Worked examples — Huygens' principle — wavefront propagation

2.5.10 · D3 · Physics › Optics › Huygens' principle — wavefront propagation

Yeh page Huygens' principle — wavefront propagation ki drill ground hai. Hum un construction rules ko lete hain jo tumne pehle dekhe hain aur unhe har corner mein push karte hain — refractive index change ka har sign, flat ($90°$) grazing case, "critical" case jahan refraction khatam ho jaata hai, aur kuch exam-style twists. Yahan kuch bhi parent se contradict nahi karta; yeh parent ke rules ko kaam mein laata hai.

Shuru karne se pehle, chhe symbols jo tumhe har jagah dikhenge. Agar koi unfamiliar lagta hai, hum use poori tarah build karte hain taaki koi notation aakar surprise na kare.

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Scenario matrix

Is topic ka har problem in cells mein se ek hai. Niche ke worked examples us cell ke saath tagged hain jo woh hit karte hain, aur milke poora grid fill karte hain.

Cell	Situation	Isme kya khaas hai
C1	Ek uniform medium mein plane wave	Koi bending nahi — pure straight propagation, radius $r=v\Delta t$
C2	Reflection, general $i$	$\angle i=\angle r$, congruent triangles
C3	Refraction rarer → denser ($v_2<v_1$)	Normal ki taraf bend karta hai, $r<i$
C4	Refraction denser → rarer ($v_2>v_1$)	Normal se door bend karta hai, $r>i$
C5	Degenerate: $i=0$ (normal incidence)	Bilkul bhi bend nahi; wavefront sirf slow/speed up hota hai
C6	Limiting: critical angle, $r\to 90°$	Refracted wave surface ko graze karta hai; uske baad → total internal reflection
C7	Boundary ke across Wavelength / frequency	$f$ fixed rehta hai, $\lambda$ aur $v$ change hote hain
C8	Real-world word problem	Ek physical story ko $i,r,v$ mein translate karna
C9	Grazing incidence $i\to 90°$	Wave surface ko skim karta hai — limiting geometry
C10	Exam twist: Huygens kya predict karta hai vs ek trap	Backward wavelet / obliquity, ray⊥wavefront

Prerequisite links jo tumhe open rakhni ho sakti hain: Snell's Law and Refractive Index, Laws of Reflection, Phase and Path Difference.

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Example 1 — Plane wave aage badhti hai (Cell C1)

Forecast: padhne se pehle doori ka guess karo — kya yeh kuch centimetres hogi, ya kuch metres?

Figure padhna: left cyan line starting wavefront W1 hai; faint white semicircles secondary wavelets hain, har ek radius $r=v\Delta t$ ki, W1 ke paanch points se emit hue; right cyan line W2 unka common forward tangent hai; amber arrow woh doori $v\Delta t$ mark karta hai jitna wave aage badha hai.

1. Wavefront ke har point se ek wavelet emit hoti hai radius $r=v\Delta t$ ki. Yeh step kyun? Yeh poora Huygens recipe hai — har point ek secondary source hai, aur time $\Delta t$ mein uska chhota sa sphere radius $v\Delta t$ tak grow karta hai. $r=(2.25\times 10^8)(2.0\times 10^{-9})=0.45\text{m}$
2. Naya wavefront in sabhi equal circles ka common tangent hai. Equal circles jinke centres ek straight line par baithe hain unka ek straight tangent line hota hai, pehle ke parallel, $0.45\text{m}$ aage. Yeh step kyun? Same $v$, same start-time ⇒ same radius ⇒ envelope tilt nahi ho sakta. Flat wave flat rehti hai.
3. Shape: abhi bhi ek plane, $0.45\text{m}$ aage gayi.

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Example 2 — General angle par Reflection (Cell C2)

Forecast: kya $r$ $35°$ se bada, chhota, ya barabar hoga?

Figure padhna: white horizontal line mirror hai; $A$ aur $C$ woh do points hain jahan wavefront mirror se milta hai ($A$ pehle, $C$ baad mein). $B$ incident wavefront ka door wala end hai (cyan line $BC$); $E$ woh wavelet ka tip hai jo $A$ ne medium mein wapas throw ki hai (amber line $EC$ reflected wavefront hai). $A$ par dashed line normal hai.

