront medium 1 se medium 2 mein cross karti hai. Wavefronts ki tilt measure karne par $i = 4 5^{\circ}$ aur $r_{ref} = 3 0^{\circ}$ milta hai. Agar $v_{1} = 2.4 \times 1 0^{8} m/s$ hai, toh $v_{2}$ find karo.

Recall Solution

Snell's law ka Huygens' form speeds ko directly relate karta hai: $\frac{s i n i}{s i n r _{ref}} = \frac{v _{1}}{v _{2}}$ $v_{2} = v_{1} \frac{s i n r _{ref}}{s i n i} = (2.4 \times 1 0^{8}) \frac{s i n 3 0 ^{\circ}}{s i n 4 5 ^{\circ}} = (2.4 \times 1 0^{8}) \frac{0.5}{0.7071}$ $v_{2} = 1.697 \times 1 0^{8} m/s$ Wave slow ho gayi (chhota refraction angle ⇒ denser medium), $v_{2} < v_{1}$ se consistent.

LEVEL 5 — Mastery

L5.1 — Backward wavelet kyun khatam ho jaati hai

Huygens' raw construction har point ko full sphere emit karne deta hai, backward direction bhi. Obliquity factor batao aur usse forward ( $θ = 0^{\circ}$ ) aur backward ( $θ = 18 0^{\circ}$ ) directions mein evaluate karo. Explain karo ki yeh backward wavefront ka kya karta hai.

Recall Solution

Fresnel–Kirchhoff obliquity factor hai $K (θ) = \frac{1}{2} (1 + cos θ) .$

Forward: $θ = 0^{\circ} \Rightarrow cos 0 = 1 \Rightarrow K = \frac{1}{2} (1 + 1) = 1$ (full amplitude).
Backward: $θ = 18 0^{\circ} \Rightarrow cos 18 0^{\circ} = - 1 \Rightarrow K = \frac{1}{2} (1 - 1) = 0$ (zero amplitude).

Toh wavelet poori tarah forward contribute karti hai aur backward kuch nahi. Isliye Huygens backward envelope discard kar sake — yeh koi fudge nahi tha, wahan amplitude genuinely vanish ho jaati hai. Poori derivation ke liye Fresnel–Kirchhoff Diffraction dekho.

L5.2 — Continuity se frequency invariance

Medium 1 mein ek wave ki do crests $T$ seconds ke antar par boundary se takraati hain. Sirf is idea se prove karo ki wave boundary par continuous hai, ki medium 2 mein crests bhi $T$ seconds ke antar par niklengi (yaani $f_{1} = f_{2}$ ).

Recall Solution

Crest wavefront ka ek specific phase point hai. Kyunki field boundary par continuous hai, medium-1 side surface pe jo bhi oscillation hoti hai woh immediately medium-2 side se match hoti hai — boundary crests store, create, ya destroy nahi kar sakti. Toh agar ek crest interface par arrive karti hai, ek crest usi instant medium 2 mein depart karti hai. Agar do consecutive crests $T$ ke antar par arrive karti hain, do consecutive crests $T$ ke antar par depart bhi zaroori hain: $f_{2} = \frac{1}{T} = f_{1} .$ Speed aur wavelength adjust hoti hain ( $λ = v / f$ ), lekin crest crossings ka rate conserved rehta hai. Yeh wahi wave-continuity argument hai jo phase continuity ke peeche hai.

L5.3 — L2 ke data ke liye glass ke andar wavelength

Apne results combine karo: vacuum wavelength $589 nm$ ki sodium light $n = 1.52$ wale glass mein enter karti hai. Glass ke andar uski wavelength find karo aur confirm karo ki $3.0 μ m$ moti slab mein in wavelengths ki poori count — ya nahi — fit hoti hai.

Recall Solution

Medium mein wavelength: $λ_{glass} = \frac{λ _{vac}}{n} = \frac{589}{1.52} = 387.5 nm .$ $3.0 μ m = 3000 nm$ slab mein wavelengths ki count: $N = \frac{3000}{387.5} = 7.742.$ Poori count nahi — toh wave slab se bahar nikalti hai glass ko skip karne wali wave se phase shifted hokar. Yeh leftover fraction (wavelength ka $0.742$ ) exactly woh optical-path idea hai jo thin-film interference power karta hai.

Recall One-line answer key

L1.1 same phase · L1.2 plane (flat) · L1.3 $r = v t$ L2.1 $n = 1.5$ · L2.2 $r = 1.0 m$ · L2.3 $r_{ref} = 19.4 7^{\circ}$ L3.1 $λ_{2} = 400 nm$ , $f$ unchanged · L3.2 $△ A B C ≅ △ A E C$ (RHS) $\Rightarrow ∠ i = ∠ r$ · L3.3 $r_{ref} = 22.6 9^{\circ}$ (bends away) L4.1 $θ_{c} = 41.8 1^{\circ}$ · L4.2 $v_{2} = 1.697 \times 1 0^{8} m/s$ L5.1 $K (0^{\circ}) = 1, K (18 0^{\circ}) = 0$ · L5.2 $f_{2} = 1/ T = f_{1}$ (crest count conserved) · L5.3 $λ_{glass} = 387.5 nm, N = 7.742$ Is page ka har numeric line upar machine-checked hai us verification block mein jo is note se attached hai.

Connections

Snell's Law and Refractive Index — yahan har refraction problem isko use karti hai.
Laws of Reflection — L3.2 $∠ i = ∠ r$ derive karta hai.
Wave Optics — Interference — L5.3 thin-film interference feed karta hai.
Phase and Path Difference — L5.2 frequency continuity.
Fresnel–Kirchhoff Diffraction — L5.1 obliquity factor.
Young's Double Slit Experiment — do slits par secondary sources.
Diffraction — Huygens edges ke around bending explain karta hai.