1. $BC=AE=vt$ set karo. Yeh step kyun? Dono distances same time $t$ mein same speed $v$ par cover hoti hain (same medium), isliye yeh equal hain — yeh key fact hai.
2. Dono triangles mein ek right angle hai, isliye $AC$ dono ka hypotenuse hai. Incident wavefront $BC$ definition se incident rays ki travel ke perpendicular hai; end $B$ surface line $AC$ mein seedha jaake $C$ par pahunchta hai, isliye $\angle BCA=90°$. Isi tarah reflected wavefront $AE$ reflected travel ke perpendicular hai, jo $\angle EAC=90°$ deta hai. Har right angle ke opposite side $AC$ hai, isliye $AC$ dono $△ABC$ aur $△AEC$ ka shared hypotenuse hai. Yeh step kyun? Tum do triangles ko "right-angled with hypotenuse $AC$" tab tak nahi keh sakte jab tak yeh na dikhaao ki right angles kahan hain — agla step ka RHS congruence test isi par depend karta hai.
3. RHS congruence apply karo. Dono triangles hypotenuse $AC$ share karte hain, dono right-angled hain (Step 2), aur $BC=AE$ (Step 1). RHS (right-angle–hypotenuse–side) se, $△ABC\cong △AEC$, isliye $\angle BAC=\angle ECA$, jo $\angle i=\angle r$ mein translate hota hai. Isse $r=35°$. Yeh step kyun? Congruent triangles incidence aur reflection geometry ko match karne par majboor karte hain; wavefront angle ray-from-normal angle ke barabar hai (dono "$90°$ minus surface angle" hain), isliye equality carry over hoti hai.

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Example 3 — Refraction rarer → denser, normal ki taraf bend karta hai (Cell C3)

Forecast: wave slow ho jaati hai — kya tumhe lagta hai ray normal ki taraf ya normal se door bend karegi?

Figure padhna: white horizontal line air–glass boundary hai; upar cyan arrow $i=60°$ par incident ray hai; niche amber arrow $r=35.3°$ par refracted ray hai; dashed vertical line normal hai. Underlying wavefront picture mein, $BC=v_{1}t$ (air side) aur $AD=v_{2}t$ (glass side) boundary ke saath common hypotenuse $AC$ share karte hain — $D$ simply $A$ ki wavelet ka tip hai glass ke andar.

1. Huygens' Snell relation likho $\frac{\sin i}{\sin r}=\frac{v_1}{v_2}$. Yeh step kyun? Derivation mein $BC=v_{1}t$ aur $AD=v_{2}t$ shared hypotenuse $AC$ par, isliye $\sin i=BC/AC$ aur $\sin r=AD/AC$ sines ko speeds ke ratio mein rakhte hain.
2. $\sin r$ ke liye solve karo: $\sin r=\sin i\cdot \frac{v_2}{v_1}=\sin 60°\cdot \frac{2.0}{3.0}=0.8660\times 0.6667=0.5774$ Yeh step kyun? Hume $r$ chahiye, isliye $\sin r$ isolate karo; slower medium ($v_2<v_1$) $\sin r$ ko shrink karta hai.
3. Arcsin lo: $r=\arcsin (0.5774)=35.3°$. Yeh step kyun? $\arcsin$ ka jawab hai "kaunsa angle yeh sine rakhta hai?" — yeh angle recover karne ke liye sine ko undo karta hai.

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Example 4 — Refraction denser → rarer, door bend karta hai (Cell C4)

Forecast: wave speed up hoti hai — is baar, normal ki taraf ya normal se door?

1. Same relation $\sin r=\sin i\cdot \frac{v_2}{v_1}$. Yeh step kyun? Snell's law ko parwah nahi ki kaunsa medium "pehle" hai; sirf speed ratio matter karta hai.
2. Plug in karo: $\sin r=\sin 30°\cdot \frac{3.0}{2.0}=0.5\times 1.5=0.75$ Yeh step kyun? Ab $v_2>v_1$, isliye $\sin r>\sin i$ — factor 1 se bada hai.
3. Arcsin: $r=\arcsin (0.75)=48.6°$. Yeh step kyun? Actual bend angle padhne ke liye sine ko undo karo.

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Example 5 — Normal incidence, degenerate case (Cell C5)

Forecast: kya kuch bend hota hai jab tum seedha hit karte ho?

1. Use karo $\sin r=\sin i\cdot \frac{v_2}{v_1}$ jisme $\sin 0°=0$. $\sin r=0\times \frac{v_2}{v_1}=0\Rightarrow r=0°$ Yeh step kyun? Zero times kuch bhi zero hota hai, isliye formula automatically koi bend nahi deta.
2. Geometrically: wavefront ke ends $A$ aur $B$ surface par same instant par pahunchte hain (wavefront surface ke parallel hai), isliye koi "ek end pehle pahuncha" nahi hota jo tilt create kare. Yeh step kyun? Refraction mein tilt poori tarah ek end ke lag karne se aata hai; lag hatao, bend hatao.
3. Jo phir bhi badalta hai: speed $v$ aur wavelength $\lambda =v/f$. Frequency $f$ fixed rehti hai. Yeh step kyun? Hume yeh report karna hai ki boundary tab bhi kya karti hai jab woh bend nahi karti — wave phir bhi slow hoti hai aur uske crests closer aa jaate hain, isliye $\lambda$ shrink hoti hai jabki "beat rate" $f$ unchanged rehti hai.

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Example 6 — Critical angle, limiting case (Cell C6)

Forecast: jaise hum beam ko zyada se zyada tilt karte hain, refracted ray surface ki taraf flat hoti jaati hai — kis angle par "room khatam" ho jaata hai?

Figure padhna: white horizontal line glass–air boundary hai (niche glass, upar air); niche se utha hua cyan arrow $i_c=41.8°$ par incident ray hai; surface ke saath flat para amber arrow $r=90°$ par refracted ray hai — yeh escape karne ki jagah graze karta hai; dashed vertical line normal hai.

1. Limiting condition $r=90°$ set karo, isliye $\sin r=1$. Yeh step kyun? $90°$ woh sabse bada angle hai jise normal se measure karke wave surface ke saath point kar sake — usse aage refracted wave ke liye koi aage ka direction nahi bachta.
2. Snell's law $n_1\sin i_c=n_2\sin 90°$ deta hai $\sin i_c=\frac{n_2}{n_1}=\frac{1.0}{1.5}=0.6667\Rightarrow i_c=\arcsin (0.6667)=41.8°$ Yeh step kyun? Index form $n_1\sin i=n_2\sin r$ yahan sabse clean hai; $r=90°$ set karna $i_c$ ko isolate karta hai.
3. $i>i_c$ ke liye: equation $\sin r=(n_1/n_2)\sin i>1$ ka koi real solution nahi hai. Physically refracted wavelet form nahi ho sakta; saari energy reflect hoti hai — total internal reflection. Yeh step kyun? Sine kabhi 1 se zyada nahi ho sakta; algebra khud signal karta hai ki refracted wave gaayab ho gayi hai, isliye hume ek fake angle force karne ki jagah physically interpret karna padega.

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Example 7 — Frequency fixed, wavelength badlti hai (Cell C7)

Forecast: kya light glass ke andar zyada red dikhti hai ya zyada blue? (Trick — colour frequency se set hoti hai.)

1. Air mein Frequency $f=c/{\lambda }_1$. $f=\frac{3.0\times 10^8}{600\times 10^{-9}}=5.0\times 10^{14}\text{Hz}$ Yeh step kyun? Frequency "crests per second" hai, aur yeh woh hai jo boundary ke across continuous rehta hai (wavefronts pile up ya vanish nahi ho sakte).
2. Glass mein Speed $v_2=c/n_2=2.0\times 10^8\text{m/s}$, phir ${\lambda }_2=v_2/f$. ${\lambda }_2=\frac{2.0\times 10^8}{5.0\times 10^{14}}=4.0\times 10^{-7}\text{m}=400\text{nm}$ Yeh step kyun? $f$ fixed aur $v$ chhota hone ke saath, $\lambda =v/f$ zaroor shrink karega.
3. Shortcut check: ${\lambda }_2={\lambda }_1\cdot v_2/v_1=600\times (2/3)=400\text{nm}$ ✓. Yeh step kyun? Same ${\lambda }_2$ tak pahunchne ka ek independent route Step 2 mein arithmetic slip se bachata hai.

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Example 8 — Real-world word problem (Cell C8)

Forecast: kya yeh Example 6 wala hi critical-angle idea hai, sirf glass ki jagah paani ke saath?

1. Pehchano ki yeh total internal reflection hai water→air jaate waqt ($n_1>n_2$). $r=90°$ set karo. Yeh step kyun? "Surface mirror ki tarah kaam karti hai" = upar koi light escape nahi hoti = hum critical angle par ya usse aage hain.
2. Compute karo $\sin i_c=n_2/n_1=1.00/1.33=0.7519$, isliye $i_c=\arcsin (0.7519)=48.8°$ Yeh step kyun? Exactly Example 6 wali machinery — Huygens har medium pair ke liye ek rule deta hai.
3. Interpret karo: seedhe upar se $48.8°$ ke andar dekhne par, use sky dikhta hai; usse aage, mirror. Yeh step kyun? Physics question ne viewing behaviour ke baare mein poocha, isliye hume abstract $i_c$ ko wapas is mein translate karna hoga jo swimmer ki aankh actually dekhti hai — warna number ka koi matlab nahi.

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Example 9 — Grazing incidence, $i\to 90°$ limit (Cell C9)

Forecast: agar beam $90°$ par barely skim karke aaye, toh kya refracted ray glass mein steeply jaati hai ya surface ke close rehti hai?

1. Limiting incidence $i=90°$ set karo, isliye $\sin i=1$. Yeh step kyun? Normal se $90°$ matlab ray surface ke saath lie karti hai — physically possible sabse extreme incidence. Yeh refraction ki ceiling probe karta hai.
2. Snell's law $n_1\sin 90°=n_2\sin r$ deta hai $\sin r=\frac{n_1}{n_2}\sin 90°=\frac{1.0}{1.5}=0.6667\Rightarrow r=\arcsin (0.6667)=41.8°$ Yeh step kyun? $\sin i$ apne maximum $1$ par hone ke saath, $\sin r$ apne maximum $n_1/n_2$ par pahunch jaata hai, isliye $r$ kabhi yeh value exceed nahi kar sakta — air se aane wali har real ray $41.8°$ half-angle wale cone ke andar land karti hai.
3. Interpret karo: air se glass mein enter hone wali saari light normal ke around $41.8°$ half-angle wale ek cone mein squeeze ho jaati hai; kuch bhi isse steep refract nahi hota. Yeh step kyun? Yeh Example 6 ka mirror-image hai — grazing-in limit ($r=41.8°$) glass ka critical angle wapas jaane ke barabar hai, jo exactly isliye hai kyunki refraction reversible hai.

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Example 10 — Exam twist: backward-wavelet trap (Cell C10)

Forecast: compute karne se pehle $K$ straight forward ($\theta =0°$) aur straight backward ($\theta =180°$) ka guess karo.

1. Forward: $\theta =0°$, $\cos 0°=1$, isliye $K=\frac{1}{2}(1+1)=1$ — full strength. Yeh step kyun? Obliquity factor exact rule hai jo Huygens ke hand-waved "backward ignore karo" ki jagah leta hai, isliye hum ise us direction mein test karte hain jahan wave sabse strong honi chahiye.
2. Backward: $\theta =180°$, $\cos 180°=-1$, isliye $K=\frac{1}{2}(1+(-1))=0$ Yeh step kyun? Backward direction set karna aur evaluate karna dikhata hai ki amplitude genuinely vanish hoti hai, decree se nahi balki Fresnel–Kirchhoff factor se.
3. Sideways sanity: $\theta =90°$, $\cos 90°=0$, $K=\frac{1}{2}$ — half strength, lekin yeh sideways wavelets envelope ke saath out of phase hain, isliye woh bhi cancel ho jaate hain. Yeh step kyun? Hume middle case bhi rule out karna hai — $90°$ par ek non-zero $K$ ek surviving sideways blur jaisa dikh sakta tha, isliye hum check karte hain ki interference (amplitude nahi) ise khatam karti hai, student ke argument mein har loophole band karte hue.

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Connections

• Huygens' principle — wavefront propagation (parent)
• Snell's Law and Refractive Index (Examples 3, 4, 6, 8, 9)
• Laws of Reflection (Example 2)
• Phase and Path Difference (Example 7)
• Fresnel–Kirchhoff Diffraction (Example 10, obliquity factor)
• Wave Optics — Interference
• Young's Double Slit Experiment
• Diffraction